Physics

Simple Harmonic Motion - 3

Session

Session Objectives

Session Objective

Problems

Class Exercise - 1

O is a point of suspension of a simple pendulum of length OA = 1.8 m. N is a nail vertically below O, where ON = 0.9 m. The bob starts from A and returns to A after a complete swing. What is the time taken to complete the swing?

A

BC

N

O

Solution

For the journey from A to B, let the timebe

AB

T 1t 2

4 4 g

1 1.82 0.67 s

4 9.8

Similarly, for tBC

1.8

' 0.92 2

BC

T ' 1 0.9t 2 0.47 s

4 4 0.8

Total time from A back to A is

AB BCt 2 t t

= 2[0.67 + 0.47]

= 2.28 s

Class Exercise - 2

Determine the period of small oscillations of a mathematical pendulum, i.e. a ball suspended by a thread = 20 cm in length, if it is located in a liquid whose density is n = 3 times less than that of the ball. The resistance of the liquid is to be neglected.

Solution

In the figure, d and are density of ball and liquid respectively.

T

Upthrust = Vpg

Weight = Vdg

Then in equilibrium,

T V ' g Vdg

or T Vg(d )

If A is the area and x is the displacement of the liquid,

then ma A x g(d )

Solution contd..

Compare this by equation a = 2x

So this represents the SHM. The time period T is given by

2 2 dT 2

g(d )g(d )

d

A g(d ) A g(d )

a x xm A d

Also d = 3

T = 1.1 s

Class Exercise - 3

A block is resting on a piston which is moving vertically with an SHM of period 1 s. At what amplitude of motion will the block and piston separate?

Solution

For the block,

mg – R = ma

or R = m(g – a)

In order to separate the block, R = 0 or a = g

mg

R Now a = 2x

g = 2x

2 2 22

T

g 9.8 9.8or x

4

= 0.248 m

Class Exercise - 4

A plank with a body of mass m placed on it starts moving straight up according to the law X = a(1 – cost), where X is the displacement from the initial position, = 11 s–1. Find the time dependence of the force that the body exerts on the plank. If a = 4.0 cm plot this dependence.

Solution

X a 1 cos t

2

22

dx d xa sin t and a cos t

dt dt

2

2d x

Now F m mgdt

2ma cos t mg

2amg 1 cos t

g

Plot is shown in the figure.

0.5

1.0

1.5

Ow t

2

Class Exercise - 5

Find the time period of a pendulum of infinite length. Assume bob to be near to surface of earth.

A B

mg

O´

R

l

O

T

Solution

Restoring force on the bob = mg sin( + )

Force on the bob = –mg{sin( + )}

force

Acceleration a gsinmass

If is small, is also small.

Hence, sin( )

AB ABa g ; also and

R

Solution contd..

AB AB

a gR

R

g ABR

2 R

So gR

Rt 2 [If is infinite]

(R )g

R

t 2g

Class Exercise - 6

In the arrangement shown in the figure, the particle m1 rotates in a radius r on a smooth horizontal surface with angular velocity 0. If m2 is displaced slightly in the vertical direction, find the time period of oscillation.

O r m1wo

m 2

Solution

Considering the equilibrium of m2, we have

21 0 2m r m g

20

1

m g

m r

Now let mass m2 is displaced downward by a distance x. Then radius of circular path of m1 decreases to (r – x). Applying the conservation of angular momentum, we have

2 21 0 1m r m (r x)

Solution contd..

2

0r

orr x

Tension is also increased as shown below.

242 0

1 3 3

m rrT m

(r x) x1

r

2

1 03 x

m r 1r

As a result, m2 gets a restoring force given by

2F T m g

2

1 0 23x

or F m r 1 m gr

Solution contd..

21 0or F 3m x

2

21 2 1 02

d xor m m 3m x

dt

221 0

21 2

3md xor x

m mdt

1 2

2

m m rSo T 2

3m g

Class Exercise - 7

Find the frequency of small oscillations of a thin uniform vertical rod of mass m and length is hinged at point O. The combined stiffness of the spring is equal to K. The mass of the spring is negligible.

Ol

Solution

A displaced position of the rod through an angle is shown in the given figure. The displacement of spring is x. Let K1 and K2 be the stiffness of the springs respectively.

O

mg

(K + K )x1 2

Considering the torques actingon the rod, we have

1 2K K x cos mg sin I2

Solution contd..

When is small, cos = 1 and sin =

2

1 2m

K K x mg2 3

x

Now

22

1 2mg m

So K K2 3

Solving it for , we get

1 22

3mg3 K K

2m

Hence, the motion is SHM.

3mg3K

2So

m

Class Exercise - 8

In the arrangement shown in thefollowing figure, the sleeve M ofmass m = 0.2 kg is fixed betweentwo identical spring whose combinedforce constant K = 20 N/m. Thesleeve can slide without friction over a horizontal bar AB. The arrangement rotates with a constant angular velocity = 4 rad/s about a vertical axis passing through the middle of the bar. Find the period of small oscillations of the sleeve?

Solution

We will analyse the problem relative to the rotating bar AB. As the acceleration of bar will be centripetal, a pseudo force will act on the sleeve away from centre and will be of magnitude m 2x.

xA Bm

If the sleeve is displaced byx1, the net force towards thecentre is

2 2F K x m x K m x

2

mPeriod 2

K m

Solution contd..

Note: If , there will be no oscillation of the sleeve. It will rush to the point B if it is displaced slightly (for K < m2) or will remain in the displaced position (for K = m2)

2K m

Class Exercise - 9

The friction coefficient between thetwo blocks shown in the figure is mand the horizontal plane is smooth.

(a) Find the time period of system, and

(b) magnitude of frictional force between the blocks.

m

M

Solution

(a) For small amplitude, the two blocks oscillate together. In this case,

K

m M

m Mor T 2

K

(b) The acceleration of the blocks at displacement x from mean position

2 Kx

a xM m

Force on upper block = ma =

mKxM m

Solution contd..

The force is provided by friction of thelower block.

Magnitude of frictional force =

mK x

M m

Class Exercise - 10

A liquid of density d is kept in a vertical U-tube of uniform cross section A. If the liquid column is slightly depressed and left, show that the resulting motion of the liquid is SHM and find the period.

Solution

Let the liquid be depressed by a height x in the right side of the U-tube. Then the liquid rises above on the left-side by a height x.

x Restoring force = Pressure × Area

= – (2xdg) × A

The motion is SHM. F x

or K = 2dgA

m m

or T 2 2K 2dgA

Thank you