4. OSCILLATION AND WAVES
THEORY
1.
INTRODUCTION
Time period (T) =
2S
Z
2S
m
as Z
k
m
k
(1)
A motion which repeats itself over and over again after a
regular interval of time is called a periodic motion.
(2)
Oscillatory or vibratory motion is that motion in which a
body moves to and fro or back and forth repeatedly about
a fixed point in a definite interval of time.
The frequency of a particle executing S.H.M. is equal to
the number of oscillations completed in one second.
(3)
Simple harmonic motion is a specific type of oscillatory
motion, in which
v
(a) partical moves in one dimension,
3.
Frequency
4.
The phase of particle executing S.H.M. at any instant is its
state as regard to its position and direction of motion at
that instant. it is measured as argument (angle) of sine in
the equation of S.H.M.
(c) net force on the particle is always directed towards
means position, and
Phase = (Zt + I)
(d) magnitude of net force is always proportional to the
displacement of particle from the mean position at that
instant.
where, k is known as force constant
Ÿ
Ÿ
Ÿ
ma = – kx
At t = 0, phase = I; the constant I is called initial phase of
the particle or phase constant.
1.2
1.
Important Relations
Position
k
x
or a Z2 x
m
where, Z is known as angular frequency.
a
d2 x
dt 2
Z2 x
This equation is called as the differential equation of
S.H.M.
If mean position is at origin the position (X coordinate)
depends on time in general as :
The general expression for x(t) satisfying the above
equation is :
x (t) = sin (Zt + I)
x (t) = A sin (Zt + I)
1.1 Some Important terms
1.
1 k
2S m
Phase
(b) particle moves to and fro about a fixed mean position
(where Fnet = 0),
So, Fnet = –kx
Z
2S
“
At mean position, x = 0
“
At extremes, x = + A, –A
2.
Velocity
Amplitude
The amplitude of particle executing S.H.M. is its maximum
displacement on either side of the mean position.
A is the amplitude of the particle.
2.
Time Period
Time period of a particle executing S.H.M. is the time taken
to complete one cycle and is denoted by T.
“
At any time instant t, v (t) = A Z cos (Zt + I)
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OSCILLATION AND WAVES
“
At any position x, v (x) = ± Z A2 x 2
“
Velocity is minimum at extremes because the particles
is at rest.
U x
1 2
kx
2
1
mA 2 Z2 sin 2 Zt I
2
i.e., v = 0 at extreme position.
“
Velocity has maximum magnitude at mean position.
|v|max = ZA at mean position.
3.
Acceleration
“
U is maximum at extremes Umax =
“
U is minimum at mean position
1 2
kA
2
Total Energy
“
At any instant t, a (t) = – Z2 A sin (Zt + I)
“
At any position x, a (x) = – Z2x
“
Acceleration is always directed towards mean position.
“
The magnitude of acceleration is minimum at mean
position and maximum at extremes.
T.E.
|a|min = 0 at mean position.
|a|max = Z2 A at extremes.
4.
Energy
1 2
kA
2
1
mA 2 Z2
2
and is constant at all time instant and at all positions.
Energy position graph
Kinetic energy
“
1
mv 2 Ÿ K
2
K
1
mZ2 A 2 x 2
2
1
mZ2 A 2 cos 2 Zt I
2
2. TIME PERIOD OF S.H.M.
To find whether a motion is S.H.M. or not and to find its
time period, follow these steps :
“
K is maximum at mean position and minimum at extremes.
“
K max
“
Kmin = 0 at extemes.
1
m Z2 A 2
2
1
kA 2 at mean position
2
(a)
Locate the mean (equilibrium) position mathematically by
balancing all the forces on it.
(b)
Displace the particle by a displacement ‘x’ from the mean
position in the probable direction of oscillation.
(c)
Find the net force on it and check if it is towards mean
position.
(d)
Try to express net force as a proportional function of its
displacement ‘x’.
Potential Energy
If potential energy is taken as zero at mean position,
then at any position x,
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OSCILLATION AND WAVES
“
111
If step (c) and step (d) are proved then it is a simple
harmonic motion.
(e)
2.1
(a)
2S
m
.
k
(c)
Combination of springs :
1.
Horizontal spring :
Let a block of mass m be placed on a smooth horizontal
surface and rigidly connected to spring of force constant
K whose other end is permanently fixed.
Mean position : when spring is at its natural length.
“
Time period : T
(b)
Vertical Spring :
2S
m
k
Springs in series
When two springs of force constant K1 and K2 are
connected in series as shown, they are equivalent to a
single spring of force constant K which is given by
Oscillations of a Block Connected to a Spring
“
m
k
Time period : T
Find k from expression of net force (F = – kx) and find time
period using T
2S
“
2.
1
K
1
1
K1 K 2
K
K1 K 2
K1 K 2
Springs in parallel
For a parallel combination as shown, the effective
spring constant is K = K1 + K2
If the spring is suspended vertically from a fixed point
and carries the block at its other end as shown, the
block will oscillate along the vertical line.
“
Mean position : spring in elongated by d
mg
k
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2.2
OSCILLATION AND WAVES
Oscillation of a Cylinder Floating in a liquid
Let a cylinder of mass m and density d be floating on
the surface of a liquid of density U. The total length of
cylinder is L.
“
2S
Time period : T
R
= 84.6 minutes
g
where, R is radius of earth.
2.5
Angular Oscillations
Instead of straight line motion, if a particle or centre of
mass of a body is oscillating on a small arc of circular
path then it is called angular S.H.M.
For angular S.H.M.,W = – kT
“
Mean position : cylinder is immersed upto A
“
Time period : T
2S
Ld
Ug
2S
Ld
U
A
g
2.3 Liquid Oscillating in a U–Tube
Ÿ
ID = –kT
Ÿ
Time period, T
2S
I
k
2.5.1 Simple Pendulum
2S
A
g
“
Time period : T
“
Time period of a pendulum in a lift :
Consider a liquid column of mass m and density U in a Utube of area of cross section A.
“
T
2S
A
(if acceleration of lift is upwards)
ga
T
2S
A
(if acceleration of lift is downwards)
g a
Second’s pendulum
Time period of second’s pendulum is 2s.
Length of second’s pendulum on earth surface 1m.
2.5.2
“
Mean position : when height of liquid is same in both
limbs.
“
Time period : T
2S
m
2AUg
2S
L
2g
where, L is length of liquid column.
2.4 Body Oscillation in tunnel along any chord of earth
“
Mean position : At the centre of the chord
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Physical Pendulum
Time period : T
2S
I
mgA
OSCILLATION AND WAVES
113
U is the density of the solid.
where, I is moment of inertia of object about point of
suspension, and
Speed of longitudinal wave in fluid is given by
l is distance of centre of mass of object from point of
suspension.
3.
1.
v
DAMPED AND FORCED OSCILLATIONS
where, B is the bulk modulus,
Damped Oscillation :
(i)
The oscillation of a body whose amplitude goes on
decreasing with time is defined as damped oscillation.
(ii)
In this oscillation the amplitude of oscillation
decreases exponentially due to damping forces like
frictional force, viscous force etc.
B
U
U is the density of the fluid.
(b)
Newton’s formula
Newton assumed that propagation of sound wave in
gas is an isothermal process. Therefore, according to
Newton, speed of sound in gas is given by v
P
U
where P is the pressure of the gas and U is the density
of the gas.
According to the Newton’s formula, the speed of sound
in air at S.T.P. is 280 m/s. But the experimental value of
the speed of sound in air is 332 ms –1. Newton could not
explain this large difference. Newton’s formula was
corrected by Laplace.
(iii)
2.
Due to decrease in amplitude the energy of the
oscillator also goes on decreasing exponentially.
Forced Oscillation :
(i)
(ii)
The oscillation in which a body oscillates under the
influence of an external periodic force are known as
forced oscillation.
Resonance : When the frequency of external force is
equal to the natural frequency of the oscillator, then
this state is known as the state of resonance. And this
frequency is known as resonant frequency.
4. WAVES
(c)
Laplace’s correction
Laplace assumed that propagation of sound wave in
gas in an adiabatic process. Therefore, according to
Laplace, speed of sound in a gas is given by
v
JP
U
According to Laplace’s correction the speed of sound
in air at S.T.P. is 331.3 m/s. This value agrees farily well
with the experimental values of the velocity of sound
in air at S.T.P.
5. WAVES TRAVELLING IN OPPOSITE DIRECTIONS
(a)
Speed of longitudinal wave
Speed of longitudinal wave in a medium is given by
v
E
U
where, E is the modulus of elasticity,
U is the density of the medium.
Speed of longitudinal wave in a solid in the form of rod
is given by
v
Y
U
where, Y is the Young’s modulus of the solid,
When two waves of same amplitude and frequency
travelling in opposite directions
y1 = A sin (kx – Zt)
y2 = A sin (kx + Zt)
interfere, then a standing wave is produced which is
given by,
y = y1 + y2
Ÿ y = 2A sin kx cos Zt
Hence the particle at location x is oscillating in S.H.M.
with angular frequency Z and amplitude 2A sin kx. As
the amplitude depends on location (x), particles are
oscillating with different amplitude.
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OSCILLATION AND WAVES
Nodes : Amplitude = 0
2A sin kx = 0
x = 0, S/k, 2S/k.......
x = 0, O/2, O, 3O/2, 2O........
Antinodes : Amplitude is maximum.
sin kx = ± 1
x = S/2k, 3S/2k
Fundamental frequency (x = 1)
x = O/4, 3O/4, 5O/4
Q0
Nodes are completely at rest. Antinodes are oscillating
with maximum amplitude (2A). The points between a
node and antinode have amplitude between 0 and 2 A.
v
2L
It is also called first harmonic.
Second harmonic or first overtone
Separation between two consecutive (or antinodes)
= O/2.
Q
Separation between a node and the next antinode=O/4.
2v
2L
The nth multiple of fundamental frequency is known
as nth harmonic or (n – 1)th overtone.
Nodes and antinodes are alternately placed.
2.
Fixed at one end
Transverse standing waves with node at fixed end and
antinode at open end are formed.
So, length of string A
2n 1
O
if there are n nodes
4
and n antinodes.
Frequency of oscillations
Ÿ Q
5.1
1.
2n 1 v
v
O
4A
It is clear from the figure that since nodes are, at rest
they don’t transfer energy. In a stationary wave, energy
is not transferred from one point to the other.
Fundamental frequency, (n = 1)
Vibrations in a stretched string
It is also called first harmonic.
Fixed at both ends.
First overtone or third harmonic.
Transverse standing waves with nodes at both ends
of the string are formed.
Q
So, length of string, A
nO
if there are (n + 1) nodes
2
and n antinodes.
Frequency of oscillations is
Ÿ Q
v
O
nv
2A
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Q0
v
4L
3v
4A
3 Q0
Only odd harmonics are possible in this case.
5.2
1.
Vibrations in an organ pipe
Open Organ pipe (both ends open)
The open ends of the tube becomes antinodes because
the particles at the open end can oscillate freely.
If there are (n + 1) antinodes in all,
OSCILLATION AND WAVES
length of tube, A
115
Let us consider net effect of two waves of frequencies
Q 1 and Q 2 and amplitude A at x = 0.
nO
2
So, Frequency of oscillations is Q
nv
2A
y1
A sin 2SQ1t
y2
A sin 2SQ 2 t
Ÿ y = y1 + y 2
Ÿ y = A sin 2SQ1t sin 2SQ 2 t
y
ª¬ 2A cos S Q1 Q 2 t º¼ sin S Q1 Q 2 t
Thus the resultant wave can be represented as a
§ Q Q2
travelling wave whose frequency is ¨ 1
© 2
·
¸ and
¹
amplitude is 2A cos S (Q1– Q2) t.
As the amplitude term contains t, the amplitude varies
periodically with time.
2.
Closed organ pipe (One end closed)
For Loud Sounds : Net amplitude = ± 2A
The open end becomes antinode and closed end
become a node.
Ÿ cos S (Q 1 – Q 2) t = ± 1
Ÿ S (Q 1 – Q 2) t = 0, S, 2S, 3S .......
If there are n nodes and n antinodes,
Ÿ t
L = (2n – 1) O/4
0,
1
2 ........
,
Q1 Q 2 Q1 Q 2
So frequency of oscillations is
Hence the interval between two loud sounds is given as :
Q
v
O
2n 1 v
1
Q1 Q 2
4L
Ÿ the number of loud sounds per second = Q 1 – Q 2
Ÿ beat per second = Q1 – Q 2
Note that Q1 – Q2 must be small (0 – 16 Hz) so that sound
variations can be distinguished.
Filling a tuning form increases its frequency of vibration.
Loading a tuning for k decreases its frequency of vibration.
6.
There are only odd harmonics in a tube closed at one end.
5.3
Waves having different frequencies
Beats are formed by the superposition of two waves of
slightly different frequencies moving in the same
direction. The resultant effect heard in this case at any
fixed position will consist of alternate loud and weak
sounds.
DOPPLER EFFECT
According to Doppler’s effect, whenever there is a relative
motion between a source of sound and listener, the apparent
frequency of sound heard by the listener is different from
the actual frequency of sound emitted by the source.
Apparent frequency,
Qc
v vL
uQ
v vs
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OSCILLATION AND WAVES
Sing Convention. All velocities along the direction S
to L are taken as positve and all velocities along the
direction L to S are taken as negative.
When the motion is along some other direction the
component of velocity of source and listener along the
line joining the source and listener is considered.
Qc
(e)
v vL
v
Q
v vL
Q i.e. v’ > Q
v
If the source and listener are approaching each other,
then v s is positive and v L is negative (figure e).
Therefore,
Special Cases :
(a)
If the source is moving towards the listener but the
listener is at rest, then vs is positive and vL = 0 (figure
a). Therefore,
Qc
(b)
(f)
If the source is moving away from the listener, but the
listener is at rest, then vs is negative and vL = 0 (figure
b). Therefore,
Qc
(c)
v
u Q i.e. Q’ > Q
v vs
Qc
v
Q
v vs
v
Q i.e. Q’ < Q
v vs
If the source is at rest and listener is moving away from
the source, the vs = 0 and vL is positive (figure c).
Therefore,
v vs
Q
§ v vL ·
¨
¸ Q i.e. Q’ > Q
© v vs ¹
If the source and listener are moving away from each
other, then vs is negative and vL is positive, (figure f).
Therefore,
Qc
(g)
v vL
v vL
Q
v vs
v vL
Q i.e. Q’ < Q
v vs
If the source and listener are both in motion in the
same direction and with same velocity, then vs = vL = v’
(say) (figure g). Therefore,
Qc
v vc
Q i.e. Q’ = Q
v vc
It means, there is no change in the frequency of sound
heard by the listerner.
Appar ent wavelength heard by the observer is
Oc
Q Qs
Q
If case the medium is also moving, the speed of sound
G G
with respect to ground is considered. i.e. Q Q m
7.
CHARACTERISTICS OF SOUND
Loudness of sound is also called level of intensity of
sound.
In decibel the loudness of a sound of intensity I is
Qc
(d)
v vL
v
Q i.e. Q’ < Q
If the source is at rest and listener is moving towards
the source, then vs = 0 and vL is negative (figure d).
Therefore,
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§ I ·
given by L = 10 log10 ¨ ¸ . (I 0 = 10 –12 w/m2)
© I0 ¹
Pitch : It is pitch depends on frequency, higher the
frequency higher will be the pitch and shriller will be
the sound.
OSCILLATION AND WAVES
117
SOLVED EXAMPLES
A = 4.00 m , Z = S rad/s, I0 = S/4
OSCILLATIONS
(a)
Example - 1
Write the displacement equation representing the following
conditions obtained in a simple harmonic motion. Amplitude
= 0.01 m, frequency = 600 Hz, initial phase = S/6.
Sol. Here A = 0.01 m, v = 600 Hz, I0 = S/6
or
Displacement at t = 1.00 s i.e.,
x = (4.00 m)cos(S × 1 + S/4) = (4.00) (–cos S/4)
= (4.00) (–0.707) = –2.83 m
(b)
Velocity at t = 1.00 s, i.e., v = – ZA sin (Zt + I0)
or
v = – (S/s)(4.00 m)sin [S × 1 + S/4]
The displacement equation of simple harmonic motion is
= –(4.00 S)(–sin S/4) m/s
y = A sin (Zt + I0) = A sin(Svt + I0)
= (4.00 × 3.14)(0.707) m/s
y = (0.01 m) sin (1200 St + S/6)
= 8.89 m/s
Example - 2
A particle executes SHM of amplitude 25 cm and time period
3 s. What is the minimum time required for the particle to
move between two points 12.5 cm on either side of the
mean position ?
Sol. With usual notation, we are given that amplitude A = 25 cm,
time period T = 3s
(c)
Acceleration,
2
a = – Z A cos (Zt + I0)
2
= – S × 4.00 cos (S × 1 + S/4)
2
2
= – (4.00 S )(–cos S/4)m/s
2
2
= 4.00 × (3.14) × 0.707 m/s
2
= 27.9 m/s
(d)
Displacement from the mean position, y = 12.5 cm
Maximum velocity, vmax = ZA × S × 4.00
= 12.6 m/s
If t is the time taken by the particle to move from the mean
position to a point 12.5 cm on any side of the mean position,
2
2
Maximum acceleration, amax = Z A = S × 4.00
2
= 39.5 m/s
§ 2S ·
§ 2S ·
t ¸ or 12.5 = 25 sin ¨ t ¸
©T ¹
© 3 ¹
y = A sin Zt = A sin ¨
or
S § 2S · S
§ 2S · 1
sin or ¨ t ¸
sin ¨ t ¸
3
2
6 © 3 ¹ 6
©
¹
or
t = 0.25 s
(e)
S
S
9S
= 2S +
=
.
4
4
4
Example - 4
A block is resting on a piston which is moving vertically
with a SHM of period 1.0 s. At what amplitude of vibration
will the block and the piston separate ? What is the
maximum velocity of the piston at this amplitude ?
Obviously, time taken by the particle to move between two
points 12.5 cm on either side of the mean position.
= 2t = 2 × 0.25 = 0.5 s.
