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Physics
Simple Harmonic Motion - 2
Session
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
kI
2
kI
IT 2
k
Simple Pendulum
resF mgsin
Ifi s small sin
ma mg
a g
x ga g x
gT 2
g
x
mg sin mg cos
T
Torsional Pendulum
k
kI
IT 2
k
k torsional cons tant.
Horizontal Vibrations of Spring
ma kx
ka x
m
mT 2
k
Same result holds good for vertical vibrations of a spring also.
Springs – Series and Parallel
K1K2Keq
seriesinSpringsK1
K1
K1
21eq
K1
K2
Keq
parallelinSpringsKKK 21eq
Questions
Illustrative Problem
Find the spring constant for the spring system shown:
K1
K2
K3
K1 K2
(a)
(b)
Solution
a) The two springs are in parallel so
b) The springs are in parallel so
eq 1 2 3K K K K
eq 1 2K K K
Illustrative Problem
The spring constant for the adjoining combination of springs is
a. K b. 2K
c. 4K d.5K2
K K
2K
Class Test
Class Exercise - 1
The appropriate graph between time period T of angular SHM of a body and radius of gyration r is
(a) (b)
(c) (d)
T
r
T
r
T
r
T
r
Solution
I
T 2C
2Mr
2C
M2 r
C
T r
Hence, answer is (a).
Class Exercise - 2
Which of the following graphs is appropriate for simple pendulum?
(a) (b)
(c) (d)
T
l
T 2
l T
l
T
l
Solution
T 2g
T So (a).
and So (b). 2T
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)
(c) (d)
3E
4
E
2
E
4
Solution
2 2 2A
1K.E. I [ – ]
2
Now 2 2A
1E I
2
A/2
2A2 2
A1
K.E. I –2 4
3
E4
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)
(c) 3K (d)
K
9
K
3
Solution
We know that for springF = Kx
If force is constant, then
1
Kx
Now let K´ be new spring constant. Then
K 9K´
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 after 3t
seconds, then n is equal to
(a) (b) 8
(c) (d) 4
1
8
1
n
1
4
Solution
31 12 8
Hence, n = 8
Hence, answer is (a).
Class Exercise - 8
The angle at which the mean position exists of a simple pendulum placed in a
car accelerating by to right is
(a) (b)
(c) (d) None of these
g
2
1 1tan
2
3
2
Solution
T cos = mg
T
mg
ma
mg
T sin2
g
2tang
–1 1tan
2
Hence, answer is (a).
Class Exercise - 9
Keg of following figure is
(a) K1 + K2 + K3 + K4 (b)
(c) (d) None of these
m
K 4
K1 K2 K3
1 2 3 4
1 2 3 4
(K K K )(K )
K K K K
1 2 3 4
11 1 1 1
K K K K
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 = Asint, x2 = Acost?
(a) (b)
(c) (d) None of these
2
2
2
Solution
X = X1 + X2
= A sint + A cost
2A sin t4
So the time period = 2
Hence, answer is (a).
Thank you
