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Physics
Session
Simple Harmonic Motion - 2
Session Objectives
Session Objective
Angular SHM
Pendulum (Simple)
Torsional Pendulum
Horizontal Vibration of a spring
Vertical Vibration of a spring
Combination of springs in series
Combination of springs in Parallel
Angular SHM
   k
 I   k
k 
   
I
2
   
 
k
I
 T  2
I
k
Simple Pendulum
Fres   mgsin 
If  is small sin 

ma   mg

a   g
x
 g
a   g      x
 
 

g
 T  2
g
T
x
mg sin mg cos
Torsional Pendulum
   k
k
 
I
T  2
I
k
k  torsional cons tant.

Horizontal Vibrations of Spring
ma   kx
k
a   x
 m
m
T  2
k
Same result holds good for vertical vibrations
of a spring also.
Springs – Series and
Parallel
K1
1
1
1


K eq K1 K2
Keq
K2
Springs
in series
K1
Keq
Keq  K1  K2
K2
Springs
in parallel
Questions
Illustrative Problem
Find the spring constant for the spring
system shown:
K1
K2
(a)
K1
K2
K3
(b)
Solution
a) The two springs are in parallel so
Keq  K1  K2
b) The springs are in parallel so
Keq  K1  K2  K3
Illustrative Problem
The spring constant for the adjoining
combination of springs is
a. K
b. 2K
c. 4K
d.
5K
2
2K
K
K
Class Test
Class Exercise - 1
The appropriate graph between time
period T of angular SHM of a body and
radius of gyration r is
T
T
(a)
(b)
r
r
T
T
(c)
(d)
r
r
Solution
T  2
I
C
Mr 2
 2
C

M
  2
 r
C

T  r
Hence, answer is (a).
Class Exercise - 2
Which of the following graphs is
appropriate for simple pendulum?
T
T2
(a)
(b)
l
T
T
(c)
l
(d)
l
l
Solution
T  2
g
T
and T 2 
So (a).
So (b).
All others are wrong except these.
Hence, answer is (a) & (b).
Class Exercise - 3
A hollow metallic bob is filled with water
and hung by a long thread. A small hole
is drilled at the bottom through which
water slowly flows out. The period of
oscillations of sphere
(a) decreases
(b) increases
(c) remains constant
(d) first increases and then decreases
Solution
As the water slowly flows out, the
centre of gravity moves down. So
the length increases and hence T
increases. After half the sphere is
empty, the centre of gravity begins
to move up. So the length
decreases and hence T decreases.
Hence, answer is (d).
Class Exercise - 4
The total energy of a simple pendulum
is E. When the displacement is half of
amplitude, its kinetic energy will be
(a) E
(b) 3E
4
(c) E
2
(d) E
4
Solution
1
K.E.  I2[A2 – 2 ]
2
Now E 
1 2 2
I A
2
1 2  2 A 2 
K.E.    I A –

A/2
2
4 

3
 E
4
Hence, answer is (b).
Class Exercise - 5
A spring has a spring constant
K. It is cut into two equal
lengths and the two cut pieces
are connected in parallel. Then
the spring constant of the
parallel combination is
(a) K
(b) 2K
(c) 4K
(d) K
2
Solution
Spring constant of cut pieces
K´ = 2K
Now parallel combination of these results
in a spring constant.
K  2K  2K
= 4K
Hence, answer is (c).
Class Exercise - 6
A spring of spring constant K is divided
into nine equal parts. The new spring
constant of each part is
(a) 9K
(b) K
9
(c) 3K
(d)
K
3
Solution
We know that for spring
F = Kx
If force is constant, then
1
K
x
 
 
K
9
Now let K´ be new spring constant. Then
 
K´
or K´ = 9K
Hence, answer is (a).
Class Exercise - 7
The amplitude of damped oscillator
becomes one-half after t seconds.
If the amplitude becomes
1
after 3t
n
seconds, then n is equal to
1
(a)
8
(b) 8
1
4
(d) 4
(c)
Solution
3
1
 1

 2
8
 
Hence, n = 8
Hence, answer is (a).
Class Exercise - 8
The angle f at which the mean position
exists of a simple pendulum placed in a
car accelerating by
g
to right is
2

3
 1 
(a) tan1 

2


(b)
(c) 
(d) None of these
2
Solution
T cos f = mg
f
T
mg
T sin f 
2
g
tan f  2
g
ma
mg
f  tan–1
1
2
Hence, answer is (a).
Class Exercise - 9
Keg of following figure is
K1
K2
K3
m
K4
(a) K1 + K2 + K3 + K4
(c)
1
1
1
1
1



K1 K 2 K 3 K 4
(K1  K 2  K 3 )(K 4 )
(b)
K1  K 2  K 3  K 4
(d) None of these
Solution
When we displace the body from its
mean position, we see the
extension in all the springs is same.
So the combination is parallel.
Hence answer is (a).
Class Exercise - 10
What is the resultant time period of a
particle, if following two SHMs in same
direction when superimposed on each
other is x1 = Asint, x2 = Acost?
(a) 2

(c)


(b) 2
2
(d) None of these
Solution
X = X1 + X2
= A sint + A cost


 2A sin  t  
4

2
So the time period =

Hence, answer is (a).
Thank you