Phase, (Zt + I0) = (S/s) × 2s +
Sol. We are given that T = 1.0 s
Further, the maximum acceleration in SHM, i.e.,
Example - 3
2
amax = Z A
A body oscillates with SHM along with X-axis. Its
displacement varies with time according to the equation :
For the block and the piston to separate,
2
amax t g or Z A t g
x = (4.00 m) cos(St + S/4)
Calculate at t = 1.00 s : (a) displacement (b) velocity
(c) acceleration (d) Also calculate the maximum speed and
maximum acceleration and (e) phase at t = 2.00 s.
Sol. By comparing the given equation with the general equation
for SHM along X-axis, i.e.,
x = Acos (Zt + I0), we get
2
or
(2S/T) A t g
or
At
or
A t 0.248 m
(9.8 m / s 2 )(1.0 s)2
39.48
or A t
gT 2
4S2
2
(as 4S = 39.48)
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OSCILLATION AND WAVES
Thus, the block and the piston separate, when
A = 0.248 m
Clearly,
§ 2 u 3.14 ·
§ 2S ·
¸ (0.248 m) = 1.56 m/s
vmax = ZA = ¨ ¸ A ¨
© T ¹
© 1.0s ¹
or
T'
= (1 + 0.005) = 1.005
T
or
T’ = (1.005)T = 1.005 × 2
(as for a second pendulum, T = 2s)
or
Since the time period has increased, the pendulum will make
lesser number of vibrations per day. In other words, it will
run slow.
Example - 5
A point particle of mass 0.1 kg is executing SHM of amplitude
0.1 m. When the particle passes through the mean position,
–3
its kinetic energy is 8 × 10 J. Obtain the equation of motion
o
of this particle if the initial phase of oscillations is 45 .
Loss in time in 2 s = 0.01 s
Loss in 1 day (i.e., 86400 s) =
Sol. Here, m = 0.1 kg, A = 0.1 m,
–3
A spring compressed by 10 cm develops a restoring force
of 10 N. A body of mass 9 kg is placed on it. What is the
force constant of the spring ? What is the depression in
the spring under the weight of the body ? What is the
period of oscillation if the body is disturbed from its
equilibrium position ? [Take g = 10 N/kg]
o
I0 = 45 = S/4
1
1
2 2
2
2
–3
Since K0 = mZ A , 2 (0.1 kg)Z (0.1 m) = 8 × 10 J
2
Z = 4 rad/s
The equation of motion of a particle executing SHM is given
by
Sol. Here, F = 10 N, y = 10 cm = 0.1 m
y = Asin(Zt + I0)
or
F 10 N
Force constant, k = y = 0.1m = 100 N/m
y = (0.1 m)sin [(4 rad/s) t + S/4]
Example - 6
Further, when F = weight of the body = 9 kg wt = 90 N,
If the length of a second pendulum is increased by 1%,
how many seconds will it lose in a day ?
F
90 N
y = k 100 N / m = 0.9 m
Sol. If l be the length of the second pendulum and T be its time
period, then
Time period, T = 2S
l
g
T = 2S
...(i)
= 2S
When length is increased by 1%, i.e., by 1/100,
new length = l + (l/100) = l (1 + 1/100)
or
If T’ be its changed time period,
T’ = 2S
or
...(ii)
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§ 3·
9
s = 2S ¨ ¸ s
© 10 ¹
100
3S
s = 1.88 s.
5
Write the values of amplitude and angular frequency in the
following simple harmonic motion :
x = 0.70 cos (180t + 0.23)
1/ 2
1 · §
1 ·
§
¨1
¸ ¨1 ¸
© 100 ¹ © 100 ¹
T' §
1 ·
1
¨
T © 200 ¸¹
T=
m
k
Example - 8
l (1 1/100)
g
From eqns. (i) and (ii),
T'
T
0.01s
2 s × 86400 s = 432 s
Example - 7
K0 (kinetic energy at mean position) = 8 × 10 J,
or
T’ = 2.01 s
where the various quantities are in SI units.
Sol. Comparing the given simple harmonic motion with the
standard SHM equation
(applying binomial theorem)
x = Acos(Zt + I0), we have A = 0.70 m, Z = 180 rad/s
OSCILLATION AND WAVES
119
Example - 9
Two simple harmonic motions are represented by the
equations y1 = 10 sin (3St + S/4) and
y2 = 5 (sin 3St +
Sol. We are given that time period of a simple pendulum on Earth,
i.e., TE = 3.5 s
Acceleration due to gravity on Earth, i.e., gE = 9.8 m/s
2
Acceleration due to gravity on Moon, i.e., gM = 1.7 m/s .
3 cos 3St)
Let TM be the time period of the simple pendulum on Moon.
Clearly,
Find the ratio of their amplitudes.
Sol. We are given that
y1 = 10 sin (3St + S/4)
y2 = 5 (sin 3St +
or
= 10[(1/2)sin 3St + ( 3 / 2) cos 3St]
= 10 [cos (S/3) sin 3St + sin (S/3) cos 3St]
...(ii)
Example - 10
How will the time period of a simple pendulum change if its
length is doubled ?
Sol. For a simple pendulum,
gE
gM
2S l / g E
9.8
1.7 = 2.4
Example - 12
A body of mass 1 kg is made to oscillate in turn on two
springs, one of force constant 1 N/m and another of 16 N/m.
Calculate the time period in each case.
Sol. For the first spring, m = 1 kg and k = 1 N/m
m
1
S
= 2S s
k
1
For the second spring, m = 1 kg, k = 16 N/m
For two pendulums of lengths l1 and l2, let their time periods
be T1 and T2.
l2
g
and T2 = 2S
l2
l1
Thus, T2 = 2S
m
1 §S·
S
¨ ¸s .
k
16 © 2 ¹
Example - 13
A spring balance has a scale that reads from 0 to 50 kg. The
length of the scale is 20 cm. A body suspended from this
spring, when displaced and released, oscillates with a
period of 0.60 s. What is the weight of the body ?
l1
Clearly, T1 = 2S
g
or T2 =
2S l / g M
TM
TE
Thus, T1 = 2S
l
g
T2
Thus, T
1
l
l
and TM = 2S
gE
gE
Clearly, TM = 2.4 TE = 2.4 × 3.5 = 8.4 s
From eqns. (i) and (ii), it is clear that the amplitudes of both
SHM’s are equal i.e., these are in the ration of 1 : 1.
T = 2S
TE = 2S
...(i)
3 cos 3St)
= 10 sin [3St + (S/3)]
2
Sol. If k is the spring constant of the spring balance,
k=
2
(as l2/ l1 = 2)
F 50 kg wt 50 u 9.8 N
= 0.2 m
0.20 m = 2450 N/m
l
Let Z be the angular frequency of the spring balance.
2 T1
2S 2S
= 10.47 rad/s
T 0.60
Thus, the time period of the simple pendulum increases by
Clearly, Z =
a factor of
If m is the mass of the body,
2.
Example - 11
The acceleration due to gravity on the surface of the Moon
2
is 1.7 m/s . What is the time-period of a simple pendulum
on the Moon if its time period on the Earth is 3.5 s ?
2
(g on Earth = 9.8 m/s )
2
Z =
k
k
or m = Z2
m
2450
(10.47)2 = 22.35 kg
Weight of the body = mg = 22.35 kg wt
= 22.35 × 9.8 N = 219 N
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120
OSCILLATION AND WAVES
Sol. With usual notations, we are given that
Example - 14
A small trolley of mass 2.0 kg resting on a horizontal
turntable is connected by a light spring to the centre of
the table. When the turntable is set into rotation at a speed
of 300 rpm, the length of the stretched spring is 40 cm. If
the original length of the spring is 35 cm, determine the
force constant of the spring.
m = 3.0 kg, k = 1200 N/m, A = 2.0 cm = 0.02 m
(a)
v=
(b)
Sol. We are given that
1 k
2S m
1
1200
20 –1
–1
or v =
s = 3.2 s
2 u 3.14
3
6.28
The maximum acceleration of the mass, i.e.,
2
frequency of rotation of the turntable,
2
2 2
amax = Z A = (2Sv) A = 4S v A
mass of the trolley, m = 2.0 kg
v=
The frequency of oscillation of the mass m is given as
2
2
2
or amax = [(39.48)(3.2) (0.02)]m/s = 8.1 m/s
(c)
The maximum speed of the mass, i.e.,
vmax = ZA = 2SvA
300
= 5 rps
60
or
length of the stretched spring,
vmax = (2 × 3.14 × 3.2 × 0.02) m/s = 0.4 m/s
Example - 16
r = 40 cm = 0.4 m
In above problem, what is
original length of the spring,
(a) the speed of the mass when the spring is compressed
by 1.0 cm ?
L = 35 cm = 0.35 m
l = r – L = 0.05 m
(b) potential energy of the mass when it momentarily
comes to rest ?
tension in the spring = centripetal force,
(c) total energy of the oscillating mass ?
extension produced in the spring,
Fc =
Sol. (a) Speed v = Z A 2 x 2
mv2
2 2
= 4S v mr
r
2Sv A 2 x 2
If k is the spring constant of the spring,
= 2 × 3.14 × 3.2 (0.02) 2 ( 0.01) 2
F = kl
= 0.35 m/s
or
F
k=
l
or
k=
(b)
4S 2 v 2 mr
l
4 u 9.87 u (5)2 u 2 u 0.4
0.05
1
mZ2 A 2
2
1
2 2
2
2 2
m(2Sv) A = 2S mv A
2
2
2
= 2 × 9.87 × 3 × (3.2) × (0.02) = 0.24 J
4
A spring of force constant 1200 N/m is mounted on a
horizontal table as shown in Figure. A mass of 3.0 kg,
attached to the free end of spring, is pulled sideways to a
distance of 2.0 cm and released.
PE of the mass when it momentarily comes to rest (at the
extreme position), i.e.,
U0 =
= 15792 N/m = 1.6 × 10 N/m
Example - 15
(as v = 3.2, A = 0.02 m, x = –0.01 m)
(c)
Total energy of the oscillating mass (E) = its potential energy
at the extreme position (U0) = 0.24 J
Example - 17
In above problem, let us take the position of the mass
when the spring is unstretched as x = 0, and the direction
from left to right is the positive direction of X-axis. Give x
as a function of time t for the oscillating mass. If at the
moment we start the stop watch (t = 0) ; the mass is
(a) at the mean position
(b) at the maximum stretched position
Determine : (a) the frequency of oscillations,
(b) the maximum acceleration of the mass,
(c) the maximum speed of the mass.
Mahesh Tutorials Science
(c) at the maximum compressed position.
In what way do these different functions for SHM differ ?
Frequency ? Amplitude ? Or initial phase ?
OSCILLATION AND WAVES
121
Sol. For SHM along X-axis,
We know that, V = Z a 2 y 2
x = Acos(Zt + I0)
(a)
...(i)
When at t = 0, x = 0,
0 = Acos(Z × 0 + I0)
or cos I0 = 0 or I0 = – S/2
x = 2 sin (20 t)
(as A = 2 cm, Z = 2Sv = 2 × 3.142 × 3.2 = 20 s )
When at t = 0, x = A,
A = Acos(Z × 0 + I0)
or
...(ii)
cos I0 = –1 or I0 = 0
2
16 a 2 9
or 9 a 2 16
2
or
16 a – 256 = 9 a – 81
or
a =
From eqn. (i) x = A cos Zt = 2cos (20t)
(c)
Case (II) 3 = Z a 2 42
2
4 Z a 9
3 Z a 2 16
–1
(b)
...(i)
Dividing (i) by (ii), we get
From eqn. (i), x = A cos(Zt – S/2) = A sin Zt
or
Case (I) 4 = Z a 2 32
2
175
= 25
7
2
or 7 a = 256 – 81 = 175
or a =
25 = 5
Substituting it in (i), we get
When at t = 0, x = –A,
4 = Z 5 2 32
–A = Acos(Z × 0 + I0)
or cos I0 = – 1 or I0 = S
or
From eqn. (i), x = Acos(Zt + S)
Z 25 9 = Z × 4
w = 4/4 = 1 rad s
–1
= –A cos Zt
When the particle is at a distance 2.5 m from the extreme
position, then its distance from the mean position,
or x = –2cos(20 t)
x = 5 – 2.5 = 2.5 m
Since, the time is to be noted from the extreme postion for
SHM therefore, we shall use the relation
Example - 18
A spring compressed by 0.1 m develops a restoring force
10 N. A body of mass 4 kg is placed on it. Deduce (i) the
force constant of the spring (ii) the depression of the spring
under the weight of the body (take g = 10 N/kg) and (iii) the
period of oscillation, if the body is disturbed.
Sol. Here, F = 10 N ; ' l = 0.1 m ; m = 4 kg
(i) k =
(ii) y =
F
10
–1
=
= 100 Nm
'l
0.1
mg
k
(iii) T = 2S
4 u10
= 0.4 m
100
m
22 4
=2×
= 1.26 s
k
7 100
x = a cos Zt
or
2.5 = 5 cos 1 × t = 5 cos t or cos t =
or
S 22
t = 3 7 u 3 = 1.048 s.
2.5 1
S
cos
5 2
3
Example - 20
A simple harmonic oscillation is represented by the
equation y = 0.40 sin (440 t + 0.61) here, y and t are in m
and s respectively. What are the values of (i) amplitude
(ii) angular frequency (iii) frequency of oscillations
(iv) time period of oscillations and (v) initial phase ?
Sol. The given equation is y = 0.40 sin (440 t + 0.61)
Comparing it with the equation of SHM y = a sin (Zt + I0)
Example - 19
A particle executing SHM along a st. line has a velocity of
–1
4 ms , when at a distance of 3 m from its mean position
–1
and 3 ms , when at a distance of 4 m from it. Find the time
it takes to travel 2.5 m from the positive extremity of its
oscillation.
–1
Sol. Here (case i), V1 = 4 ms ; y1 = 3m ;
–1
(case ii), V2 = 3 ms ; y2 = 4m.
We have, (i) Amplitude,
a = 0.40 m
(ii) Angular frequency,
Z = 440 Hz
Z
440
(iii) Frequency of oscillations, v = 2 S 2 u 22 / 7 = 70 Hz.
(iv) Time period of oscillations, T =
1 1
= 0.0143
v 70
(v) Initial phase angle, I0 = 0.61 rad.
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122
OSCILLATION AND WAVES
Example - 21
A particle executes SHM of time period 10 s. The
displacement of particle at any instant is given by the
relation x = 10 sin Zt (in cm). (i) Find the velocity of body
2 s after it passes through the mean position and (ii) the
acceleration 2 s after it passes the mean position.
2S 2 S
–1
rad/s ; amplitude, r = 10 cm.
Sol. Here ; T = 10 s ; Z =
T 10
When t = 2 s, V = 10 ×
2S
§ 2S ·
cos ¨ u 2 ¸ = 2 S cos (0.4 S) =
10
© 10 ¹
or
2
2
2
1
1
2 2
2
2 2
mZ (a –y ) = m Z y
2
2
?
2
2
a – y = y or y = a /2
1
2 2
mZ y
2
or y = a/ 2 = 0.71 a
Example - 24
A mass of 1 kg is executing SHM which is given by,
x = 6.0 cos (100 t + S/4) cm. What is the maximum kinetic
energy ?
Sol. Here, m = 1 kg.
1.942 cm/s.
(ii)
and P.E. =
As, K.E. = P.E.
Velocity of the particle at any instant t is given by V = r Z cos Zt
(i)
1
2 2
2
mZ (a –y )
2
Sol. K.E. =
The given equation of SHM is x = 6.0 cos (100 t + S/4)
Acceleration at any time t is given by
Comparing it with the equation of SHM x = a cos (Zt + I),
§ 2S ·
§ 2S ·
2
2
A = –Z r sin Zt = – ¨ ¸ × 10 × sin ¨ u 2 ¸ = –3.754 cm/s
© 10
¹
© 10 ¹
2
–1
we have, a = 6.0 cm = 6/100 m and Z = 100 rad s
2
Max. kinetic energy =
In fact, the acceleration is 3.754 cm/s and is directed towards
the mean position.
Example - 22
A particle executes simple harmonic motion with a time
period of 16 s. At time t = 2 s, the particle crosses the mean
–1
position while at t = 4 s, its velocity is 4 ms . Find its
amplitude of motion.
1
1
2
= m(aZ) = × 1 ×
2
2
1
2
m(Vmax)
2
2
§ 6
·
u100 ¸ = 18 J
¨
100
©
¹
Example - 25
An 8 kg body performs SHM of amplitude 30 cm. The
restoring force is 60 N, when the displacement is 30 cm.
Find (a) time period (b) the acceleration, P.E and K.E., when
displacement is 12 cm.
Sol. Here, T = 16 s ; At t = 2 s, y = 0 and at t = 4 s ;
–1
V = 4 ms ; a = ?
For simple harmonic motion, y = a sin Zt = a sin
Sol. Here, m = 8 kg ; a = 30 cm = 0.30 m ;
2S
t
T
(a)
F = 60 N ; y = 0.30 m
When t = 4 s, the time taken by particle to travel from the
mean position to a given position = 4 – 2 = 2 s. The
displacement,
F 60
–1
F = ky or k = y 0.30 = 200 Nm
§ 2S ·
y = a sin ¨ u 2 ¸ = a sin
© 16 ¹
As Z =
§S·
¨4¸
© ¹
a
2
...(i)
Velocity, V = Z a 2 y 2
?
S a
§ 2S · 2 2
u
4 = ¨ ¸ a a / 2
8
2
© 16 ¹
or
32 2
a=
= 14.4 m.
S
k
m
200
–1
= 5 rad/s
8
2 S 2 u 22 44
Time period, T = Z 7 u 5 35 = 1.256 s
(b)
Here, y = 12 cm = 0.12 m
2
Example - 23
A particle executes SHM of amplitude a. At what distance
from the mean position is its K.E. equal to its P.E. ?
