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Oscillation Of Pendulum - Practice Questions & MCQ

Edited By admin | Updated on Sep 18, 2023 18:34 AM | #JEE Main

Quick Facts

Simple pendulum is considered one the most difficult concept.
61 Questions around this concept.

Solve by difficulty

EASY

A child swinging on a swing in sitting position stands up, then the time period of the swing will :

increase

decrease

remains same

increases if the child is long and decreases if the child is short.

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The bob of a simple pendulum is a spherical hollow ball filled with water. A plugged hole near the bottom of the oscillating bob gets suddenly unplugged. During observation, till water is coming out, the time period of oscillation would

remain unchanged

increase towards a saturation value

first increase and then decrease to the original value

First decrease and then increase to the original value

View Solution

Concepts Covered - 4

Simple pendulum

An ideal simple pendulum consists of a heavy point mass body suspended by a weightless, inextensible
and perfectly flexible string from rigid support about which it is free to oscillate.

The time period of oscillation of simple pendulum (T)-

When the bob is displaced to position B, through a small angle from the vertical as shown in the below figure.

Then Bob will perform SHM and its time period is given as

$T=2\pi \sqrt{\frac{l}{g}}$

where

m=mass of the bob

l = length of pendulum

g = acceleration due to gravity.

key points

1. The time period of a simple pendulum is independent of the mass of the bob.

I.e If the solid bob is replaced by a hollow sphere of the same radius but different mass, the time period remains
unchanged.

2. $T \ \alpha \ \sqrt{l}$
where l is the distance between the point of suspension and center of mass of bob and is called effective length.

3. The period of a simple pendulum is independent of amplitude as long as its motion is simple harmonic.

Oscillation of Pendulum in different situations-part 1

Pendulum in a lift

1.The time period of a simple pendulum , If the lift is at rest or moving downward /upward with constant velocity.

$T= 2\pi \sqrt{\frac{l}{g}}$

where

$l=$ the length of pendulum

$g=$ acceleration due to gravity.

2.The time period of a simple pendulum, If the lift is moving upward with constant acceleration a

$T= 2\pi \sqrt{\frac{l}{g+a}}$

where

$l=$ the length of pendulum

$g=$ acceleration due to gravity.

$a=$ acceleration of pendulum.

3. The time period of simple pendulum If the lift is moving downward with constant acceleration a

$T= 2\pi \sqrt{\frac{l}{g-a}}$

where

$l=$ the length of pendulum

$g=$ acceleration due to gravity.

$a=$ acceleration of pendulum.

4. The time period of a simple pendulum , If the lift is moving downward with acceleration a =g

$T= 2\pi \sqrt{\frac{l}{g-g}}= \infty$

It means there will be no oscillation in a pendulum as here $g_{eff}=0$

Similarly in the case of a satellite and at the center of the earth the $g_{eff}=0$ so in these cases, effective acceleration becomes zero and
the pendulum will stop.

5. The time period of a simple pendulum whose point of suspension moving horizontally with acceleration 'a'

For the above figure $g_{eff}= (g^{2}+a^{2})^{ \frac{1}{2}}$

$T= 2\pi \sqrt{\frac{l}{(g^{2}+a^{2})^{ \frac{1}{2}}}}$

Where

$l=$ the length of pendulum

$g=$ acceleration due to gravity.

$a=$ acceleration of pendulum.

6. The time period of simple pendulum accelerating down an incline

In this case $g_{eff}= gcos\theta$

$T=2\pi \sqrt{\frac{l}{g\cos \Theta }}$

where

$l=$ the length of pendulum

$g=$ acceleration due to gravity.

$\Theta=$ angle of inclination

Oscillation of Pendulum in different situations-part 2

1.The time period of the pendulum in a liquid

If we immerse a simple pendulum in a liquid, the bob of the pendulum will experience a buoyant force in an upward direction in
addition to the other forces such as gravity and tension.

If bob a simple pendulum of density $\sigma$ is made to oscillate in some fluid of density $\rho$ (where $\rho < \sigma$ ).

Then the buoyant force is given as $F_B=V \rho g$

As buoyant force will oppose its weight therefore $F_{net}=mg_{eff}=mg-F_B$

And for the above figure let bob is displaced for a small displacement x and is at an angle $\theta$ with the verticle.

$\begin{array}{l}{\text { For small displacement } x \text { of the bob, restoring force }} \\ {\qquad F_{\text {rest }}=(m g-V \rho g) \sin \theta=-(m g-V \rho g) \frac{x}{l}} \\ {\text { and acceleration }=-\left(g-\frac{V \rho g}{m}\right) \frac{x}{l}}\end{array}$

$\begin{array}{l}{\text { On comparing with standard equation of SHM, } a=-\omega^{2} x, \text { we }} \\ {\text { get }} \\ {\qquad \omega=\sqrt{\frac{\left(g-\frac{V \rho g}{m}\right)}{l}}=\sqrt{\frac{g}{l}\left(1-\frac{\rho}{\sigma}\right)}} \\ {\text { and } T=2 \pi \sqrt{\frac{\ell}{g\left(1-\frac{\rho}{\sigma}\right)}}}\end{array}$

2. The time period of the Second's pendulum

Second, ’s Pendulum: It is that simple pendulum whose time period of vibrations is two seconds.

Putting T=2 sec in $T=2\pi \sqrt{\frac{l}{g}}$ we get the Length of a second’s pendulum is nearly 1 meter on the earth's surface.

Pendulum of large length but small amplitude

If the length of the pendulum is comparable to the radius of the earth

then $T= 2\pi \sqrt{\frac{1}{g\left ( \frac{1}{l}+\frac{1}{R} \right )}}$

where

$l=$ length of pendulum

$g=$ acceleration due to gravity.

$R=$ Radius of earth

Various cases

$\text { A. If } l<<R, \text { then } \frac{1}{l}>>\frac{1}{R} \quad \text { so } \quad T=2 \pi \sqrt{\frac{l}{g}}$

$\\ \text { B. If } l>>R\ \ ( or \ \ l\rightarrow \infty) \ then \ \ \frac{1}{l} < \frac{1}{R}\quad \\ \text { \ \ \ \ so } T=2 \pi \sqrt{\frac{R}{g}}=2 \pi \sqrt{\frac{6.4 \times 10^{6}}{10}} \cong 84.6 \text { minutes }\\ \text { \ \ \ \ and it is the maximum time period which an oscillating simple pendulum can have}$