Physics. Simple Harmonic Motion - 2 Session Session Objectives

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