Mahesh Tutorials Science
2
–2
? Acceleration, A = Z y = (5) × 0.12 = 3.0 ms
P.E. =
1 2 1
2
ky = × (200) × (0.12) = 1.44 J
2
2
K.E. =
1
1
2
2
2
2
k (a – y ) = × 200 × [(0.30) – (0.12) ] = 7.56 J
2
2
OSCILLATION AND WAVES
123
–1
Sol. (a) Here m = 200 g = 0.2 kg, k = 90 Nm
Example - 26
Two identical springs of spring constant k are attached to
a block of mass m and to fixed supports as shown in Figure.
Show that when the mass is displaced from its equilibrium
position on either side, it executes a simple harmonic
motion. Find the period of oscillation.
Sol. Let the mass m be displaced by a small distance x to the
right as shown in Figure. Due to it, the spring on the left
hand side gets stretched by length x and the spring on the
right hand side gets compressed by length x. The forces
acting on the mass due to springs are
F1 = – kx towards left hand side
?
k 90
–1
–1
= 450 Nm kg
m 0.2
b2
4m 2
(b)
As,
k
b2
!!
, therefore, Z’ =
m
4m 2
T=
2S
m
22 0.2
2S
2u
| 0.3 s.
Z'
k
7 90
If T is the time, when amplitude drops to half value then
amplitude of the damped oscillations at time t is
–bt/2m
when t = T, a = x0/2
?
x0
–bT/2m
= x0 e
2
or
2=e
or
T=
=
Therefore, total restoring force on mass m is
(c)
As time period, T = 2S
2.3026 u 2 u 0.2 u 0.3010
0.04
If T’ is the time, when mechanical energy drops to half its
mechanical energy E0, then mechanical energy E of the
damped oscillations at an instant t is given by
–bt/m
E0
–bT’/m
= E0e
2
bT’/m
or
2=e
or
loge2 =
or
T’ = 2.3026 log102 ×
or
T’ = 2.3026 × 0.3010 ×
Example - 27
For the damped oscillator shown in Figure, the mass m of
–1
the block is 200 g, k = 90 Nm and the damping constant b
–1
is 40 gs . Calculate (a) the period of oscillation, (b) time
taken for its amplitude of vibrations to drop to half of its
initial value and (c) the time taken for its mechanical energy
to drop to half its initial value.
bT
2m
When t = T’, E = E0/2 then
Here, interia factor = mass of the block = m
m
2S
2k
or log e 2 =
2m
2.3026 u 2m
loge2 =
log102
b
b
E = E0 e
Comparing (i) with the relation, F = – kx, we have Spring
factor, K = 2 k
inertia factor
spring factor
bT/2m
= 6.93 s.
...(i)
Here –ve sign shows that force F is directed towards the
equilibrium position O and F v x. Therefore, if the mass m is
left free, it will execute linear SHM.
k
b2
k
|
m 4m 2
m
a = x0 e
F2 = – kx towards left hand side
F = F1 + F2 = – kx + (– kx) = –2 kx
(0.04) 2
2 –2
4 u (0.2)2 = 0.01 kg s
bT '
m
m
b
0.2
= 3.46 s.
0.04
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124
OSCILLATION AND WAVES
(a)
WAVES
The reflected sound is travelling in air.
Therefore if Oa is the wavelength of the reflected wave,
Example - 28
An observer standing at the sea coast observes 54 waves
reaching the coast per minute. If the wavelength of the
wave is 10m, find the speed. What types of waves did he
observe? Explain
Oa
(b)
Sol. Since 54 waves reach the coast in 1 min (i.e., 60 s),
frequency of the waves, i.e., v =
54
= 0.9 Hz
60
As O = 10 m, speed of the waves
Xw
v
1486 m / s
–3
= 1.49 × 10 m
106 s 1
Example - 31
v = Q O = (0.9 × 10) m/s = 9 m/s
A travelling harmonic wave on a string is described by :
y = 7.5 sin (0.0050 x + 12t + S/4)
The waves are transverse. Strictly speaking, these waves
are ripples, which are neither entirely transverse nor
longitudinal. This is due to the reason that the water particles
while moving up and down, also move back and forth
horizontally. Thus, water waves are wrongly classified as
transverse waves.
(a) What are the displacement and velocity of oscillation
of a point at x = 1 cm and t = 1 a ? Is this speed equal to
the speed of wave propagation ?
(b) Locate the points of the string which have the same
transverse displacements and velocity as the x = 1 cm
point at t = 2 s, 5 s, 11 s.
Example - 29
Sol. We are given that
y = 7.5 sin (0.0050 x + 12t + S/4)
(a)
Sol. Here, speed of sound in air,
v = 336 m/s
340 m / s
–4
10 6 s 1 = 3.4 × 10 m
Further, the transmitted wave is travelling through water
and as such wavelength of the transmitted wave, i.e.,
Ow
2
How far does the sound travel in air when a tuning fork of
frequency 560 Hz makes 30 vibrations ? Given speed of
sound in air = 336 m/s.
Xa
v
...(i)
At x = 1 cm and t = 1 s, displacement,
y = 7.5 sin (0.0050 × 1 + 12 × 1 + S/4)
or
y = 7.5 sin (12.7904 rad)
or
180q ·
§
y = 7.5 sin ¨12.7904 u
¸
S ¹
©
frequency of the tuning fork, v= 560 Hz
Time taken to complete 30 vibrations, i.e.,
t=
30 3
s
560 56
= 7.5 sin 732.83°
y = 7.5 sin (720° + 12.83°)
Distance travelled by sound,
§ 3 ·
s = vt = (336 m/s) ¨ s ¸ = 18 m
© 56 ¹
= 7.5 sin 12.83°
or
y = (7.5 × 0.2215) cm = 1.666 cm
Speed of oscillation at a point,
Example - 30
A bat emits ultrasonic sound of frequency 1000 kHz in air.
If this sound meets a water surface, what is the wavelength
of : (a) the reflected sound (b) the transmitted sound ?
Speed of sound in air = 340 m/s and in water = 1486 m/s.
Sol. We are given that frequency of the ultrasonic sound,
vp
or
dy
dt
v p = 7.5 × 12 × cos (0.0050 x + 12t + S/4)
= 90 cos (0.0050 x + 12t + S/4)
When x = 1 cm and t = 1 s,
6
v = 1000 kHz = 10 Hz
speed of sound in air, va = 340 m/s
speed of sound in water,
vw = 1486 m/s
Mahesh Tutorials Science
v p = (90 cos 12.83°) cm/s
= (90 × 0.9751) cm/s = 87.75 cm/s
The general equation of the travelling harmonic wave
(travelling towards left) is
OSCILLATION AND WAVES
y = A sin [Zt + kx + I0]
125
...(ii)
Sol. We are given that
Comparing eqns. (i) and (ii), we get
frequency of the fundamental mode, v = 45 Hz
A = 7.5 cm, Z = 12 rad/s
linear density of the wire, P = 4 × 10 kg/m
k = 0.0050 rad/cm
mass of the wire, M = 3.5 × 10 kg
Speed of wave propagation,
Clearly, length of the wire,
Z
12 rad / s
v = k 0.0050 rad / cm = 2400 cm/s = – 24 m/s
L = M/P = (3.5 × 10 kg)/ (4 × 10 kg/m) = 0.875 m
–2
–2
–2
(a)
–2
When the string is vibrating in its fundamental mode,
L = O/2
(Negative sign has been taken as the wave travels towards left)
or O = 2L = (2 × 0.875) m = 1.75 m
We thus find that vp (particle speed) is not equal to v (wavespeed*).
(b)
S
2S
2S
,O
As k =
= 12.6 m
O
k 0.0050
If v is the speed of transverse wave on the string,
X = vO = (45 × 1.75) m/s = 79 m/s
(b)
All points located at distances, x = nO (where n = + 1, +2, + 3, ...)
from the point x = 1 cm have the same displacement and velocity.
Example - 32
X=
or
A uniform rope of length 12 m and mass 6 kg hangs
vertically from a rigid support. A block of mass 2 kg is
attached to the free end of the rope. A transverse pulse of
wavelength 0.06 m is produced at the lower end of the
rope. What is the wavelength of the rope when it reaches
the top of the rope ?
Sol. Tension at the lower end, T1 = 2 kg wt
Tension at the top end T2 = 2 kg wt + 6 kg wt = 8 kg wt
Speed of the wave at the lower end,
X1 = T1 / P
If T is the tension in the string,
T
P
2
or T = X P
2
–2
T = (79) (4 × 10 ) = 248 N
Example - 34
One end of a l ong stri ng of l inear mass d ensi ty
–3
= 8.0 × 10 kg/m is connected to an electrically driven
tuning fork of frequency 256 Hz. The other end passes
over a pulley and is tied to a pan containing a mass of
90 kg. The pulley end absorbs all the incoming energy so
that reflected waves at this end have negligible amplitude.
At t = 0, the left end (fork end) of the string x = 0 has zero
transverse displacement (y = 0) and is moving along
positive y-direction. The amplitude of the wave is 5.0 cm.
Write down the transverse displacement y as a function
of x and t that describes this wave on the string.
Speed of the wave at the top end,
X2 =
Sol. We are given that,
T2 / P
X1
Thus, X
2
tension in the string, T = (90 × 9.8) N = 882 N
T1
T2
2
8
1
2
mass per unit length of the string,
–3
P = 8.0 × 10 kg/m
frequency of the wave, v = 256 Hz
or
X2 = 2X1 or vO2 = 2 (vO1) or O2 = 2O1
(as frequency remains the same)
As O1 = 0.06 m,
amplitude of the wave, A = 5.0 cm = 0.05 m
Since the wave propagation along the string is a transverse
travelling wave, the velocity of the wave is given by
O2 = 2 (0.06 m) = 0.12 m
Example - 33
A wire stretched between two rigid supports vibrates in its
fundamental mode with a frequency of 45 Hz. The mass of
–2
–2
the wire is 3.5 × 10 kg and its linear density is 4 × 10 kg/m.
What is : (a) speed of a transverse wave on the string and
(b) the tension in the string ?
X=
T
P
882 N
2
= 3.32 × 10 m/s
8 u103 kg / m
3
Now, Z = 2Sv = (2 × 3.14 × 256) rad/s = 1.61 × 10 rad/s
Z
Z 1.61u10 3 rad / s
,
k
As v = k
X 3.32 u10 2 m / s = 4.84 rad/m
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126
OSCILLATION AND WAVES
As the wave travels towards the positive direction of Xaxis, the equation of the wave is :
Example - 36
Calculate the speed of sound in a liquid of density
3
9
2
8000 kg/m and bulk modulus 2 × 10 N/m .
y = Asin(Zt – kx)
where x, y and A are in metres. Thus,
3
3
Sol. Here, U = 8000 kg/m
y = 0.05 sin (1.61 × 10 t – 4.84 x)
9
2
and B = 2 × 10 N/m
Example - 35
A uniform rope of mass 0.1 kg and length 2.45 m hangs
from a ceiling.
Thus, X =
(a) Find the speed of transverse wave in the rope at a point
0.5 m distant from the lower end.
= 500 m/s
(b) Calculate the time taken by a transverse wave to travel a
full length of the rope.
B
U
2 u109 N / m 2
8000 kg / m 3
Example - 37
10
For aluminium, the bulk modulus of elasticity is 7.5 × 10
2
3
3
N/m and density is 2.70 × 10 kg/m . Deduce the speed of
longitudinal waves in aluminium.
Sol. (a) If L is the length of the rope and m is its mass, then mass
per unit length of the rope, i.e.,
Sol. We are given that
P=
m
L
10
Tension at a point distant y from the lower end, i.e.,
If X is the speed of the transverse wave set up in the rope,
3
3
7.5 u1010 N / m 2
2.70 u103 kg / m3
3
= 5.3 × 10 m/s.
Example - 38
Pyg
P
X= T/P
B
U
Thus, X =
T = weight of the rope of length y = Pyg
yg
The prongs of a tuning fork A, originally in unison with
tuning fork B are filed. Now the tuning forks on being
sounded together produce 2 beats/s. What is frequency
of A after filing, if frequency of B is 250 Hz ?
When y = 0.5 m,
X=
2
B = 7.5 × 10 N/m , U = 2.7 × 10 kg/m
(0.5 m)(9.8 m / s 2 ) = 2.2 m/s
Sol. On filing, the frequency of tuning fork A increases.
(b)
dy dy
,
As X =
dt dt
dy
yg
or dt =
Integrating both sides within proper limits, we get
t
1
³ dt
L
³
g
0
0
dy
1
= (250 + 2) Hz = 252 Hz
Example - 39
L
³y
g
y
The new frequency of A after filing
gy
1/ 2
A sitar wire and a tabla when sounded together give four
beats/s. What do we conclude from this ? As the tabla
membrane is tightened, the beat rate may decrease or
increase. Explain.
dy
0
L
or
1 ª y1 / 2 º
«
»
g ¬ (1/ 2) ¼ 0
t=
2
=
g
[ L 0] 2
L
g
2.45m
As L = 2.45 m, t = 2
9.8m / s2
or
§ 1·
t = ¨¨ 2 4 ¸¸ s = 1 s
©
¹
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Sol. Let v1 and v2 be the frequencies of the sitar wire and the
tabla membrane respectively. Since in this case 4 beats/s are
heard,
v1 = v2 + 4
On tightening the tabla membrane, its frequency increases.
Now there are two possibilities :
(i)
If v1 < v2, the number of beats increases on tightening the
membrane.
(ii)
If v1 > v2, the number of beats decreases on tightening the
membrane.
OSCILLATION AND WAVES
127
Example - 40
Example - 42
A tuning fork of unknown frequency gives 4 beats/s. With
another fork of frequency 310 Hz, it gives the same number
of beats/s when loaded with wax. Find the unknown
frequency.
Two travelling waves of equal amplitudes and equal
frequencies move in opposite directions along a string.
They interfere to produce a standing wave having the
equation
y = A cos kx sin Zt
Sol. Let us name the tuning fork of unknown frequency as A and
the tuning fork of known frequency i.e., 310 Hz as B. When
A and B are sounded together, 4 beats are heard per second.
–1
where A = 1.0 mm, k = 1.57 cm and Z = 78.5 s
Find : (a) the velocity of the component waves (b) the
node closest to the origin (in the region x > 0) and (c) the
antinode closest to the origin (x > 0) and (d) the amplitude
of the particle at x = 2.33 cm.
Thus, frequency of A, i.e.,v is either (310 + 4) Hz = 314 Hz
or
v = (310 – 4) Hz = 306 Hz
In case, the frequency of A is 306 Hz, on loading it with wax,
its frequency decreases and may become 305 Hz, 304 Hz,
303 Hz etc. In that case, the number of beats given by it
per second when sounded with B will be (310 – 305) = 5,
(310 – 304) = 6, (310 – 303) = 7 etc. i.e., the number of beats/
s is more than 4. Thus, v = 306 Hz.If v = 314 Hz, on loading
A with wax, its frequency may decrease to 306 Hz and then
it will give (310 – 306) = 4 beats/s with B. Thus, the unknown
frequency is 314 Hz.
Sol. (a) The standing wave is formed by the superposition of
the waves y1 = (A/2) sin (Zt – kx) and
y2 = A/2 sin (Zt + kx) as y = y1 + y2
The wave velocity (magnitude) of either wave, i.e.,
Z 78.5s 1
X = k 1.57 cm 1 = 50 cm/s
(b)
For a node, y = 0, i.e., cos kx = 0
Example - 41
or kx =
Calculate the speed of sound in a gas in which two
wavelength 204 cm and 208 cm produce 20 beats in 6 second.
(c)
Let speed of sound in the gas = X cm/s
For an antinode, |cos kx| = 1
or kx = S (for the smallest positive value of x)
Frequency of one wave = v0
or x =
Frequency of second wave = v2
Number of beats produced per second,
vb =
20
6
(d)
S
= 2 cm
k
The amplitude of vibration of the particle at x is given by
|A cos kx|.
For the given point (x = 2.33 cm),
amplitude = Acos kx = (1.0 mm)
As v1 – v2 = vb,
X
X 20 §
X
and v2
¨ as v1
204 208 6 ©
O1
S
(for the smallest positive value of x)
2
S / 2 3.14 / 2
or x = k 1.57 cm 1 1cm
Sol. Here, wavelength of one wave, O1 = 204 cm
Wavelength of the second wave O2 = 208 cm
–1
–1
cos [(1.57 cm )(233 cm)]
X·
¸
O2 ¹
= (1.00 mm) cos (3.658 rad) = 0.875 mm
[cos (3.658 rad) = cos (3.658 × 57°)
= cos 209° = cos (180° + 29°)
or
or
X (208 204) 20
204 u 208
6
§ 20 · § 204 u 208 ·
X ¨ ¸¨
¸ cm/s
4
© 6 ¹©
¹
= 35360 cm/s = 353.6 m/s
= – cos 29° = –0.875
= 0.875 in magnitude]
Example - 43
A 3.6 g string of a sonometer is 64 cm long. What should
be the tension in the string in order that it may vibrate in
2 segments with a frequency of 256 Hz ?
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128
OSCILLATION AND WAVES
Sol. Here, l = 64 cm = 0.64 m,
–3
–2
(On shortening the length, the frequency of the wire
increases)
–3
P = 3.6 × 10 kg/64 × 10 m = 5.6 × 10 kg/m,
v1 = 256 Hz
According to the law of length,
Since the string vibrates in 2 segments,
vL = v’L’
v1 = 2v
or
1 T
256 = 2 ×
2A P
or
v × 73 = (v + 3) (72.5) = 72.5 v + 217.5
or
73 v – 72.5 v = 217.5 or 0.5 v = 217.5
v = 435 Hz
Example - 46
or
1
T
256 =
0.64 (5.6 u103 )
or
T = (256 × 0.64) (5.6 × 10 ) N = 150 N
2
A string vibrates with a frequency 200 Hz. Its length is
doubled and its tension is altered until it begins to vibrate
with frequency 300 Hz. What is the ratio of the new tension
to the original tension ?
–3
Example - 44
A wire of length 1.5 m is stretched by a force of 44 N. The
3
diameter of the wire is 2 mm and its density is 1.4 g/cm .
Calculate the frequency of the fundamental note emitted
by it.
