Oscillation Of Pendulum MCQ - Practice Questions with Answers

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
I = length of the 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. $T \propto \sqrt{l}$
where I is the distance between the point of suspension and the center of mass of the 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.

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 the 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 a 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 the 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 $g_{e f f}=0$
Similarly in the case of a satellite and at the center of the earth the $g_{e f f}=0$ 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 $g_{e f f}=\left(g^2+a^2\right)^{\frac{1}{2}}$

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

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 $g_{\text {eff }}=g \cos \theta$
$T=2 \pi \sqrt{\frac{l}{g \cos \Theta}}$
where
$l=$ the length of pendulum
$g=$ acceleration due to gravity.
$\Theta=$ angle of inclination

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 $\rho<\sigma_{\text {) }}$.
Then the buoyant force is given as $F_B=V \rho g$
As buoyant force will oppose its weight therefore $F_{n e t}=m g_{e f f}=m g-F_B$

For the above figure let bob be displaced for a small displacement $\times$ and is at an angle $\theta$ with the verticle.
For small displacement $x$ of the bob, restoring force

$$
F_{\text {rest }}=(m g-V \rho g) \sin \theta=-(m g-V \rho g) \frac{x}{l}
$$

and acceleration $=-\left(g-\frac{V \rho q}{m}\right) \frac{x}{l}$
On comparing with a standard equation of SHM, $a=-\omega^2 x$, we get

$$
\omega=\sqrt{\frac{\left(g-\frac{V \rho g}{m}\right)}{l}}=\sqrt{\frac{g}{l}\left(1-\frac{\rho}{\sigma}\right)}
$$

and $T=2 \pi \sqrt{\frac{\ell}{g\left(1-\frac{\rho}{\sigma}\right)}}=\sqrt{\frac{g}{l}\left(1-\frac{\rho}{\sigma}\right)}$
and $T=2 \pi \sqrt{\frac{\ell}{g\left(1-\frac{\rho}{\sigma}\right)}}$
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

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

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

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

where
$l=$ length of pendulum
$g=$ acceleration due to gravity.
$R=$ Radius of earth
- Various cases
A. If $l \ll R$, then $\frac{1}{l} \gg \frac{1}{R} \quad$ so $\quad T=2 \pi \sqrt{\frac{l}{g}}$
B. If $l \gg R($ or $l \rightarrow \infty)$ then $\frac{1}{l}<\frac{1}{R}$ so $T=2 \pi \sqrt{\frac{R}{g}}=2 \pi \sqrt{\frac{6.4 \times 10^6}{10}} \cong 84.6$ minutes and it is the maximum time period which an oscillating simple pendulum can have
C. If $l=R \quad$ so $\quad T=2 \pi \sqrt{\frac{R}{2 g}} \cong 1$ hour