4. OSCILLATIONS
A particle performs S.H.M of period 12
A particle excuting S.H.M of amplitude 2.
seconds and amplitude 8 cm. If initially
5 cm and period of 2 s. Find the speed
the particle is at the positive extremity,
of the particle at a point where its
how much time will it take to cover a
acceleration is half of its maximum
distance of 6 cm from the extreme
value.
position.
Given :
Solution :
A
=
5 cm
=
5 × 10–2 m,
T
=
12 s,
T
=
2 s.
A
=
8
cm
a max
Aω 2
a
=
=
2π
π
2
2
∴
ω
=
=
rad/s
T
6
To Find :
When the particle covers a distance of 6
v
=
?
cm from the positive extremity, its
Formula :
displacement from the mean position is
x
=
8–6
=
2 cm
v
=
ω A 2 – x2
From
the
equation
of
S.H.M
Solution :
x
=
A cos ωt
Since
2
[from
extreme position]
a
=
ωx
1.
∴
=
a max
2
=
Aω 2
2
π 
∴ cos  t  =
6 
x
=
A
2
∴
v
=
ω A 2 – x2
=
A2
ω A2 –
4
a
ω2x
∴
∴
v
=
2
=
3
Aω
ω
2
2π
3
× 5 × 10–2 ×
T
2
2π 

∵ ω = T 


=
∴
v
Oscillations
=
=
2 × 3.14
3
× 5 × 10–2 ×
2
2
–2
13.6 × 10 m/s
13.6 cm/s
π
t
6
=
π 
8 cos  t 
6 
0.25
=
cos–1 (0.25)
=
75052′
∴
π
t
6
=
75052′ ×
∴
t
=
6 × 75.52
180
=
∴
t
=
75.52
30
2.517 s
π
1800
(in rad)
MAHESH TUTORIALS SCIENCE
3.
.. 19
When the displacement in S.H.M is
1 rd
of the amplitude, what fraction of
3
total energy is kinetic energy and what ∴
fraction is potential energy ?
Given :
4.
A
x
=
3
To Find :
K.E
T.E
P.E
T.E
Formula :
=
?
=
?
=
1
kA2
2
ii)
K.E
=
1
k(A2 – x2)
2
=
1  2  A 2 
k A –   
2 
 3  

=
1  2 A2 
k A –

2 
9 
1
8A
k×
2
9
=
1
2  8 
 2 kA  9  
 

1

kA 2 
 2

=
8
9
Since, P.E
=
T.E – K.E
=
T.E – K.E
T.E
∴
P.E
T.E
=
1–
K.E
T.E
∴
2
=
K.E
T.E
∴
P.E
T.E
=
1
9
8
9
The displacement of a particle
performing linear S.H.M is given by
x=
∴
=
5π 

6 sin  3πt +
 metre
6 

=
To Find :
A
=
?
n
=
?
α
=
?
Formula :
x
=
A sin (ω
ωt +α
α)
Solution :
Comparing formula with given equation
We have,
A
=
6m
ω
=
3π
π rad/s
ω
=
2π
πn
Solution :
∴
1–
x
T.E
K.E
T.E
=
5π 
6 sin  3 πt +
 meter. Find
6 