Sol. Let v1 and v2 be the frequencies in the two cases and L1, L2
and T1, T2 be respectively the corresponding lengths of the
string and tensions under which it is vibrating.
v1 =
1 T1
,
2L1 P
v2 =
1
2L2
Sol. With ususal notation, we are given that
L = 1.5 m, T = 44 N
–3
D = 2 mm = 2 × 10 m
3
3
U = 1.4 g/cm = 1400 kg/m
We know that v =
=
=
T2
P
(For the same string, P in both the cases remains same)
1
T
LD SU
1
44
Hz
3
1.5(2 u10 ) (22 / 7) u1400
v1
Thus, v
2
or
100
Hz = 33.3 Hz.
3
Example - 45
When the wire of a sonometer is 73 cm long, it is in
resonance with a tuning fork. On shortening the wire by
0.5 cm, it makes 3 beats with the same fork. Calculate the
frequency of the tuning fork.
Sol. We are given that
L (original length of the wire) = 73 cm
L’ (shortened length of the wire) = (73 – 0.5) cm = 72.5 cm
Let v be the frequency of the tuning fork.
This also is the frequency of the wire when its length is L as
it is then in resonance with the tuning fork. Let v’ be the
frequency of the wire when its length is L’. Since it then
produces 3 beats/s with the tuning fork, v’ = v + 3
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L2
L1
T1
T2
T
200
2 1
300
T2
(as v1 = 200 Hz, v2 = 300 Hz and L2 = 2L1 or L2/L1 = 2)
or
T
2
T
4
T
2 1 or
4 1 or 2 9
T1
3
T2 9
T2
Thus, the new tension is 9 times the original tension.
Example - 47
A wire of length 108 cm produces a fundamental note of
frequency 256 Hz, when stretched by a weight of 1 kg. By
how much its length should be increased so that its pitch
is raised by a major tone, if it is now stretched by a weight
of 4 kg ?
Sol. Case I. Frequency of the fundamental note, v = 256 Hz
length of the wire, L = 108 cm
stretching force, T = 1 kg f
OSCILLATION AND WAVES
129
Case II. Frequency of the major tone emitted
i.e., v’ =
harmonics are present whereas in the case of a closed organ
pipe only odd harmonics are present.
9
§9·
v ¨ ¸ 256 Hz = 288 Hz
8
©8¹
Example - 49
Find the ratio of length of a closed pipe to that of the open
pipe in order that the second overtone of the former is in
unison with fourth overtone of the latter.
stretching force, T = 4 kg f
Let the length be increased by x.
Increased length of the wire, i.e.,
Sol. Let v and v’ be the fundamental frequencies of the open and
the closed pipes respectively of lengths L and L’.
L’ = L + x = (108 + x) cm
As v =
1 T
,
2L P
Clearly, v =
v
v
and v’ =
2L
4L '
...(i)
where v is the speed of sound.
for Case I, 256 =
Second overtone of the closed pipe = 5v’
1
1kg f
2 u108
P
Fourth overtone of the open pipe = 5v
Since the two notes are to be in unison,
and for Case II,
5v’ = 5v or v’ = v
1
4 kg f
288 =
2 u (108 x)
P
Thus, from eqns. (i) and (ii),
Dividing, we get
v
v
4L ' 2L
256 108 x
288 108 u 4
or
or
8 108 x 108 x
or 972 + 9x = 1728
9 108 u 2
216
or
9x = 1728 – 972 = 756 cm or x =
4L’ = 2L or
L' 2 1
or L’ : L :: 1 : 2
L 4 2
Example - 50
3
A brass rod (density 8.3 g/cm ), 3 m long is clamped at the
centre, It is excited to give longitudinal vibrations and the
frequency of the fundamental note is 600 Hz. Calculate the
velocity of sound in the rod and its Young’s modulus.
756
= 84 cm
9
Sol. Here, density of the brass rod,
Thus, the length should be increased by 84 cm.
3
3
length of the rod, L = 3 m
A pipe of length 2L open at both ends has the same
frequency as another pipe of length L closed at one end.
Prove this. Also state if the sounds will be identical for the
two pipes.
frequency of the note produced, v = 600 Hz
Let the wavelength of vibration in the rod = O
As the rod is clamped at its centre, the centre is a node and
free ends are antinodes.
Sol. Fundamental frequency of open pipe of length 2L, i.e.,
v
v = 2 u 2L
3
U = 8.3 g/cm = 8.3 × 10 kg/m
Example - 48
v
4L
Thus, O = 2 L = 6m
If speed of sound in brass is X, then
...(i)
X = vO
Fundamental frequency of a closed pipe of length L, i.e.,
v
v’ =
4L
From eqns. (i) and (ii), v = v’
or
...(ii)
The sound (i.e., its quality) which depends upon the number
of harmonics and their relative intensities will not be the
same in the two cases. In case of an open organ pipe, all the
3
X= (600 Hz) (6 m) = 3.6 × 10 m/s
Let Young’s modulus of the material of the rod = Y
As X=
Y/U ,
2
Y=X ×U
3
2
3
3
11
2
= (3.6 × 10 m/s) (8.3 × 10 kg/m ) = 1.1 × 10 N/m .
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130
OSCILLATION AND WAVES
From eqns. (i) and (ii),
Example - 51
A steel rod 100 cm long is clamped at its middle. The
fundamental frequency of longitudinal vibrations of the rod
is given to be 2.53 kHz. What is the speed of sound in steel?
v’ – v’’ = (1 + Xs/X)v – (1 – Xs/X)v =
or
Sol. Here, length of the steel rod, L = 100 cm = 1m
fundamental frequency of the longitudinal vibrations,
Example - 53
If Ois the wavelength of the wave generated,
Consider a source moving towards an observer at the speed
of Xs = 0.95X. Deduce the observed frequency if the original
is 500 Hz. (Think what would happen if Xs>X. Jet planes
moving faster than sound are now so common). Here, Xis
the speed of sound.
O/2 = L or O= 2L = 2m
Therefore, speed of sound in steel rod, i.e.,
3
X = vO = (2.53 × 10 )(2 m)
3
= 5.06 × 10 m = 5.06 km/s
Sol. We are given that v = 500 Hz, Xs = 0.95 v.
Aliter : For a rod vibrating longitudinally,
Since the observer is at rest and the source is moving towards
the observer, the apparent frequency,
fundamental frequency,
or
or
X= 5.06 × 10 m/s = 5.06 km/s
v=
§ X · §
·
X
500
v’ = ¨ X X ¸ v ¨ X 0.95 X ¸ 500 =
0.05
©
¹
s ¹
©
3
or
Example - 52
Two tuning forks with natural frequencies of 340 Hz each
move relative to a stationary observer. One fork moves
away from the observer while the other moves towards
him at the same speed. The observer hears beats of
frequency 3 Hz. Find the speed of the tuning fork. Speed
of sound in air = 340 m/s.
Sol. Let Xs be the speed of source and X be that of the sound
waves. Let v be the frequency of the tuning fork. If v’ is its
apparent frequency when it moves towards the stationary
observer,
v’ = 10,000 Hz
If Xs >X(X–Xs) is negative and as such v’ is negative which
has no meaning. Thus, Doppler formula is applicable so
long as the velocity of the source does not exceed the
velocity of the wave. In case of jet planes which move with
speed greater than the sound, a shock wave is formed whose
wavefront is a cone with the plane at its apex. The semi–1
vertical angle of the cone is sin (X/Xs)
Example - 54
What is the speed of the observer for whom a note is
10 per cent lower than the emitted frequency ?
Sol. As the apparent frequency (v’) is less than emitted frequency
(v), the observer must move away from the source.
§ X ·
1
–1
v’ = ¨ (X X ) ¸ v (1 X / X) v = (1 – XS/X) v
©
S ¹
S
If Xis the speed of sound and Xo that of the observer, then
Applying binomial theorem (as XS/X < 1), we get
§ Xs
v’ = ¨1 X
©
·
¸v
¹
v’ =
...(i)
·
1
v = (1 + X /X)–1v
¸v
s
(1
X
S / X)
¹
Applying binomial theorem (since Xs/X< 1), we get
v’’ = (1 – Xs/X)v
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...(ii)
X Xo
v
X
As the apparent frequency is 10% lower than the emitted
frequency,
If v’’ is the apparent frequency of the tuning fork when it
moves away from the stationary source,
§ X
v’’ = ¨ X X
s
©
(v ' v '')v 3u 340 m / s
2v
2 u 340 = 1.5 m/s
(as v’ – v’’ = 3, X = 340 m/s and v = 340 Hz)
3
v = 2.53 kHz = 2.53 × 10 Hz
v
2L
3 –1
X= 2vL = 2 (2.53 × 10 s )(1 m)
Xs =
2 Xs v
X
v’ = v –
10
90
v
v = 0.9 v
100
100
or
§ X Xo
0.9 v = ¨
© X
or
0.9 X = X–X0
·
¸v
¹
§ X Xo ·
or 0.9 = ¨
¸
© X ¹
or X0 = X– 0.9 X= 0.1 X
Thus, speed of the observer is (1/10)th of the speed of sound.
OSCILLATION AND WAVES
131
EXERCISE - 1 : BASIC OBJECTIVE QUESTIONS
Derivation of equation of SHM
1.
6.
Which of the following equations does not represent a simple
harmonic motion ?
when it is in its mean position. If the amplitude of oscillations
(a) y = a sin Z t
period of its oscillation in second is
(b) y = b cos Z t
(a) S/5
(b) 2 S
(c) y = a sin Z t + b cos Z t
(c) 5 S
(d) 20 S
(d) y = a tan Z t
2.
is 25 cm and the mass of the particle is 5.12 kg, the time
7.
oscillating simultaneously from two opposite extreme
(a) y = 0.1 cos (3 t + S/4) (b) y = 0.1 sin (3 t + S/4)
(a)
24
s
5
(b)
12
s
5
(c)
24
s
11
(d)
12
s
11
positions. After how much time they will be in the same
phase ?
(d) y = 0.1 cos (4 t + S/4)
The displacement of a particle is represented by the equation:
§S
·
y = 3 cos ¨ 2Zt ¸
4
©
¹
8.
The motion of the particle is :
(a) simple harmonic with period 2S/Z
–2
(b) 4 × 10 J
–2
(d) 16 × 10 J
(a) 2 × 10 J
(c) periodic but not simple harmonic
(c) 8 × 10 J
9.
Phase, Displacement, Velocity, Acceleration, Kinetic, Potential
Energy as a time and positon
5.
The force constant of a weightless spring is 16 N/m. A body
of mass 1.0 kg suspended from it is pulled down through
5 cm and then released. The maximum kinetic energy of the
system (spring + body) will be
(b) simple harmonic with period S/Z
(d) non-periodic
4.
Two pendulums of time period 3 s and 8 s respectively starts
A particle of mass 0.1 kg is executing SHM of amplitude 0.1 m.
When the particle passes through the mean position, its
–3
K.E. is 8 × 10 J. The equation of motion of the particle,
o
when the initial phase of oscillation is 45 , is
(c) y = 0.1 sin (4 t + S/4)
3.
The kinetic energy of a particle executing S.H.M. is 16 J
A particle is performing S.H.M. along X-axis with amplitude
4 cm and time period 1.2 sec. The minimum time taken by the
particle to move from x = + 2 cm to x = + 4 cm and back again
is given by
(a) 0.6 s
(b) 0.4 s
(c) 0.3 s
(d) 0.2 s
–2
–2
The time taken by a particle executing S.H.M. of period T to
move from the mean position to half the maximum
displacement is
(a) T/2
(b) T/4
(c) T/8
(d) T/12
10. A particle is executing simple harmonic motion with
frequency f. The frequency at which its kinetic energy
changes into potential energy is
The displacement of an oscillating particle varies with time
(a) f/2
(b) f
S§ t 1·
(in seconds) according to the equation y = sin ¨ ¸ ,
2 © 2 3¹
(c) 2 f
(d) 4 f
wher y is in cm. The maximum acceleration of the particle is
approximately
–2
(b) 1.81 cm s
–2
–2
(a) 0.62 cm s
(c) 3.62 cm s
–2
(d) 5.2 cm s
11.
A particle executes simple harmonic motion between x = –A
and x = + A. The time taken for it to go from 0 to A/2 is T1 and
to go from A/2 to A is T2. Then
(a) T1 < T2
(b) T1 > T2
(c) T1 = T2
(d) T1 = 2 T2
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132
OSCILLATION AND WAVES
12. A simple pendulum is oscillating without damping, Figure.
When the displacement of the bob is less than maximum, its
G
acceleration vector a is correctly shown in
(a)
(b)
(a)
(b)
(c)
(d)
16. The relation between acceleration and displacement of four
particles are given below :
(a) ax = + 2x
2
(c) ax = – 2x
(c)
(d)
13. For a particle executing simple harmonic motion, the
displacement x is given by x = A sin Zt. Identify the graph,
Figure which represents the variation of potential energy
(PE) as a function of time t and displacement x
2
(b) ax = + 2x
(d) ax = – 2x
which one of the particles is executing simple harmonic
motion ?
17. A particle executing SHM has a maximum speed of 30 cm/s
2
and a maximum acceleration of 60 cm/s . The period of
oscillation is :
(a) S s
(b)
S
s
2
(c) 2S s
(d)
S
s
4
Spring Block Oscillation
18. A force of 6.4 N stretches a vertical spring by 0.1 m. The
mass that must be suspended from the spring so that it
oscillates with a period of (S/4) second is
(a) (S/4) kg
(b) 1 kg
(c) (S) kg
(d) 10 kg
19. A mass of 1 kg attached to the bottom of a spring has a
certain frequency of vibration. The following mass has to
be added to it in order to reduce the frequency by half.
(a) I, III
(b) II, III
(a) 1 kg
(b) 2 kg
(c) I, IV
(d) II, IV
(c) 3 kg
(d) 4 kg
14. The potential energy of a harmonic oscillation of mass 2 kg
in its mean position is 5 J. If its total energy is 9 J and its
amplitude is 0.01 m, its time period of oscillation will be
(a) (S/10) s
(b) (S/20) s
(c) (S/50) s
(d) (S/100) s
15. A mass m is performing linear simple harmonic motion, then
which of the following graph represents correctly the variation
of acceleration a corresponding to linear velocity v ?
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20. Two bodies M and N of equal masses are suspended from
two separate massless springs of spring constant k1 and k2
respectively. If the two bodies oscillate vertically such that
their maximum velocities are equal, the ratio of the amplitude
of M to that of N is
(a) k1/ k2
(b)
k1 / k 2
(c) k2 / k1
(d)
k 2 / k1
OSCILLATION AND WAVES
21.
133
The period of oscillation of a mass m suspended from a
spring is 2 seconds. If along with it another mass of 2 kg is
also suspended, the period of oscillation increases by one
second. The mass m will be
(a) 2 kg
(b) 1 kg
(c) 1.6 kg
(d) 2.6 kg
Angular SHM
22.
–1
A pendulum bob has a speed of 3 ms at its lowest position.
The pendulum is 0.5 m long. The speed of the bob, when the
o
length makes an angle of 60 to the vertical will be
–2
(if g = 10 ms )
–1
(a) (1/3) ms
–1
2T
(b)
5
5T
2
2
5
(c)
2T
(d)
5T
Miscellaneous SHM problems
24.
Four simple harmonic motions ; x 1 = 8 sin Zt ; x2 = 6 sin
(Zt + S/2) ; x3 = 4 sin (Zt + S) and x4 = 2 sin (Zt + 3 S/2) are
superimposed on each other. The resulting amplitude and
its phase difference with x1 are respectively
–1
(b) 4 2 , S/2
–1
(d) 4 2 , S/4
(a) 20, tan (1/2)
(c) 20, tan (2)
25.
(c) 15 rad/s
(d) None of these
28. A body is moving in a room with a velocity of 20 m/s
perpendicular to the two walls separated by 5 m. There is no
frictrion and the collisions with the walls are elastic. The
motion of the body is
(c) periodic and simple harmonic
(d) periodic with variable time period
29. A particle of mass m moving along the x-axis has a potential
2
energy U(x) = a + bx where a and b are positive constants.
It will execute simple harmonic motion with a frequency
detemined by the value of
(a) b alone
(b) b and a alone
(c) b and m alone
(d) b, a and m alone
30. A metre stick swinging in vertical plane about a fixed
horizontal axis passing through its one end undergoes small
oscillation of frequency f0. If the bottom half of the stick
were cut off, then its new frequency of small oscillation
would become.
The displacement equation of a particle is
x = 3 sin 2 t + 4 cos 2 t
The amplitude and maximum velocity will be respectively
26.
(b) 14 rad/s
(b) periodic but not simple harmonic
(d) 3 ms
A man measures time period of a simple pendulum inside a
stationary lift and find it to be T. If the lift starts accelerating
upwards with an acceleration g/4, then the time period of
pendulum will be
(a)
(a) 13 rad/s
(a) not periodic
–1
(c) 2 ms
23.
–1
(b) (1/2) ms
27. A block of mass 1 kg hangs without vibrating at the end of
a spring whose force constant is 200 N/m and which is
attached to the ceiling of an elevator. The elevator is rising
with an upward acceleration of g/3 when the acceleration
suddenly ceases. The angular frequency of the block after
the aceceleration ceases is
(a) 5, 10
(b) 3, 2
(c) 3, 4
(d) 4, 2
The displacement of a particle varies with time according to
the relation : y = asin Zt + bcos Zt.
Choose the correct statement.
(a) f0
(b)
(c) 2f0
(d) 2 2 f0
31. A physical pendulum is positioned so that its centre of
gravity is above the suspension point. When the pendulum
is released it passes the point of stable equilibrium with an
angular velocity Z. The period of small oscillations of the
pendulum is
(a) The motion is oscillatory but not SHM.
(b) The motion is SHM with amplitude (a + b).
2
(a)
4S
Z
(b)
2S
Z
(c)
S
Z
(d)
S
2Z
2
(c) The motion is SHM with amplitude (a + b )
(d) The motion is SHM with amplitude
2 f0
a b .
