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Senior Inter Physics
Waves
Very Short Answer Questions (2 Marks)
Q:
Distinguish between Longitudinal Waves and Transverse Waves.
A:
Longitudinal Waves
Transverse Waves
1. In this wave, particles of 1. In this wave, particles
the medium vibrate along
of the medium vibrate
the direction of wave
perpendicular to the
propagation.
direction of wave
propagation.
2. Examples: Sound waves,
Waves on springs.
2. Examples:
Light
waves, Waves on
3. When a Longitudinal
strings under tension.
wave travels through a
.
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medium compressions 3. When a Transverse
and Rarefactions are
wave travels through a
formed alternatively.
medium crests and
troughs are formed
alternatively.
4. Longitudinal waves are
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possible in solids, liquids
and gases, as they need 4. Transverse waves are
bulk
modulus
of
possible in solids only,
elasticity.
as they need shear
modulus of elasticity.
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5. Longitudinal waves do 5. Transverse waves can
undergo polarisation.
not undergo polarisation.
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R-25-11-14
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Q:
Explain the differences in the formation of Stationary Waves and Beats.
A:
Stationary Waves
Beats
1. These are formed due
to the superposition of
two sound waves of
slightly
different
frequencies travelling
in the same direction.
1. These waves are
formed due to the
superposition of two
identical progressive
waves of same kind
travelling in the
opposite directions
along the same line. 2. Energy is transported
in
the
medium.
2. Energy
is
not
Redistribution
of
transported in the
energy takes place
medium.
with respect to time.
Q:
What is the ratio of the frequencies of harmonics in an air column of same
length in (i) a closed pipe and (ii) an open pipe?
A:
(i) In closed pipes, only odd harmonics are possible and the frequencies ratio is
1 : 3 : 5 : 7 : ..
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(ii) In open pipes, all harmonics are possible and frequencies of harmonics are
in the ratio 1 : 2 : 3 : 4 : ....
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Q:
What is Doppler effect? Write its limitation.
A:
The apparent change in frequency of sound due to relative motion between the
listener and source of sound is called Doppler effect. Doppler effect is
applicable when the velocities of the source of sound and listener are much less
than the velocity of sound in the surrounding medium.
Q:
A string of mass 2.50 kg is under a tension of 200 N. The length of the stretched
string is 20.0 m. If the transverse jerk is caused at one end of the string, how
long does the disturbance take to reach the other end?
l l = 20.0 m
Time taken, t = 
v

T
V= 
µ where T = 200 N
A:
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√
2.50
25
m
µ =  =  =  kg ms−1
20.0
200
l
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V=
√(