amplitude, frequency and the phase
constant of the motion
Given :
i)
K.E
P.E
T.E
8
9
ω
2π
n
=
1.5 Hz
Phase constant,
n
=
α
=
=
3π
2π
5π
rad
6
5.
The period of oscillation of simple
pendulum increases by 20 % when its
length increased by 44 cm. Find its
i)
intial length
ii)
initial period
Given :
∴
T2
T1
=
l2
=
120
=
100
l1 + 0.44
6
5
Oscillations
MAHESH TUTORIALS SCIENCE
.. 20
To Find :
i)
l1
ii)
T2
Formula :
T
=
=
?
=
?
2π
l
g
∴
T1
l1
g
2π
Also, T1 =
=
2 × 3.14
=
2.006 s
1
9.8
Solution :
T1
=
2π
l1
g
T2
=
2π
l2
g
∴
T2
T1
∴
6
T1
5
T1
∴
6
5
=
l2
l1
=
l1 + 0.44
l1
=
l1 + 0.44
l1
Squaring both sides, we get
6.
A clock regulated by a seconds
pendulum keeps correct time. During
summer the length of the pendulum
increases to 1.01 m. How much will the
clock gain or lose in one day ?
(g = 9.8 m/s2)
Given :
l
=
1.01 m
g
=
9.8 m/s2
To Find :
Time lost or gain per day in summer
∆T =
?
Formula :
T
=
l1 + 0.44
l1
36
25
=
l+
∴
36
–1
25
=
0.44
l1
∴
∴
36 – 25
25
=
0.44
l1
∴
11
25
=
0.44
l1
∴
∴
l1
=
∴
l1
=
Oscillations
T
0.44
l1
25 × 0.44
11
1m
2π
l
g
Solution :
Substituting the given values,
36
25
∴
=
∴
=
2 × 3.14 ×
=
6.28
1.01
9.8
1.01
9.8
=
2.017 s
T
=
2.017 s
The period of a seconds pendulum is 2 s
ie, pendulum clock will lose its period.
loss in period = 2.017 – 2
= 0.017 s
ie, 0.017 s is lost in 2.017 s
Loss in period per day
∆T
=
24 × 3600 × 0.017
2.017
∆T
=
728.21 s
MAHESH TUTORIALS SCIENCE
.. 21
7.
An object performing S.H.M with mass
of 0.5 kg, force constant 10 N/m and
amplitude 3 cm.
i)
What is the total energy of object ?
ii)
What is its maximum speed ?
iii) What is the speed at x = 2 cm ?
iv) What are kinetic and potential
energies when x = 2 cm ?
Given :
m
=
0.5 kg
k
=
10 N/m
∴
A
=
3 cm = 3 × 10–2 m,
ω2
k
m
=
∴
ω
To Find :
i)
ii)
iii)
iv)
v)
Formula :
=
=
10
0.5
20 rad/sec
=
=
=
=
=
?
?
? at x
?
? at x
i)
T.E.
=
ii)
vmax
=
1
kA2
2
ωA
iii)
v
=
ω A 2 – x2
iv)
P.E.
=
v)
K.E.
Solution :
=
= 2cm
= 2cm
1
kx2,
2
T.E. – P.E.
T.E. =
ii)
vmax
vmax
20 × 3 × 10–2
=
= 0.1342 m/s
iii)
v
=
(
1
kx2,
2
P.E.
=
1
× 10 × (2 × 10–2)2
2
P.E.
Since,
K.E.
=
2 × 10–3 J
=
=
=
T.E. – P.E.
4.5 × 10–3 – 2 × 10–3
2.5 × 10–3 J
K.E.
T
=
l
g
2π
Solution :
From formula
∴
T2
=
g
=
4π2
l
g
4π 2 l
T2
2
20 ×
3 × 10
=
A simple pendulum is used in physics
laboratory experiment to obtain
experimental value for gravitational
acceleration g. A student measures the
length of pendulum 0.51 m, displaces it
o
10 from equilibrium position and
released it. Using a stopwatch, the
student determines that period of
pendulum is 1.44 s. Determine the
experimental value of the gravitational
acceleration.
Given :
l
=
0.51 m
T
=
1.44 s
To Find :
g
=
?
Formula :
1
× 10 × (3 × 10–2)2
2
T.E. = 4.5 × 10–3 J
i)
v)
P.E.
8.
= 20
T.E
vmax
v
K.E
P.E
iv)
–2 2
∴
v
=
20 ×
∴
v
=
0.1 m/s
) (
– 2 × 10
∴
g
=
∴
g
=
–2 2
)
4 × ( 3.14 ) × 0.51
( 1.44 )2
9.699 m/s2
5 × 10–4
Oscillations
MAHESH TUTORIALS SCIENCE
.. 22
9.
A particle executes S.H.M with
amplitude of 10 cm and period of 10 s.
Find the
i)
velocity
ii)
acceleration of the particle at a
distance 5 cm from the equilibrium
position.
Given :
A
=
10 cm, T
=
10 s
To Find :
i)
v
=
?
ii)
aat x = 5 cm =
?
Formula :
i)
ii)
Solution :
v
=
a
=
± ω A 2 – x2
– ω2 x
v
=
±ω A 2 – x2
v
=
±
2π
T
∴
∴
v
v
v
a
A body describes S.H.M in a path 0.12 m
long. Its velocity at the centre of the line
is 0.12 m/s. Find the period, and
magnitude of velocity at a distance
3 × 10–2 m from the central position.
Given :
2A
∴
A
vmax
x
=
To Find :
T =
Formula :
i) vmax
=
±
2π
T
=
±
2π
×5
T
=
=
=
±π 3
± 5.442 cm/s
– ω2 x
=
 2π 
– 
 ×5
 T 
( 10 )2 – ( 5 )2
=
a
=
–
?
=
ωA
=
ω A 2 – x2
ωA
2π
T
=
2
∴
T
=
∴
T
=
2π
2
3.142 s
v
=
ω A 2 – x2
v
=
2π
T
=
2π
0.0036 – 0.0003
T
=
=
=
2 0.0033
2 × 0.0574
0.1149 m/s
v
( 10 )2
=
0.12
0.06
=
π
( 0.06 )2
11.
–
(
=
3 × 10 –2
2
)
2
A particle executes S.H.M with a period
8 s. Find the time in which half the total
energy is potential.
Given :
T =
8s
1
T.E
2
Oscillations
=
∴
4π × 5
π2
5
?, v
vmax
A
3
20π 2
–
=
100
– 1.974 cm/s2
3 × 10–2 m
=
2
=
0.12 m,
0.06 m,
0.12 m/s,
ω
( 10 )2 – ( 5 )2
2
a
=
=
=
ii) v
Solution :
vmax =
2π 

∵ ω = T 


∴
10.
=
P.E
MAHESH TUTORIALS SCIENCE
.. 23
To Find :
t
=
?
i)
T.E
=
1
kA2
2
ii)
P.E
=
1
kx2
2
Formula :
Solution :
Since
1
T.E =
P.E
2
From formula (i) and (ii)
∴
1
1
× kA2
2
2
=
1 2
kx
2
∴
1
kA2
4
=
1 2
kx
2
∴
1 2
A
2
=
x2
∴
x
=
x
=
∴
A sin ωt
=
∴
sin ωt
=
∴
 2π 
sin 
t
 T 
=
∴
 2π 
sin 
t
 8 
=
∴
π
sin   t
4
=
∴
π
 t
4
=
Also,
∴
∴
π
 t
4
t
=
=
A
2
A sin ωt
A
2
1
2
1
2
1
2
1
2
 1 
sin–1 

 2
π
4
1 sec
Oscillations