2
2
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OSCILLATION AND WAVES
Waves
Standing Wave in organ Pipe
Wave Parameter
38. A pipe closed at one end porudces a fundamental note of
412 Hz. It is cut into two equal length, the fundamental notes
produced by the two pieces are
32. A string of 5.5 m length has a mass 0.035 kg. If the tension in
the string is 77 N, then the speed of wave on the string is :
(a) 77 m/s
(b) 102 m/s
(a) 206 Hz, 412 Hz
(b) 206 Hz, 824 Hz
(c) 110 m/s
(d) 164 m/s
(c) 412 Hz, 824 Hz
(d) 824 Hz, 1648 Hz
Superposition (Interference)
–2
33. A string of length 0.4 m and mass 10 kg is tightly clamped
at the ends. The tension in the string is 1.6 N. Identical wave
pulses are produced at one end at equal intervals of time 't.
The minimum value of 't which allows constructive
interference between successive pulses is :
(a) 0.05 s
(b) 0.10 s
(c) 0.20 s
(d) 0.40 s
Standing Wave
34. A standing wave having 3 nodes and 2 antinodes is formed
o
two atoms having a distance of 1.21 A between them. The
wavelength of standing wave is
(b) 300 Hz
(c) 240 Hz
(d) 480 Hz
40. A vehicle with a horn of frequency n is moving with a
velocity of 30 m/s in a direction perpendicular to the straight
line joining the observer and the vehicle. The observer
perceives the sound to have a frequency (n + n1). If velocity
of sound in air is 300 m/s, n1 would be
(a) n1 = 10 n
(b) n1 = 0
o
(c) n1 = 0.1 n
(d) n1 = – 0.1 n
(b) 2.42 A
o
(d) 3.63 A
(c) 6.05 A
(a) 200 Hz
o
o
(a) 1.21 A
39. An open pipe is suddenly closed at one end with the result
that the frequency of third harmonic of the closed pipe is
found to be higher by 100 Hz than fundamental frequency
of the open pipe. The fundamental frequency of the open
pipe is
35. A standing wave consisting of 3 nodes and 2 antinodes is
formed between the two atoms having a distance of 1.21 Å
between them. The wavelength of the standing wave is :
(a) 1.21 Å
(b) 2.42 Å
(c) 6.05 Å
(d) 3.63 Å
41. An open pipe is in resonance in 2nd harmonic with frequency
f1. Now one end of the tube is closed and frequency is
increased to f2 such that the resonance again occurs in nth
harmonic. Choose the correct option.
(a) n = 3, f2 =
3
f
4 1
(b) n = 3, f2 =
5
f
4 1
(c) n = 5, f2 =
5
f
4 1
(d) n = 5, f2 =
3
f
4 1
Standing Wave on Stretched string
36. A string is stretched between fixed points separated by 75.0 cm.
It is observed to have resonant frequencies of 420 Hz and
315 Hz. There are no other resonant frequencies between
these two. Then the lowest resonance frequency for this
string is
(a) 1.05 Hz
(b) 1050 Hz
(c) 10.5 Hz
(d) 105 Hz
–4
37. A stretched string of length 1 m and mass 5 × 10 kg, fixed
at both ends, is under a tension of 20 N. If it is plucked at
points situated at 25 cm from one end, it would vibrate with
a frequency :
42. If the length of a closed organ pipe is 1 m and velocity of
sound is 330 m/s, then the frequency of 1st overtone is :
(a) 4 (330/4) Hz
(b) 3 (330/4) Hz
(c) 2 (330/4) Hz
(d) none of these
43. An open organ pipe of length l vibrates in its fundamental
mode. The pressure variation is maximum :
(a) at the two ends
(b) at the distance l/2 inside the ends
(a) 400 Hz
(b) 200 Hz
(c) at the distance l/4 inside the ends
(c) 100 Hz
(d) 256 Hz
(d) at the distance l/8 inside the ends
Mahesh Tutorials Science
OSCILLATION AND WAVES
135
Resonance Column Experiment
44.
A pipe closed at one end and open at the other end resonates
with sound waves of frequencies 135 Hz and also 165 Hz
but not with any wave of frequency intermediate between
these two. The frequency of the fundamental note is :
(a) 30 Hz
(b) 15 Hz
(c) 60 Hz
(d) 7.5 Hz
Sonometer experiment and Tuning Fork
45.
A pipe closed at one end and open at the other end resonates
with sound waves of frequency 135 Hz and also 165 Hz but
not with any wave of frequency intermediate between these
two. Then the frequency of the fundamental note is
(a) 30 Hz
(b) 15 Hz
(c) 60 Hz
(d) 7.5 Hz
Beats
46.
Two wires are fixed on a sonometer. Their tensions are in the
ratio 8 : 1, their lengths are in the ratio 36 : 35, the diameters
are in the ratio 4 :1 and densities are in the ratio 1 : 2. If the
–1
note of the higher pitch has a frequency 360 s , the frequency
of beats produced is
(a) 5 s
–1
(b) 10 s
(d) 20 s
Two sources A and B are sounding notes of frequency
680 Hz. A listener moves from A to B with a constant velocity
–1
u. If speed of sound is 340 ms , what should be the value of
u so that he hears 10 beats/s ?
–1
(b) 3.0 ms
–1
(d) 3.5 ms
(a) 2.0 ms
(c) 2.5 ms
49.
50.
(c) 205 Hz
(d) 215 Hz
51. A source of frequency v gives 5 beats/s when sounded
with a frequency 200 Hz. The second harmonic of source
gives 10 beats/s when sounded with a source of frequency
420 Hz. The value of v is :
(a) 200 Hz
(b) 210 Hz
(c) 205 Hz
(d) 195 Hz
Dopplers Effect
52. When a source is going away from a stationary observer
with a velocity equal to velocity of sound in air, then the
frequency heard by the observer will be
(a) same
(b) double
(c) half
(d) one third
53. A car sounding its horn at 480 Hz moves towards a high wall
–1
at a speed of 20 ms , the frequency of the reflected sound
heard by the man sitting in the car will be nearest to ; (speed
of sound 330 m/s)
(a) 480 Hz
(b) 510 Hz
(c) 540 Hz
(d) 570 Hz
–1
(c) 15 s
48.
(b) 210 Hz
–1
–1
47.
(a) 200 Hz
–1
–1
Two sound waves with wavelength 5.0 m and 5.5 m
respectively, each propagate in a gas with velocity 330 m/s.
We expect the following number of beats/sec.
(a) 6
(b) 12
(c) 0
(d) 1
Three sound waves of equal amplitudes have frequencies
(v – 1), v and (v + 1). They superpose to give beats. The
number of beats per second will be :
(a) 4
(b) 3
(c) 2
(d) 1
A tuning fork vibrating with a sonometer having 20 cm wire
produces 5 beats/s. The beat frequency does not change if
the length of the wire is changed to 21 cm. The frequency of
the tuning fork must be :
54. A siren placed at a railway platform is emitting sound of
frequency 5 k Hz. A passenger sitting in a moving train A
records a frequency of 5.5 k Hz, while the train approaches
the siren. During his return journey in a different train B, he
records a frequency of 6.0 k Hz, while approaching the same
siren. The ratio of the velocity of train B to that of train A is
(a) 242/252
(b) 2
(c) 5/6
(d) 11/6
55. A car is moving towards a high cliff. The driver sounds a
horn of frequency f. The reflected sound heard by the driver
has a frequency 2 f. If v be the velocity of sound, then the
velocity of the car, in the same velocity units would be
(a) v/4
(b) v/2
(c) V/ 2
(d) v/3
56. A police car with a siren of frequency 8 kHz is moving with
uniform velocity of 36 km/h towards a tall building which
reflects the sound waves. The speed of sound in air is 320 m/s.
The frequency of the siren heard by the car driver is :
(a) 8.50 kHz
(b) 8.25 kHz
(c) 7.75 kHz
(d) 7.50 kHz
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136
OSCILLATION AND WAVES
57. Two trains, each moving with a velocity of 30 m/s, cross
each other. One of the trains gives a whistle whose frequency
is 600 Hz. If the speed of sound is 330 m/s, the apparent
frequency for passengers sitting in the other train before
crossing would be :
(a) 600 Hz
(b) 630 Hz
(c) 920 Hz
(d) 720 Hz
58. A whistle producing sound waves of frequencies 9500 Hz
and above is approaching a stationary person with speed v
m/s. The velocity of sound in air is 300 m/s. If the person
can hear frequencies upto a maximum of 10,000 Hz, the
maximum value of v upto which he can hear the whistle is :
(a) 15 / 2 m/s
(b) 15 m/s
(c) 30 m/s
(d) 15 2 m/s
62. A transverse wave is described by the equation
y
x·
§
y0 sin 2S ¨ ft ¸
O¹
©
The maximum particle velocity is equal to four times the
wave velocity if
(a) O
Sy0
4
(b) O
Sy0
2
(c) O
Sy 0
(d) O
2 Sy 0
63. Wave pulse on a string shown in figure is moving to the
right without changing shape. Consider two particles at
positions x1 = 1.5 m and x2 = 2.5 m. Their transverse velocities
at the moment shown in figure are along directions
59. A vehicle with a horn of frequency v is moving with a
velocity of 30 m/s in a direction perpendicular to the straight
line joining the observer and the vehicle. The observer
perceives the sould to have a frequency v + v1. If the velocity
of sound in air is 300 m/s, v1 would be :
(a) v1 = 10 v
(b) v1 = 0
(a) positive y-axis and positive y-axis respectively
(c) v1 = 0.1 v
(d) v1 = –0.1 v
(b) negative y-axis and positive y-axis respectively
60. The path difference between the two waves
y1
(a)
(c)
(d) negative y-axis and negative y-axis respectively
2Sx ·
§
a1 sin ¨ Zt ¸
O ¹
©
and y 2
64. A progressive wave is given by
y = 3 sin 2S[(t/0.04) – (x/0.01)]
2Sx
§
·
I ¸ is
a 2 cos ¨ Zt O
©
¹
O
I
2S
(b)
2S §
S·
¨I ¸
O ©
2¹
(d)
Where x, y are in cm and t in s. The frequency of wave and
maximum acceleration will be
O §
S·
I ¸
¨
2S ©
2¹
2S
I
O
61. The equation of a wave travelling on a string is
y
(c) positive y-axis and negative y-axis respectively
S§
x·
4sin ¨ 8t ¸
2©
8¹
3
2
(a) 100 Hz, 4.7 × 10 cm/s
3
2
4
2
4
2
(b) 50 Hz, 7.5 × 10 cm/s
(c) 25 Hz, 4.7 × 10 cm/s
(d) 25 Hz, 7.5 × 10 cm/s
65. Which of the following is not true for the progressive wave
y
x ·
§ t
4sin 2S ¨
¸
© 0.02 100 ¹
if x and y are in centimetres, then velocity of wave is
Where x and y are in cm and t in seconds.
(a) 64 cm/sec in –ve x-direction
(a) The amplitude is 4 cm
(b) 32 cm/sec in –ve x-direction
(b) The wavelength is 100 cm
(c) 32 cm/sec in +ve x-direction
(c) The frequency is 50 Hz
(d) 64 cm/sec in +ve x-direction
(d) The velocity of propagation is 2 cm/s
Mahesh Tutorials Science
OSCILLATION AND WAVES
66.
67.
137
An increase in intensity level of 1 dB implies an increase in
density of (given antilog10 0.1 = 1.2589)
(a) 1 %
(b) 3.01 %
(c) 26 %
(d) 0.1 %
68. A source of sound emits 200S W power which is uniformly
distributed over a sphere of radius 10 m. What is the
loudness of sound on the surface of the sphere?
The intensity level of two sounds are 100 dB and 50 dB.
What is the ratio of their intensities?
1
(b) 10
5
(d) 10
(a) 10
(c) 10
(a) 70 dB
(b) 74 dB
(c) 80 dB
(d) 84 dB
3
10
EXERCISE - 2 : PREVIOUS YEARS AIEEE/JEE MAINS QUESTIONS
OSCILLATIONS
1.
5.
Two identical springs of spring contant k are connected
in series and parallel as shown in figure. A mass M is
suspended from them. The ratio of their frequencies of
vertical oscillation will be :
(1997)
Which one of the following statement is not correct for a
particle executive SHM ?
(1999)
(a) Acceleration of the particle is maximum at the mean
position
(b) Restoring force is always directed towards a fixed point
(c) Total energy of the particle always remains the same
(d) Restoring force is maximum at the extreme position.
6.
2.
3.
4.
(a) 1 : 2
(b) 2 : 1
(c) 4 : 1
(d) 1 : 4
A particle executes simple harmonic motion with an angular
2
velocity of 3.5 rad/s and maximum acceleration 7.5 m/s .
The amplitude of oscillation will be :
(1999)
(a) 0.53 cm
(b) 0.28 cm
(c) 0.61 cm
(d) 0.36 cm
If the time period of oscillation of mass m suspended from
a spring is 2s, the time period of mass 4m will be :(1998)
A spring is vibrating with frequency under same mass. If it
is cut into two equal pieces and same mass is suspended
then the new frequency will be :
(1999)
(a) 2s
(b) 3s
(a) n 2
(b) n/ 2
(c) 4s
(d) 5s
(c) n/2
(d) n
7.
If a simple pendulum oscillates with an amplitude of 50 mm
and time period of 2 s then its maximum velocity is :
(a) 0.6 m/s
(b) 0.16 m/s
(c) 0.4 m/s
(d) 0.32 m/s
(b) 0.1556 s
(c) 0.1256 s
(d) 0.1356 s
In arangement given in figure, if the block of mass m is
displaced, the frequency is given by :
(1999)
(1998)
A horizontal platform with an object placed on it is
executing simple harmonic motion in the vertical direction.
The amplitude of oscillation is 3.92 × 10-3 m. What should
be the least period of these oscillations, so that the object
is not detached from the platform ?
(1999)
(a) 0.145 s
8.
(a) n
1 § k1 k 2 ·
2S ¨© m ¸¹
(b) n
1 § m ·
¨
¸
2S © k1 k 2 ¹
(c) n
1 § m ·
¨
¸
2S © k1 k 2 ¹
(d) n
1 § k1 k 2 ·
2S ©¨ m ¸¹
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138
9.
10.
OSCILLATION AND WAVES
Simple pendulum is executing simple harmonic motion with
time period T. If the length of the pendulum is increased
by 21%, then the increase in the time period of the
pendulum of the increased length is :
(2001)
(a) 22%
(b) 13%
(c) 50%
(d) 10%
The frequency of oscillator of the springs as shown in
figure will be :
(2001)
12.
Assertion: In SHM, the motion is to and fro and periodic.
Reason : Velocity of the particle, v
Z a x2
(where x is displacement)
13.
(2002)
(a) A
(b) B
(d) D
(e) E
(c) C
Assertion : The time period of pendulum on satellite
orbiting the earth is infinity.
Reason: Time period of a pendulum is inversely
proportional to
14.
1
2S
k1 k 2 m
1
(b) 2S
k1k 2
k1 k 2 m
(a)
(c)
k1k 2
15.
(a) A
(b) B
(d) D
(e) E
(c) C
Assertion : The amplitude of an oscillating pendulum
decreases gradually with time.
(a) A
(b) B
(d) D
(e) E
(c) C
Assertion : Water in a U-tube executes SHM, the time
period for mercury filled up to the same height in the Utube be greater than that in case of water.
Reason : The amplitude of an oscillating pendulum goes
on increasing.
(2007)
k
m
ASSERTION REASON
(a)
If both assertion and reason are true and reason is a correct
explanation of assertion.
(b)
If both assertion and reason are true but reason is not a
correct explanation of assertion.
(c)
If assertion is true but reasons is false.
(d)
If both assertion and reason are false.
(e)
If assertion is false but reason is true.
11.
Assertion : In simple harmonic motion, the velocity is
maximum when the acceleration is minimum.
Reason : Displacement and velocity of SHM differe in
phase by S/2.
(1999)
(a) A
(b) B
(d) D
(e) E
Mahesh Tutorials Science
(2002)
Reason : The frequency of the pendulum decreases with
time.
(2003)
1 k
2S m
(d) 2S
g.
16.
(a) A
(b) B
(d) D
(e) E
(c) C
Two springs of force constants k and 2k are connected to
a mass as shown below. The frequency of oscillation of
the mass is :
(a)
1 k
2S m
(b)
1 2k
2S m
(c)
1 3k
2S m
(d)
1 m
2S k
(c) C
OSCILLATION AND WAVES
17.
139
motion is
Two springs are connected to a block of mass M placed
on a frictionless surface as shown below. If both the springs
have a spring constant k, the frequency of oscillation of
block is :
(2004)
22.
(2007)
(a) 0.5 S
(b) S
(c) 0.707 S
(d) zero
A coin is placed on a horizontal platform which undergoes
vertical simple harmonic motion of angular frequency Z .
The amplitude of oscillation is gradually increased. The
coin will leave contact with the platform for the first time
(2008)
(a) at the mean position of the platform
(b) for an amplitude of g/Z2
1
2S
k
M
(b)
1
(c)
2S
2k
M
1
(d)
2S
(a)
18.
19.
20.
1
2S
k
2M
(c) for an amplitude of g2/Z2
(d) at the highest position of the platform
23.
M
k
(b) sin2 Zt
(c) sin Zt sin 2 Zt
(d) sin Zt sin 2 Zt
(b) T/8
(c) T/12
(d) T/2
A mass of 2.0 kg is put on a flat pan attached to a vertical
spring fixed on the ground as shown in the figure. The
mass of the spring and the pan is negligible. When pressed
slightly and released the mass executes a simple harmonic
motion. The spring constant is 200 N/m. What should be
the minimum amplitude of the motion, so that the mass
gets detached from the pan ?
(Take g = 10 m/s2)
(b) a periodic, but not simple harmonic motion with a period
SZ
(c) a simple harmonic motion with a period 2SZ
The partcial executes simple harmonic oscillation with an
amplitude a. The period of oscillation is T. The minimum
time taken by the particle of travel half of the amplitude
from the equilbrium position is
(2007)
(a) T/4
(2008)
(a) a periodic, but not simple harmonic motion with a period
2SZ
Which of the following functions represents a simple
harmonic oscillation ?