200
200
 =  = 40 ms−1
5
25

200
t
e
)
n
.
a
20
1
∴ t =  =  seconds
40
2
h
b
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Short Answer Questions (4 Marks)
t
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Q:
Describe a procedure for measuring the velocity of sound in a stretched
string.
A:
A uniform wire is stretched between two rigid supports of a Sonometer. It is
tuned to resonance by adjusting the distance between the bridges. When the
system of a vibrating tuning fork of known frequency (f) is pressed on the
Sonometer board, the paper rider placed on the wire at its centre is
automatically thrown off at correct resonance.
λ
If 'l' is the distance between the bridges, when resonance occurs, then l = −
2
(length of a loop) ⇒ λ = 2l
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.
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The velocity of Sound, V = f λ
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⇒ V = f × 2l = 2 lf.
Q:
What do you understand by "resonance"? How would you use resonance
to determine the velocity of sound in air?
A:
Resonance: When the frequency of external periodic force acting on an
oscillator is equal to the natural frequency of the oscillator, it oscillates with
maximum amplitude. This phenomenon is called resonance.
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Experiment: When natural frequency of air column coincides with the
frequency of vibrating tuning fork, the air column would be in resonance with
the tuning fork and the air column is called resonating air column.
Consider a uniform cylindrical tube which is open at one end and closed at the
other end. The length of air column in the closed tube can be changed by
changing the water level in the tube. When a vibrating tuning fork of known
frequency 'f' Hz is placed close to the opened end of the tube, the air column in
the tube begins to vibrate. Due to reflection of waves at water surface in the tube
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and superposition of waves stationary waves are formed in the air column with
node at water surface and antinode at free end of the tube.
The first and second modes of vibrating air column in resonance with the
vibrating tuning fork are shown in the figure.
f Hz
f Hz
AN
↓ AN
↓
l1
λ
−
4
↑
↑
- N- -- -- ----
n
.
a
U
U
t
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h
b
i
↓
↓
.N ↑ λ/4
AN ↓
l2
↑
Fundamental (or)
First Harmonic
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up
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.
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λ .... (1)
l1 = −
4
-- N
- -- --- ---
↑
λ/2
Third Harmonic (or)
First overtone
λ
λ
3λ
l2 =  +  =  .... (2)
4
2
4
3λ
λ
2λ
From (1) & (2), l2 − l1 =  −  = 
4
4
4
λ
⇒ l2 − l1 = −
⇒ λ = 2 (l2 − l1)
2
∴ Velocity of Sound in air is, V = f λ
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⇒ V = f × 2 (l2 − l1)
.
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b
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t
⇒ V = 2f (l2 − l1)
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Writer: G.V. Chandra Sekhar
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Senior Inter Physics
Waves
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Q: What is Doppler Effect? What is Doppler Shift? Obtain an expression for
the apparent frequency of sound heard
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(i) When the sound source is in motion with respect to an observer at rest.
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(ii) When the observer is in motion with respect to a source at rest.
(8 Marks)
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A: Doppler Effect: The phenomenon of apparent change in the frequency of sound
a
n
heard by a listener due to relative motion between the source of sound and the
listener is called Doppler Effect.
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Doppler Shift: The difference between apparent frequency of sound heard by the
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listener (observer) and actual frequency of sound produced by the sound source
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is called Doppler Shift.
.
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(i) Source in motion and observer at rest:
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t
Consider a sound source 'S' which is moving with a constant velocity 'Vs' by
emitting sound having a frequency 'n' and an observer 'O' is stationary in a frame
a
r
p
in which the medium is also at rest.
Let the speed of sound wave of angular frequency 'ω' and time period 'T0' both
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measured by this observer, be 'V'.
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Assume that the observer has a detector that counts every time a wave crest
reaches it.
As shown in the figure, at time t = 0 the source is at point 'S1' located at a
distance 'L' from the observer who is resting at 'O', and source emits a crest.
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This reaches the
l
observer at time t1 = .
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S1 (Vs→)
O
S2
l
( l + Vs.T0)
← Vs.T0 →
At time t = T0 the source has moved a distance Vs. T0 and is at point 'S2',
located at a distance (l + Vs. T0) from the observer.
At 'S2' the source emits a second crest and this reaches the observer at a time
(l + Vs.T0 )
t2 = T0 + 
V
At time 'nT0', the source emits its (n + 1)th crest and it reaches the observer at a
time,
t
e
n
(l
+ nVsT0)
tn + 1 = nT0 + 
V
.
a
h
Hence in a time interval (tn + 1 − t1), the detector with the observer counts 'n'
crests and the observer records the period of the wave as 'T' which is given by
[
b]
i
t
l
(l + nVsT0)
nT0 + 
− 
V
V
(tn + 1 − t1)
T =  = 
n
n
a
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p
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a
n.Vs.T0
nT0 + 
V
⇒ T = 
n
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Vs.T0
⇒ T = T0 + 
V
Vs
⇒ T = T0 1 + 
V
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1
1
But T =  and T0 = 
n'
n
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−1
Vs
1
⇒ n' = n.  = n 1 + 
Vs
V
1+ 
V
Vs −1
Vs
If Vs << V, then 1 +  _∼ 1 − 
V
V
(
(
)
(
)
) (
)
n
.
a
∴ Apparent frequency of sound as heard by the observer is
(
)
(
)
Vs
V − Vs
n' = n 1 −  ⇒ n' = n  
V
V
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b
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★ If the source is approaching the stationary observer, then
(
V − (−Vs)
n' = n    
V
(
)
)
p
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d
V+V
⇒ n' = n   s
V
a
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.
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ii) Observer in motion and Source at rest:
Consider an observer who is moving with a constant velocity 'V0' towards a
stationary source of sound emitting sound having frequency 'n'.
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O1 V0
→
Sound
Wave
(
)
S (Vs = 0)
O2
(V + V0)
l
The time interval between the arrival of first and the (n + 1)th crests is,
nV0T0
tn + 1 − t1 = n T0 − 
V0 + V
∴ Time period of waves as measured by the observer is,
nV0T0
n T0 − 
tn + 1 − t1
V0 + V
T =  = 
n
n
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V0 T0
⇒ T = T0 − 
V0 + V
(
(
(
)
)
V0
⇒ T = T0 1 − 
V0 + V
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.
a
V0 + V − V0
= T0 
V0 + V
V
= T0 
V + V0
1
1
⇒  = 
n'
n
(
(
)
V

V + V0
V + V0
⇒ n' = n 
V
)
)
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t
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∴ Apparent frequency of Sound as heard by the observer in this case is given by
( .ee)
V + V0
n' = n 
V
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★ If the observer is moving away from the source at rest, then
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(
)
(
)
.
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V + (−V0)
V −V0
n' = n 
⇒ n' = n 
V
V
b
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t
Q: Two Sitar strings 'A' and 'B' playing the note 'Dha' are slightly out of tune
and produce beats of frequency 5 Hz. The tension of string 'B' is slightly
increased and the beat frequency is found to decrease to 3 Hz. What is the
original frequency of 'B', if the frequency of 'A' is 247 Hz?
(4 Marks)
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A: Let original frequency of B = fB Hz.
fA = 247 Hz (data)
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Increase in tension of 'B' increases its frequency. If fB > fA, increase in fB should
have resulted in an increase in beat frequency. But the beat frequency is found to
decrease.
So, fB < fA ⇒ fA − fB = 5 Hz
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⇒ 247 − fB = 5
⇒ fB = 247 − 5 = 242 Hz
Writer: G.V. Chandra Sekhar
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