(2005)
(a) sin Zt cos Zt
The function sin2 Z t represents
(d) a simple harmonic motion with a period SZ.
24.
25.
A particle of mass m is executing oscillations about the
origin on the x-axis. Its potential energy is U(x) = k [x]3,
where k is a positive constant. If the amplitude of oscillation
is a, then its time period T is :
(2008)
(a) proportional to 1/ a
(b) independent of a
(c) proportional to a
3/2
(d) proportional to a .
Assertion : The periodic time of a hard spring is less as
compared to that of a soft spring.
(2009)
Reason : The periodic time depends upon the spring
constant, and spring constant is large for hard spring.
(a) If both ASSERTION and REASON are true and
reason is the correct explanation of the assertion.
(2007)
(b) If both ASSERTION and REASON are true but reason
is not the correct explanation of the assertion.
(c) If ASSERTION is true but REASON is false.
(d) If both ASSERTION and REASON are false.
(e) If ASSERTION is false but REASON is true.
(AFMC Questions)
(a) 8.0 cm
(b) 10.0 cm
(c) Any value less than 12.0 (d) 4.0 cm
21.
The phase difference between the instantaneous velocity
and acceleration of a particle executing simple harmonic
26.
Energy of simple harmonic motion depends upon :(1997)
(a) 1/Z2
(b) Z
(c) a2
(d) 1/a2
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140
27.
28.
OSCILLATION AND WAVES
(a) first increase then decrease
A particle is executing simple harmonic motion with an
angular frequency of 2 rad/s. The velocity of the particle
at 20 mm displacement, when the amplitude of the motion
is 60 mm is :
(1998)
(a) 131 mm/s
(b) 118 mm/s
(c) 113 mm/s
(d) 90 mm/s
(b) remains constant
(c) decrease
(d) increase
A particle is executing a simple hormonic motion. Its
Two springs of spring constant 1500 N/m and 3000 N/m
respectively are stretched with the same force. They will
have the potential energies in the ratio of :
(2001)
maximum acceleration is D and maximum velocity is E ,
(a) 1 :2
(b) 1 : 4
then its time period of vibration will be :
(c) 4 : 1
(d) 2 : 1
33.
(1998)
34.
(a)
(c)
29.
30.
2SE
E2
(b) 2
D
D
D
E
(d)
E2
D
A point mass m is suspended at the end of a massless wire
of length L and cross-sectional area A. If Y is the Young’s
modulus of the wire, then the frequency of the oscillation
for the simple harmonic oscillation along vertical direction:
(a)
1
2S
LA
mY
(b)
1
2S
LAm
Y
(c)
1
2S
YA
mL
(d)
1
2S
mY
AL
35.
(b) 30%
(c) 21%
(d) 10%
Two light springs of force constants k1 and k2 and a block
of mass m are in one line. AB on a smooth horizontal table
such that one end of each spring is fixed to a required support
and other end is attached to block of mass m kg as shown in
figure. The frequency of vibration is :
(2002)
A body is vibrating in simple harmonic motion with an
amplitude of 0.06 m and frequency of 15 Hz. The maximum
velocity and acceleration of the body is :
(1999)
(a) n
1
2S
k1 k2
m
(b) n
(c) n
1
2S
k1k 2
m
(d) None of these
1
2S
k1 k2
m
(c) 6.82 m/s and 7.62 × 102 m/s2
A simple pendulum of length has a maximum angular
displacement T . The maximum kinetic energy of the bob
of mass m will be :
(2002)
(d) 5.65 m/s and 5.32 × 102 m/s2
(a) mgl (1– cos T )
(b) mgl cos T
A particle executes simple harmonic motion of amplitude
A. At what distance from the mean position its kinetic
energy is equal to its potential energy ?
(2000)
(c) 2 mgl
(d) mgl
36.
(b) 8.90 m/s and 8.21 × 102 m/s2
32.
(a) 50%
(1998)
(a) 9.80 m/s and 9.03 × 102 m/s2
31.
A simple pendulum is executing simple harmonic motion
with a time period T. If the length of the pendulum is
increased by 21% the increase in the time period of the
pendulum of increased length is :
(2001)
(a) 0.81 A
(b) 0.71 A
(c) 0.61 A
(d) 0.51 A
A hollow sphere filled with water forms the bob of a simple
pendulum. A small hole at the bottom of the bob allows the
water to slowly flow out as it is set into small oscillation and its
period of oscillation is measured. The time period will :(2000)
Mahesh Tutorials Science
37.
The amplitude of the vibrating particle due to super
position of two simple harmonic motions of
y1
S·
§
sin ¨ Zt ¸ and y2
3¹
©
(a) 2
(c)
sin Zt will be : (2002)
(b)
2
(d) 1
3
OSCILLATION AND WAVES
38.
141
Work done by a simple pendulum in one complete
oscillation is :
(2002)
(a)
1
2S
1
8
(c) mg
(b) v sin
45.
In case of a forced vibration, the resonance wave becomes
very sharp when the
(2007)
T
(a) applied periodic force is small
2
(b) quality factor is small
(c) damping force is small
(d) zero
(d) restoring force is small
39.
40.
41.
42.
Pendulum after some time becomes slow in motion and
finally stops due to :
(2003)
(a) air friction
(b) earth’s gravity
(c) mass of pendulum
(d) none of the above
A period of oscillation of a simple pendulum is T in a
stationary lift. If the lift moves upwards with acceleration
of 8g, the period will :
(2004)
(a) remain the same
(b) decreases by T/2
(c) increase by T/3
(d) none of the above
46.
S
100t x , where x and y are in metre and time
2
is in second. The period of the wave in second will be
y = 5 sin
47.
(b) 0.01
(c) 1
(d) 5
(2008)
A body executes simple harmonic motion under the action
4
s. If the force is changed to
5
F2 it executes simple harmonic motion with time period
The length of second’s pendulum is 1m on earth. If mass
and diameter of a planet is doubled than that of earth, then
its length becomes :
(2005)
3
s. If both forces F1 and F2 act simultaneously in the
5
same direction on the body, its time period will be (2009)
(a) 1m
(b) 2 m
(a)
12
s
25
(b)
24
s
25
(c) 0.5 m
(d) 4m
(c)
35
s
24
(d)
15
s
12
What effect occurs on the frequency of a pendulum if it is
taken from the earth’s surface to deep into a mine ?
(a) Increases
(2005)
CBSE-PMT Questions
48.
(c) First increase then decreases
(d) No effect
44.
(a) 0.04
of force F1 with a time period
(b) Decreases
43.
The equation of a simple harmonic wave is given by
A simple pendulum is made of a body which is a hollow
sphere containing mercury suspended by means of a wire.
If a little mercury is drained off, the period of pendulum
will :
(2006)
(a) remain unchanged
(b) increase
(c) decrease
(d) become erratic
The time period of a simple pendulum in a stationary train
is T. The time period of a mass attached to a spring is also
T. The train accelerates at the rate 5m/s2. If the new time
periods of the pendulum and spring by T p and T s
respectively, then
(2007)
49.
50.
Two simple pendulums of length 0.5 m and 2.0 m
respectively are given small linear displacement in one
direction at the same time. They will again be in the same
phase when the pendulum of shorter length has completed
oscillations :
(1998)
(a) 5
(b) 1
(c) 2
(d) 3
A mass m is vertically suspended from a spring of negligible
mass; the system oscillates with a frequency n. What will
be the frequency of the system, if a mass 4 m is suspended
from the same spring ?
(1998)
(a) n/4
(b) 4 n
(c) n/2
(d) 2 n
The time period of a simple pendulum is 2s. If its length is
increased by 4 times, then its period becomes : (1999)
(a) Tp = Ts
(b) Tp > Ts
(a) 16 s
(b) 12 s
(c) Tp < Ts
(d) cannot be predicted
(c) 8 s
(d) 4 s
Lakshya Educare
142
51.
OSCILLATION AND WAVES
Two
simple
harmonic
motions
given
by
S·
§
x A sin Zt G and y A sin ¨ Zt G ¸ act
2¹
©
on a particle simultaneously; then the motion of particle
will be :
(2000)
57.
(a) circular anti-clockwise (b) circular clockwise
(c) elliptical anti-clcokwise (d) elliptical clcokwise
52.
53.
In SHM restoring force is F = – kx, where k is force
constant, x is displacement and A is amplitude of motion,
then total energy depends upon :
(2001)
(a) k, A and M
(b) k, x, M
(c) k, A
(d) k, x
(b) ± a
(c) ± 2a
(d) ± 1
A particle of mass m oscillates with simple harmonic motion
between points x1 and x2, the equilibrium position being O.
Its potential energy is plotted. It will be as given below in
the graph :
(2003)
(a)
(b)
(c)
(d)
Two springs A and B have force constant k A and kB
such that k B = 2 k A . The four ends of the springs are
stretched by the same force. If energy stored in spring A is
E, then energy stroed in spring B is :
(2001)
54.
a
2
(a) r
(a) E/2
(b) 2E
(c) E
(d) 4 E
A mass is suspended separately by two springs of spring
constant k1 and k2 in successive order. The time periods of
oscillations in the two cases are T1 and T2 respectively. If
the same mass be suspended by connecting the two
springs in parallel, (as shown in figure) then the time period
of oscillations is T. The correct relations is
(2002)
58.
59.
The potential energy of a simple harmonic oscillator when
the particle is half way to its end point is :
(2003)
(a)
1
E
4
(b)
1
E
2
(c)
2
E
3
(d)
1
E
8
Two springs of spring constants k1 and k2 are joined in
series. The effective spring constant of the combination
is given by :
(2004)
(a)
k1k 2
(c) k1 + k2
60.
(b) (k1 + k2)/2
(d) k1k2/(k1 + k2)
The circular motion of a particle with constant speed is :
(a) simple harmonic but not periodic
(2005)
(b) periodic and simple harmonic
(c) neither periodic nor simple harmonic
55.
56.
(a) T 2
T12 T22
(b) T 2
(c) T 1
T11 T21
(d) T
T12 T22
T1 T
2
(d) periodic but not simple harmonic
61.
When a damped harmonic oscillator completes 100
oscillations, its amplitude is reduced to 1/3 of its initial
value. What will be its amplitude when it completes 200
oscillations ?
(2002)
(a) 1/5
(b) 2/3
(c) 1/6
(d) 1/9
The displacement of particle between maximum potential
energy position and maximum kinetic energy position in
simple harmonic motion is :
(2002)
Mahesh Tutorials Science
62.
A particle executing simple harmonic motion of amplitude
5 cm has maximum speed of 31.4 cm/s. The frequency of
its oscillation is :
(2005)
(a) 3 Hz
(b) 2 Hz
(c) 4 Hz
(d) 1 Hz
A particle executes simple harmonic oscillation with an
amplitude a. The period of oscillation is T. The minimum
time taken by the particle to travel half of the amplitude
from the equilibrium position is :
(2007)
(a) T/4
(b) T/8
(c) T/12
(d) T/2
OSCILLATION AND WAVES
63.
143
A mass of 2.0 kg is put on a flat pan attached to a vertical
spring fixed on the ground asshown in the figure. The
mass of the spring and the pan is negligible. When pressed
slightly and released the mass executes a simple harmonic
motion. The spring constant is 200 N/m. What should be
the minimum amplitude of the motion, so that the mass gets
detached from the pan ? (Take g = 10 m/s2)
(2007)
68.
69.
A block is resting on a piston which is moving vertically
with SHM of period 1 s. At what amplitude of motion will
the block and piston separate ?
(2009)
(a) 0.2 m
(b) 0.25 m
(c) 0.3 m
(d) 0.35 m
Two simple harmonic motions are given by x = Asin (Zt+G)
S·
§
and y Asin ¨ Zt G ¸ act on a particle simultaneously,,
2¹
©
then the motion of particle will be
(2009)
(a) circular anti-clockwise (b) elliptical anti-clockwise
(c) elliptical clockwise
70.
(a) 8.0 cm
(b) 10.0 cm
(c) Any value less than 12.0 cm
(d) 4.0 cm
64.
Two simple harmonic motions of angular frequency 100
and 1000 rad s 1 have the same displacement amplitude.
The ratio of their maximum acceleration is
65.
(a) 1 : 10
(b) 1 : 102
(c) 1 : 103
(d) 1 : 104
(2008)
71.
(d) circular clockwise
The phase difference between the instantaneous velocity
and acceleration of a particle executing simple harmonic
motion is
(a) 0.5 S
(b) S
(c) 0.707 S
(d) zero
If ks and kp respectively are effective spring constant in
series and parallel combination of spring as shown in figure,
find
ks
kp
(2010)
A point performs simple harmonic oscillation of period T
and the equation of motion is given by
x
a sin Zt S /6 . After the elapse of what fraction
of the time period the velocity of the point will be equal to
half of its maximum velocity ?
(2008)
66.
(a) T/8
(b) T/6
(c) T/3
(d) T/12
A simple pendulum performs simple harmonic motion about
x = 0 with an amplitude a and time period T. The speed of
the pendulum at x =
67.
a
will be
2
(2009)
(a)
Sa 3
2T
(b)
Sa
T
(c)
3S2 a
T
(d)
Sa 3
T
Which one of the following equations of motion
represents simple harmonic motion ?
(2009)
(a) 9/2
(b) 3/7
(c) 2/9
(d) 7/3
72. Two particles are oscillating along two close parallel straight
lines side by side, with the same frequency and amplitudes.
They pass each other, moving in opposite directions when
their displacement is half of the amplitude. The mean
positions of the two particles lie on a straight line
perpendicular to the paths of the two particles. The phase
difference is :
(CBSE, 2011)
(a) S
(b)
S
6
(c) 0
(d)
2S
3
(a)Acceleration = –k0x + k1x2 (b) Acceleration = –k (x + a)
(c) Acceleration = k (x + a) (d) Acceleration = kx
where k, k0, k1 and a are all positive
Lakshya Educare
144
OSCILLATION AND WAVES
79.
WAVES
73.
74.
A hospital has an ultrasonic scanner to locate the tumours in
a tissue. The operating frequency of the scanner is 4.2 MHz.
The speed of sound in a tissue is 1.7 km/s. The wavelength
of sound in the tissue is close to
(CBSE-1995)
77.
(a) lA = 4lB, regardless of masses
(CBSE-2000)
(b) lB = 4lA, regardless of masses
(c) 0.8 mm
(d) 0.4 mm
(c) MA = 2MB, lA = 2lB
A pulse of a wavetrain travels along a stretched string and
reaches the fixed end of the string. It will be reflected back
with
(CBSE-1997)
(d) MB = 2MA, lB = 2lA
80.
Equations of two progressive waves are given by y1=a sin
(Zt + I1) and y2 = a sin (Zt + I2). If amplitude and time
period of resultant wave is same as that of both the waves,
then (I1 – I2) is
(CBSE-2001)
(c) a phase change of 180° with no reversal of velocity
(a)
S
3
(b)
(d) the same phase as the incident pulse but with velocity
reversed
2S
3
(c)
S
6
(d)
S
4
A vehicle, with a horn of frequency n is moving with a
velocity of 30 m/s in a direction perpendicular to the straight
line joining the observer and the vehicle. The observer
perceives the sound to have a frequency n+n1. Then (if
the sound velocity in air is 300 m/s)
(CBSE-1998)
81.
A wave enters water from air. In air frequency, wavelength,
intensity and velocity are n1, O1, l1 and v1 respectively. In
water the corresponding quantities are n2, O2, l2 and v2
respectively; then
(CBSE-2001)
(a) n1 = 10n
(b) n1 = 0
(a) l1 = l2
(b) n1 = n2
(c) n1 = 0.1n
(d) n1 = – 0.1n
(c) v1 = v2
(d) O1 = O2
A standing wave having 3 nodes and 2 antinodes is formed
between two atoms having a distance 1.21 Å between
them. The wavelength of the standing wave is
(a) 1.21 Å
(b) 1.42 Å
(c) 6.05 Å
(d) 3.63 Å
(c)
2u
O
(b)
u
2O
(d)
1
(c) n
1 1
1
...
n1 n 2 n 3
Mahesh Tutorials Science
x·
§
y a sin ¨ 100t ¸ , where x and y are in metre and t is
10
©
¹
83.
(b) 10 m/s
(c) 100 m/s
(d) 1000 m/s
The frequency of a vibrating wire is n. If tension doubled,
density is halved and diameter is doubled, then new
frequency will be
(CBSE-2001)
(a) n
(b)
O
u
(c) 2 n
(d) 4n
1
n
n12 n 22 n 32 ...
1
1
1
...
n1
n2
n3
84.
(CBSE-2001)
(a) 0.1 m/s
u
O
(b) n 2
(d)
The equation of a wave is given by
in second; then velocity of wave is
A sonometer wire when vibrated in full length has
frequency n. Now it is divided by the help of bridges into
a number of segments of lengths l1, l2, l3 ... When vibrated
these segments have frequencies n1, n2, n3... Then the
correct relation is
(CBSE-2000)
(a) n = n1 + n2 + n3 +...
82.
(CBSE-1998)
Two sources are at a finite distance apart. They emit
sounds of wavelength O. An observer situated between
them on line joining the sources approaches one source
with speed u. Then the number of beats heard/sec by
observer will be
(CBSE-2000)
(a)
78.
2n B , then
(b) 4 mm
(b) the same phase as the incident pulse with no reversal
of velocity
76.
frequencies of their vibrations and n A
(a) 8 mm
(a) a phase change of 180° with velocity reversed
75.
Two strings A and B have lengths lA and lB and carry
masses MA and MB at their lower ends, the upper ends
being supported by rigid supports. If nA and nB are their
n
2
A wave of amplitude A = 0.2 m, velocity v = 360 m/s and
wavelenght 60 m is travelling along positive x-axis, then
the correct expression for the wave is :
(CBSE-2002)
x
(a) y 0.2sin 2S§¨ 6t ¸· (b) y
60 ¹
©
x ·
§
0.2sin S ¨ 6t ¸
60
©
¹
x·
§
(c) y 0.2sin2S ¨ 6t ¸
60
©
¹
x ·
§
0.2sin S ¨ 6t ¸
60 ¹
©
(d) y
OSCILLATION AND WAVES
85.
145
A whistle revolves in a circle with an angular velocity
90.
(Z = 20 rad/s) using a string of length 50 cm. If the actual
minimum frequency heard by the observer far away from
the centre is : (velocity of sound v = 340 m/s)
(CBSE-2002)
86.
(a) 385 Hz
(b) 374 Hz
(c) 394 Hz
(d) 333 Hz
91.
An observer moves towards a stationary source of sound
with a speed 1/5th of the speed of sound. The wavelength
and frequency of the source emitted are O and f
respectively. The apparent frequency and wavelength
recorded by the obsever are respectivley : (CBSE-2003)
(a) f, 1.2 O
87.
88.
89.
(d) 1.2f, O
(a) 2 : 1
(b) 1 : 2
(c) 1 : 1
(d) 3 : 1
A number of tuning forks are arranged in order of increasing
in order of increasing frequency and any two successive
tuning forks produce 4 beats/s when sounded together. If
the last tuning fork has a frequency octave higher than
that of the first tuning fork and the frequency of the first
tuning fork is 256 Hz. Then the number of tuning forks is :
(CPMT-1998)
(b) 0.8f, 0.8 O
O
(c) 1.2f,
1.2
An engine blowing whistle is moving towards a stationary
observer with a speed of 110 m/s. What will be the ratio of
the frequencies of whistle when engine is approaching
and receding from the observer? (Speed of sound, v = 330
m/s) :
(CPMT-1998)
92.
(a) 63
(b) 64
(c) 65
(d) 66
In a travelling wave
2·
§
y 0.1sin S ¨ x 330t ¸
3¹
©
A source emitting a note of frequency 240 Hz is moving
towards an observer with a speed of 20 m/s. If the observer
also moves with a speed of 20 m/s towards the source, the
apparent frequency heard by observer is : (velocity of
sound = 340 m/s)
(CPMT-1995)
The phase difference between x1 = 3 m and x 2 = 3.5 m is :
(a) S/2
(b) S
(a) 270 Hz
(b) 240 Hz
(c) 3S/2
(d) 2 S
(c) 268 Hz
(d) 360 Hz
(CPMT-2000)
93.
A tube closed at one end produces a note of frequency
512 Hz. If the tube is opened at both ends, the frequency
will become :
(CPMT-1995)
(a) 512 Hz
(b) 256 Hz
(c) 128 Hz
(d) 1024 Hz
(a) light and sound waves in air
(b) sound waves in air and string waves
(c) light waves and string waves
(d) all of them
Two vibrating tuning forks produce waves given by
y1
4sin 500 St ,
y2
2sin 506St
Consider transverse waves in a stretched string light waves
and sound waves in air. The polarisation can be observed
for :
(CPMT-2000)
94.
The equation of a progressive wave is
y = 0.3 sin (314t - 1.57x)
If they are held near the ear of a person, the person will
hear :
(CPMT-1995)
the velocity of the wave :
(a) 100 m/s
(b) 200 m/s
(a) 3 beats/s with intensity ratio of maxima to minima equal
to 9
(c) 300 m/s
(d) 400 m/s
(b) 3 beats/s with intensity ratio of maxima to minima equal
to 2
(c) 6 beats/s with intensity ratio of maxima to minima equal
to 2
(d) 6 beats/s with intensity ratio of maxima to minima equal
to 9
95.
(CPMT-2001)
The equation of a stationary wave is given by
y = 0.4 sin 160 St cos
S
x
16
Where t is in second, x and y in cm, separation between
successive nodes is :
(CPMT-2001)
(a) 32 cm
(b) 16 cm
(c) 8 cm
(d) 4 cm
Lakshya Educare
146
96.
97.
98.
OSCILLATION AND WAVES
An air column of closed pipe of length 20 cm resonates
with a tuning fork of frequency 500 Hz, the speed of sound
is :
(CPMT-2001)
(a) 100 m/s
(b) 200 m/s
(c) 400 m/s
(d) 1000 m/s
103.
y1 = 2a sin (Zt kx)
y 2 = 2a sin (Zt kx T)
The amplitude of the medium particle will be :
The frequency of vibrating air column in closed organ
pipe is n. If its length be doubled and radius halved, its
frequency will be nearly :
(CPMT-2002)
(a) n
(b) n/2
(c) 2n
(d) 4n
(CBSE-2004)
104.
th
T
2
(d)
2a cos
(a) 9/11
(b) 11/10
(c) 10/11
(d) 10/9
A source and observe are approaching each other with 50
(a) 300 cycle/s
(b) 320 cycle/s
(c) 340 cycle/s
(d) 330 cycle/s
Distance between successive compression and
(a) 180 Hz
(b) 45 Hz
(c) 120 Hz
(d) 90 Hz
105.
106.
A car is moving towards a high cliff. The car driver sounds
a horn of frequency f. The reflected sound heard by the
driver has a frequency 2f. If v be the velocity of sound,
then the velocity of the car, in the same velocity units, will
be :
(CBSE-2004)
(a) v / 2
(b) v/3
(c) v/4
(d) v/2
(a)
1
s
20
(b)
1
min
20
(c)
19
s
20
(d)
19
min
20
Which of the following expressions is that of a simple
harmonic progressive wave?
(CBSE-2006)
(a) a sin Zt
(b) a sin ( Zt ) cos k x
(c) a sin ( Zt kx )
(d) a cos k x
Two viberating running forks produce progressive waves
of beats produced per minute is :
107.
(a) 360
(b) 180
(c) 3
(d) 60
(CBSE-2005)
Two sound waves with wavelengths 5.0 m and 5.5 m
respectivley, each propagate in a gas with velocity 330
m/s. We expect the following number of beats/second :
(CBSE-2006)
ª
º
§ x ·
10 sin «100t ¨ ¸ 0.5» m
50
© ¹
¬
¼
6
ª
§ x ·º
y 2 106 cos «100t ¨ ¸ » m
© 50 ¹ ¼
¬
A train approaches a stationary observer, the velocity of
given by y1 = 4 sin 500 St and y 2 = 2 sin 506 St . Number
The phase difference between two waves, represented by
y1
T
2
1
of the velocity of sound. A sharp blast is
20
blown with the whistle of the engine at equal intervals of
a second. The interval between the successive blasts as
heard by the observer is :
(CBSE-2005)
rarefactions is 1 m and velocity of sound is 360 ms 1 .
Find frequency.
(CPMT-2003)
102.
2a cos T
train being
ms 1 velocity. What will be original frequency if the
observer receives 400 cycle/s ? (v = 350 m/s) (CPMT-2003)
101.
(b)
A source is approaching a stationary observer with
§1·
100.
(a) 2a cos T
(c) 4a cos
velocity ¨ ¸ that of sound. Ratio of observed and real
© 10 ¹
frequencies will be :
(CPMT-2003)
99.
Equations of motion in the same direction are given by :
108.
(a) 12
(b) zero
(c) 1
(d) 6
Each of the two strings of length 51.6 cm and 49.1 cm are
tensioned separately by 20 N force. Mass per unit length
of both the strings is same and equal to 1 g 1 . When both
where x is expressed in metres and t is expressed in
seconds, is aproximately :
(CBSE-2004)
the strings vibrate simultaneoulsy the number of beats is
(a) 1.07 rad
(b) 2.07 rad
(a) 5
(b) 7
(c) 0.5 rad
(d) 1.5 rad
(c) 8
(d) 3
Mahesh Tutorials Science
(CBSE-2009)
OSCILLATION AND WAVES
109.
147
The driver of a car travelling with speed 30 ms 1 towards
(c) v = v1 + v2 + v3
a hill sounds a horn of frequency 600 Hz. If the velocity of
1
(d) v
1
sound in air is 330 ms , the frequency of reflected sound
as heard by driver is
(CBSE-2009)
110.
(a) 550 Hz
(b) 5505 Hz
(c) 720 Hz
(d) 500 Hz
114. Two sources of sound placed close to each other, are
emitting progressive waves given by
y1 = 4 sin 600 St and y2 = 5 sin 608 St. An observer located
near these two sources of sound will hear :
A wave in a string has an amplitude of 2 cm. The wave
travels in the +ve direction of x axis with a speed of
128 ms
1
(CBSE, 2012)
(a) 4 beats per second with intensity ratio 81 : 1 between
waxing and waning.
and it is noted that 5 complete waves fit in 4 m
length of the string. The equation describing the wave is
(b) 4 beats per second with intensity ratio 25 : 16 between
waxing and waning
(CBSE-2009)
(a) y = (0.02) m sin (7.85x + 1005t)
(c) 8 beats per second with intensity ratio 25 : 16 between
waxing and waning
(b) y = (0.02) m sin (15.7x - 2010t)
(c) y = (0.02) m sin (15.7x + 2010t)
(d) 8 beats per second with intensity ratio 8 : 1 between
waxing and waning
(d) y = (0.02) m sin (7.85x - 1005t)
111. The equation of a simple harmonic wave is given by
y = 3 sin
115.
S
(50t – x)
2
where x and y are in metres and t is in seconds. The ratio of
maximum particle velocity to the wave velocity is : (2012)
(a)
2
S
3
(b) 2S
(c)
3
S
2
(d) 3 S
(d) may be greater than, less or equal to f depending on
the factors like speed of train, length of train and radius of
circular track
116.
(a) 3000 Hz
(b) 3500 Hz
(c) 4000 Hz
(d) 5000 Hz
113. When a string is divided into three segments of length l1, l2
and l3, the fundamental frequencies of these three segments
are v1, v2 and v3 respectively. The original fundamental
frequency (v) of the string is :
(CBSE, 2012)
1
If O1 , O 2 and O3 are the wavelengths of the waves giving
resonance with the fundamental, first and second
overtones respectivley of a closed organ pipe. Then the
ratio of wavelengths O1 : O 2 : O3 is
117.
(a) 1 : 3 : 5
(b) 1 : 2 : 3
(c) 5 : 3 : 1
1 1
(d) 1: :
3 5
(CPMT-2009)
Two sources are at a finite distance apart. They emit sound
of wavelength O . An observer situated between them on
line joining the sources, approaches towards one source
with speed u, then the number of beats heard per second
by observer will be
1
(a)
v
v1
(b)
v
v1 v 2 v3
v2
(b) equal to f
(c) is greater than f
–1
1
A train is moving with a constant speed along a circular
track. The engine of the train emits a sound of frequency
f. The frequency heard by the guard at rear end of the train
is
(CPMT-2009)
(a) less than f
112. A train moving at a speed of 220 ms towards a stationary
object, emits a sound of frequency 1000 Hz. Some of the
sound reaching the object gets reflected black to the train
as echo. The frequency of the echo as detected by the driver
–1
of the train is : (Speed of sound in air is 330 ms ) (2012)
1
1
1
1
v1 v 2 v3
v3
(a)
2u
O
(b)
u
O
(c)
u
2O
(d)
O
u
Lakshya Educare
148
118.
OSCILLATION AND WAVES
Two points are located at a distance of 10 m and 15 m from
the source of oscillation. The period of oscillation is 0.05 s
and the velocity of the wave is 300 ms 1 . What is the
phase difference between the oscillations of two points?
S
(a)
3
2S
(b)
3
S
(d)
6
(c) S
119.
(c) B leads by
(d) B leads by S and C lags by S
124.
(CBSE-2008)
The wave discribed by y = 0.25 sin (10Sx 2St) , where x
and y are in metre and t in second, is a wave travelling
along the
(CBSE-2009)
125.
(a) –ve x direction with frequency 1 Hz
(b) +ve x direction with frequency S Hz and wavelength
O = 0.2 m
(c) –ve x direction with frequency 1 Hz and wavelength
O = 0.2 m
126.
121.
122.
123.
An observer moves towards a stationary source of sound,
with a velocity one-fifth of the velocity of sound. What is
the percentage increase in the apparent frequency?
(a) Zero
(b) 0.5%
(c) 5%
(d) 20%
127.
(CPMT-2008)
A sound absorber attenuates the sound level by 20 dB.
The intensity decreases by a factor of
(CPMT-2008)
(a) 1000
(b) 10000
(c) 10
(d) 100
Reverberation time does not depend upon (CPMT-2007)
(a) temperature
(b) volume of room
(c) size of window
(d) carpet and curtain
128.
Three progressive waves A, B and C are shown in the
figure.
129.
With respect to A, the progressive wave (CPMT-2007)
(a) B lags by
S
S
and C leads by
2
2
(b) B lags by S and
Mahesh Tutorials Science
S
leads by S
2
A source of unknown frequency produces 12 beats with a
source of frequency 240 Hz and 8 beats with another
source of frequency 260 Hz. The unknown frequency will
be :
(AIIMS-1994)
(a) 292 Hz
(b) 272 Hz
(c) 268 Hz
(d) 252 Hz
A tube closed at one end contianing air produces
fundamental note of frequency 512 Hz. If the tube is opened
at both the ends, the fundamental note will be :
(a) 1280 Hz
(b) 1024 Hz
(c) 768 Hz
(d) 512 Hz
(AIIMS-1995)
A star emitting radiation at a wavelength of 5000Å is
approaching earth with a velocity of 1.5 u 106 m / s . The
change in wavelength on the earth will be : (AIIMS-1995)
(d) –ve x directon with amplitude 0.25 m and wavelength
O = 0.2 m
120.
S
S
and C lags by
2
2
130.
(a) 2.5Å
(b) 100 Å
(c) Zero
(d) 25Å
A sources of sound is moving with a velocity 50 m/s
towards a stationary observer. The observer measures the
frequency of the source as 1000 Hz. What will be the
apparent frequency of the source when it is moving away
from the observer after crossing him? The velocity of sound
in the medium is 350 m/s :
(AIIMS-1995)
(a) 1330 Hz
(b) 1140 Hz
(c) 750 Hz
(d) 850 Hz
If tension in the string is increased from 1 kN to 4 kN, other
factors remaining unchanged the frequency of second
harmonic will :
(AIIMS-1996)
(a) become four time
(b) be doubled
(c) remain unchanged
(d) be halved
A which temperature velocity of sound (at 270qC ) double:
(a) 327°C
(b) 927°C
(c) 54°C
(d) 1899°C
(AIIMS-1997)
An objective producing a pitch of 1200 Hz is moving with
a velocity of 50 m/s towards a stationary person. The
velocity of sound is 350 m/s. The frequencey of sound
heard by the stationary person is :
(AIIMS-1997)
(a) 1250 Hz
(b) 1050 Hz
(c) 700 Hz
(d) 1400 Hz
OSCILLATION AND WAVES
131.
132.
133.
The air column in a pipe which is closed at one end will be
in resonance with a vibrating tuning fork at a frequency
260 Hz. The length of the air column is : (AIIMS-1997)
136.
(b) 40 cm
(a) 12.5 cm
(b) 35.75 cm
(c) 50 cm
(d) 62.5 cm
(c) 31.73 cm
(d) 62.5 cm
Standing waves are produced in 10 m long stretched string.
If the string vibrates in 5 segments and wave velocity is 20
m/s, then its frequency will be :
(AIIMS-1998)
(a) 5 Hz
(b) 2 Hz
(c) 10 Hz
(d) 2 Hz
(b) v
U
P
(d) v
(A) If both ASSERTION and REASON are true and
reason is the correct explanation of the assertion.
(B)
2 P
(AIIMS-1998)
3 U
(D) If both ASSERTION and REASON are false.
(E)
140.
2P
P
Energy is not carried by which of the following wave?
If ASSERTION is false but REASON is true.
Assertion : Resonance is a special case of forced vibration
in which the natural frequency of vibration of the body is
the same as the impressed frequency and the amplitude of
forced vibration is maximum.
(a) Progressive
(AIIMS-1999)
(b) Electromagnetic
Reason : The amplitude of forced vibration of a body
increases with an increase in the frequency of the externally
impressed periodic force.
(AIIMS-1994)
(c) Transverse
(d) Stationary
(a) A
(b) B
(d) D
(e) E
If the vibrations of a string are to be increased by a factor
of two, then tension in the string should be made :
(c) C
(a) twice
(b) four times (AIIMS-1999)
Assertion : In a stationary wave three is no transfer of
energy.
(c) eight times
(d) half
Reason : The distance between two consecutive nodes or
141.
A transverse wave passes through a string with the
equation
where x is in metre and t in second. The maximum velocity
of the particle in wave motion is :
(AIIMS-2000)
138.
If both ASSERTION and REASON are true but reason
is not the correct explanation of the assertion.
(C) If ASSERTION is true but REASON is false.
Newton’s formula for the velocity of sound in gas is :
P
U
(AIIMS-2002)
ASSERTION REASON
antinodes is O / 2 .
y = 10 sin S (0.02 x - 2.00t)
137.
A string in a musical instrument is 50 cm long and its
fundamental frequency is 800 Hz. If the frequency of 1000
Hz is to be produced, then required length of string is :
(a) 37.5 cm
(c) v
135.
139.
(Speed of sound wave = 330 m/s)
(a) v
134.
149
(AIIMS-1994)
(a) A
(b) B
(d) D
(e) E
(c) C
(a) 100 m/s
(b) 63 m/s
Assertion : In everyday life the Doppler’s effect is
observed readily for sound waves than light waves.
(c) 120 m/s
(d) 161 m/s
Reason : Velocity of light is greater than that of sound.
142.
The tension in a piano wire is 10 N. The tension in a piano
wire to produce a node of double frequency is :
(a) A
(b) B
(a) 20 N
(b) 40 N
(d) D
(e) E
(c) 10 N
(d) 120 N
(AIIMS-2001)
Two sound waves have phase difference of 60q , then
they will have the path difference of :
(AIIMS-2001)
(a) 3 O
(c)
O
6
(b)
O
3
(d) O
143.
(AIIMS-1995)
(c) C
Assertion : Intensity of sound wave does not change when
the listener moves towards or away from the stationary
source.
Reason : The motion of listener causes the apparent change
in wavelength.
(AIIMS-1995)
(a) A
(b) B
(d) D
(e) E
(c) C
Lakshya Educare
150
144.
OSCILLATION AND WAVES
Assertion : Ratio waves can be polarised.
152.
Reason : Sound waves in air are longitudinal in nature.
(a) A
(d) D
145.
(AIIMS-1996)
(c) C
(b) B
(e) E
Assertion : Sound travels faster in solids than gases.
153.
Reason : Solids posses greater density than gases.
(a) A
(d) D
146.
(b) B
(e) E
147.
148.
149.
150.
151.
(d) 0.25 s
The equation y = 4 + 2 sin (6t - 3x) represents a wave
motion. Then wave speed and amplitude, respectivley are
(AIIMS-2009)
(d) wave speed 1/2 unit, amplitude 5 unit
154.
A closed organ pipe of length 1.2 m vibrates in its first
overtone mode. The pressure variation is maximum at
(a) 0.4 m from the open end
(c) C
(AIIMS-2009)
(b) 0.4 m from the closed end
(c) Both (a) and (b)
Reason : The speed of sound in gases is proportional to
the square of pressure.
(AIIMS-2008)
(b) B
(e) E
(c) 4 s
(c) wave speed 4 unit, amplitude 1/2 unit
Assertion : The change in air pressure effects the speed
of sound.
(a) A
(d) D
(b) 75 s
(b) wave speed 2 unit, amplitude 2 unit
Reason : Wavelength is the distance between two nearest
particles in phase.
(AIIMS-2002)
(b) B
(e) E
(a) 2500 s
(a) wave speed 1 unit, amplitude 6 unit
(AIIMS-2000)
(c) C
wavelength
Assertion : Speed of wave = time period
(a) A
(d) D
A stationary boat is rocked by waves whose crests are
100 m apart velocity is 25 m/s. The boat bounces up once
in every :
(AIIMS-2006)
(d) 0.8 m from the open end
155.
(c) C
Assertion : Ocean waves hitting a beach are always found
to be nearly normal to the shore.
A resonance in an air column of length 40 cm resonates
with a tuning fork of frequency 450 Hz. Ignoring end
correction, the velocity of sound in air is (AIIMS-2009)
-1
(b)720 ms
-1
(d) 820 ms
(a) 102 ms
(c) 620 ms
156.
-1
-1
A vehicle with a horn of frequency n, is moving with a
Reason : Ocean waves hitting a beach are assumed as
plane waves.
(AIIMS-2008)
velocity of 30 ms 1 in a direction perpendicular to the
straight line joining the observer and the vehicle. The
(a) A
(d) D
observer receives the sound to have a frequency n n1 .
(b) B
(e) E
(c) C
Then (if the sound velocity in air is 300 ms 1 )
An organ pipe closed at one end has fundamental
frequency of 1500 Hz. The maximum number of tones
generated by this pipe which a normal person can hear is?
(a) 4
(b) 13
(c) 6
(d) 9
(AIIMS-2000)
157.
Two tuning forks P and Q when set viberating, given 4
beats/s. If a prong of the fork P is filed, the beats are
reduced to 2 beats/s. What is frequency of P, if that of Q is
250 Hz ?
(AIIMS-2006)
(a) 246 Hz
(b) 250 Hz
(c) 254 Hz
(d) 252 Hz
(a) n1 10n
(b) n1
0
(c) n1
(d) n1
0.1n
0.1n
(AIIMS-2009)
Assertion : To distinct beats, difference in frequencies of
two sources should be less than 10.
Reason : More the number of beats per second more difficult
to hear them.
(AIIMS-2009)
(a) A
(b) B
(d) D
(e) E
(c) C
A point source emits sound equally in all directions in a
non-absorbing medium. Two points P and Q are at distance
of 2 m and 3 m respectively from the source. The ratio of
the intensities of the waves at P and Q is :
A string is stretched between fixed pionts separated by
75.0 cm. It is observed to have resonant frequencies of
420 Hz and 315 Hz. There are no other resonant frequencies
between these two. Then, the lowest resonant frequency
for this string is
(AIIMS-2008)
(a) 9 : 4
(b) 2 : 3
(a) 105 Hz
(b) 1.05 Hz
(c) 3 : 2
(d) 4 : 9
(c) 1050 Hz
(d) 10.5 Hz
Mahesh Tutorials Science
158.
(AIIMS-2006)
OSCILLATION AND WAVES
159.
160.
161.
151
A person speaking normally produces a sound intensity
of 40 dB at a distance of 1 m. If the threshold intensity for
reasonable audibility is 20 dB, the maximum distance at
which he can be heard clearly is
(AIIMS-2008)
(a) 4 m
(b) 5 m
(c) 10 m
(d) 20 m
167.
fundamental frequency f in air. If half of the length is
dipped vertically in water. The fundamental frequency of
the air column will be :
(AFMC-1997)
(a)
A sings with a frequency n and B sings with a frequency
1/8 that of A. If the energy remains the same and the
amplitude of A is a then amplitude of B is : (AFMC-1994)
(a) 16 a
(b) 8 a
(c) 2 a
(d) a
168.
(AFMC-1994)
162.
163.
164.
165.
(b) 7.06 km/s
(c) 6.06 km/s
(d) 5.06 km/s
169.
Two open in organ pipes of length 25 cm and 25.5 cm
produce 10 beat/s their fundamental mode. The velocity
of sound is :
(AFMC-1995)
(a)255 m/s
(b) 303 m/s
(c) 375.5 m/s
(d) 350 m/s
(b) 2 f
(d)
f
2
A wave frequency 100 Hz is sent along a string towards a
fixed end. When this wave travels back, after reflecton, a
node is formed at a distance of 10 cm from the fixed end of
the string. The speeds of incident (and reflected) waves
are :
(AFMC-1997)
(a) 48 m/s
(b) 20 m/s
(c) 10 m/s
(d) 15 m/s
A source of sound is travelling with a velocity 40 km/h
towards an observer and emits sound of frequency 2000
Hz. If velocity of sound is 1220 km/h. Then the apparent
frequency heard by an observer is :
(AFMC-1997)
(a) 1980 Hz
(b) 1950 Hz
(c) 2068 Hz
(d) 2080 Hz
The frequency of a radar is 780 MHz. The frequency of the
reflected wave from aeroplane is increased by 2.6 kHz.
The velocity of aeroplane is :
(AFMC-1996)
A glass rod of 20 cm long is clamped at the middle. It is set
into the longitudinal vibration. If the emitted sound
frequency is 4000 Hz, the velocity of sound in glass will
be:
(AFMC-1997)
(a) 0.25 km/s
(b) 0.5 km/s
(a) 2800 m/s
(b) 3200 m/s
(c) 1 km/s
(d) 2 km/s
(c) 1600 m/s
(d) 2000 m/s
Resonance is a special case of :
170.
(AFMC-1996)
(a) undamped vibration
(b) damped vibration
(c) forced vibration
(d) natural vibration
The effect of a sound pulse lasts on the ear for 0.1 s. Thus,
for a person to hear the echo of a sharp sound the minimum
distance of the reflector in terms of speed of sound v is :
171.
(a)
(c)
v
40
v
20
(b)
v
5
(d)
v
10
A whistle giving out 450 Hz approaches a stationary
observer at a speed of 33 m/s, the frequency heard by the
observer is : (Speed of sound = 330m/s) (AFMC-1996)
The ratio of speed of sound in helium and hydrogen gases
at the same temperautre is :
(AFMC-1998)
(a)
42 : 25
(c) 42 : 25
172.
(AFMC-1996)
166.
3f
2
(c) f
A steel rod 100 cm long is clamped at its middle. The
fundamental frequency of longitudinal vibration of the
rod is 2.53 kHz. The speed of sound in steel will be :
(a) 8.06 km/s
A cylindrical resonance tube, open at both ends, has a
173.
(b)
25 : 42
(d) 25 : 42
The driver of a car travelling with speed 30 m/s towards a
hill sounds a horn of frequency 600 Hz. If the velocity of
sound in air is 330 m/s. The frequency of reflected sound
as heard by the driver is :
(AFMC-1998)
(a) 500 Hz
(b) 550 Hz
(c) 720 Hz
(d) 555 Hz
On producing the waves of frequency 1000 Hz in a Kundt’s
tube, the total distance between 6 successive nodes is 85
cm. Speed of sound in the gas filled in the tube is :
(AFMC-1999)
(a) 500 Hz
(b) 517 Hz
(a) 300 m/s
(c) 429 Hz
(d) 409 Hz
(c) 340 m/s
(b) 350 m/s
(d) 330 m/s
Lakshya Educare
152
174.
OSCILLATION AND WAVES
Which of the following are not the transverse wave?
182.
Which one proves transverse nature of waves :
(AFMC-1999)
(AFMC-2003)
(a) sound waves in the air (b) visible light waves
(c) X-rays
175.
176.
177.
178.
179.
(d) J - rays
A man fires a bullet standing between two cliffs. First echo
is heard after 3 s and second echo is heard after 5 s. If the
velocity of sound is 330 m/s. Then the distance between
cliffs is :
(AFMC-2000)
(a) 660 m
(b) 990 m
(c) 1320 m
(d) 1950 m
183.
(b) 16.28 m
(c) 32.50 m
(d) 8.25 m
(b) 77 m/s
(c) 110 m/s
(d) 150 m/s
185.
186.
(AFMC-2003)
Two similar waves are propagating in x-axis and another
one in y-axis. When they colloide to superimpose, the
resultant wave will be :
(AFMC-2003)
(a) elliptical
(b) hyperbolic
(c) straight line
(d) parabolic
Two closed organ pipes of length 100 cm and 101 cm long
give 16 beats in 20 s when each pipe is sounded in its
fundamental mode, calculate the velocity of sound :
(a) 303 ms 1
(b) 332 ms 1
(c) 323 ms 1
(d) 300 ms 1
Apparatus used to find out velocity of sound in gas is :
(AFMC-2004)
(a) 49 : 1
(b) 25 : 7
(a) Melde’s experiment
(b) Kundt’s tube
(c) 10 : 9
(d) 4 : 3
(c) Quincke’s tube
(d) none of these
A standing wave is represented by
187.
where t is in seconds and x is in metres. Then the velocity
of the wave is :
(AFMC-2002)
(a) 10 2 m/s
(b) 104 m / s
(c) 1 m/s
(d) 10 4 m / s
188.
If the equation of a sound wave is given as
Then wavelength of this wave is :
(a) 0.1 unit
(b) 0.2 unit
(c) 0.4 unit
(d) 0.5 unit
If the temperature of atmosphere is increased, the following
character of sound waves is effected :
(AFMC-2004)
(a) phase
(b) frequency
(c) speed
(d) none of these
A string of length 0.4 m and mass 102 kg is tightly clamped
at its ends. The tension in the string is 1.6 N. Identical
wave pulses are produced at one end in equal intervals of
time 't . The minimum value of 't , which allows
y = 0.00155 sin (62.8 x + 316t)
181.
Doppler’s efffect in sound is due to :
(AFMC-2003)
y = a sin (100 t) cos (0.01x)
180.
(d) All of these
(c) relative motion of source and observer
(d) none of the above
184.
Two waves, whose intensities are 9 : 16 are made to
interfere. The ratio of maximum and minimum intensities in
the interference pattern is :
(AFMC-2001)
(c) Interference
(b) motion of observer
A 5.5 m length of string has a mass of 0.035 kg. If the
tension in the string is 77 N. Then the speed of a wave on
the string will be :
(AFMC-2000)
(a) 102 m/s
(b) Diffraction
(a) motion of source
The speed of a wave in a medium is 650 m/s. If 4000 waves
are passing through a point in the medium in 1.67 min,
then its wavelength will be :
(AFMC-2000)
(a) 25.16 m
(a) Polarisation
constructive interference between successive pulses, is :
(AFMC-2002)
(a) 0.05 s
(b) 0.40 s
(c) 0.20 s
(d) 0.10 s
(AFMC-2004)
Which of the following waves have the maximum
wavelength ?
(AFMC-2002)
The time of reverberation of a room A is 1 s. What will be
the time (in seconds) of reverberation of a room, having all
the dimensions double of those of room A?
(a) Infrared rays
(b) UV-rays
(a) 2
(b) 4
(c) Radio waves
(d) X-rays
(c) 1/2
(d) 1
Mahesh Tutorials Science
189.
(AFMC-2006)
OSCILLATION AND WAVES
190.
153
A band playing music at a frequency f is moving towards
193.
The loudness and pitch of a sound note depends on
a wall at a speed vb . A motorist is following the band with
(a) intensity and frequency
a speed vm . If v be the speed of the sound, the expression
(b) frequency and number of harmonics
for beat frequency heard by motorist is
(c) intensity and velocity
v vm
(a) v v f
b
2vb ( v v m )
(c)
v 2 vb2
(AFMC-2009)
(d) frequency and velocity
v vm
(b) v v f
b
f
(AFMC-2008)
194.
The fundamental frequency of a closed pipe is 220 Hz. If
1
of the pipe is filled with water, the frequency of the first
4
2vm ( v vb )
f
(d) v 2 v m2
overtone of the pipe now is
191.
A tuning fork A produces 4 beats s 1 with another tuning
fork B of frequency 320 Hz. On filing one of the prongs of
A, 4 beats s1 are again heard when sounded with the
same fork B. Then, the frequency of the form A before
filing is
(AFMC-2008)
192.
195.
(AFMC-2007)
(a) 220 Hz
(b) 440 Hz
(c) 880 Hz
(d) 1760 Hz
The intensity of sound increases at night due to
(a) increase in density of air
(a) 328 Hz
(b) 316 Hz
(b) decrease in density of air
(c) 324 Hz
(d) 320 Hz
(c) low temperature
(d) None of these
The equation of a simple harmonic wave is given by y = 5
S
sin (100t x) , where x and y are in metre and time is in
2
second. The period of the wave in second will be
(AFMC-2008)
(AFMC-2007)
196.
In case of a forced vibration, the resonance wave becomes
very sharp when the
(AFMC-2007)
(a) applied periodic force is small
(b) quality factor is small
(a) 0.04
(b) 0.01
(c) damping force is small
(c) 1
(d) 5
(d) restoring force is small
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154
OSCILLATION AND WAVES
ANSWER KEY
EXERCISE - 1 : BASIC OBJECTIVE QUESTIONS
1. (d)
2. (c)
3. (b)
4. (b)
5. (a)
6. (a)
7. (b)
8. (a)
9. (d)
10. (c)
11. (a)
12. (c)
13. (b)
14. (d)
15. (d)
16. (d)
17. (a)
18. (b)
19. (c)
20. (d)
21. (c)
22. (c)
23. (a)
24. (d)
25. (a)
26. (d)
27. (b)
28. (b)
29. (c)
30. (b)
31. (a)
32. (c)
33. (a)
34. (a)
35. (a)
36. (d)
37. (b)
38. (d)
39. (a)
40. (b)
41. (c)
42. (b)
43. (b)
44. (b)
45. (b)
46. (b)
47. (c)
48. (a)
49. (a)
50. (c)
51. (c)
52. (c)
53. (c)
54. (b)
55. (d)
56. (a)
57. (d)
58. (b)
59. (b)
60. (b)
61. (d)
62. (b)
63. (b)
64. (d)
65. (d)
66. (c)
67. (c)
68. (d)
EXERCISE - 2 : PREVIOUS YEARS AIEEE/JEE MAINS QUESTIONS
1. (a)
2. (c)
3. (b)
4. (c)
5. (a)
6. (c)
7. (a)
8. (a)
9. (d)
10. (b)
11. (b)
12. (b)
13. (a)
14. (c)
15. (d)
16. (c)
17. (b)
18. (a)
19. (c)
20. (b)
21. (a)
22. (b)
23. (b)
24. (a)
25. (a)
26. (c)
27. (c)
28. (a)
29. (c)
30. (d)
31. (b)
32. (a)
33. (a)
34. (d)
35. (b)
36. (a)
37. (b)
46. (a)
38. (d)
47. (a)
39. (a)
48. (c)
40. (c)
49. (c)
41. (c)
50. (d)
42. (b)
51. (b)
43. (b)
52. (c)
44. (c)
53. (a)
45. (c)
54. (b)
55. (d)
56. (b)
57. (c)
58. (a)
59. (d)
60. (d)
61. (d)
62. (c)
63. (b)
64. (b)
65. (d)
66. (d)
67. (b)
68. (b)
69. (d)
70. (a)
71. (c)
72. (d)
73. (d)
74. (a)
75. (b)
76. (a)
77. (a)
78. (c)
79. (d)
80. (b)
81. (b)
82. (d)
83. (a)
84. (c)
85. (b)
86. (c)
87. (a)
88. (d)
89. (a)
90. (a)
91. (c)
92. (a)
93. (c)
94. (b)
95. (b)
96. (c)
97. (b)
98. (d)
99. (a)
100. (a)
101. (b)
102. (a)
103. (c)
104. (c)
105. (a)
106. (b)
107. (d)
108. (b)
109. (c)
110. (d)
111. (c)
112. (d)
113. (d)
114. (a)
115. (b)
116. (d)
117. (a)
118. (b)
119. (c)
120. (d)
121. (d)
122. (a)
123. (c)
124. (d)
125. (b)
126. (d)
127. (c)
128. (b)
129. (d)
130. (d)
131. (c)
132. (a)
133. (a)
134. (d)
135. (b)
136. (b)
137. (b)
138. (c)
139. (b)
140. (c)
141. (b)
142. (a)
143. (e)
144. (a)
145. (b)
146. (a)
147. (d)
148. (a)
149. (c)
150. (a)
151. (c)
152. (c)
153. (b)
154. (a)
155. (b)
156. (b)
157. (c)
158. (a)
159. (c)
160. (b)
161. (d)
162. (a)
163. (b)
164. (c)
165. (c)
166. (a)
167. (c)
168. (b)
169. (c)
170. (c)
171. (b)
172. (c)
173. (c)
174. (a)
175. (c)
176. (b)
177. (a)
178. (a)
179. (d)
180. (a)
181. (c)
182. (a)
183. (c)
184. (c)
185. (c)
186. (b)
187. (c)
188. (d)
189. (a)
190. (c)
191. (b)
192. (a)
193. (a)
194. (c)
195. (a)
196. (c)
Dream on !!
[\]^[\]^
Mahesh Tutorials Science