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 $a=g$

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

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 $g_{\text{eff }}=g\cos \theta$
$T=2\pi \sqrt{\frac{l}{g\cos \mathrm{\Theta }}}$
where
$l=$ the length of pendulum
$g=$ acceleration due to gravity.
$\mathrm{\Theta }=$ 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 $F_{net}=m{g}_{eff}=mg-F_B$

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

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

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

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

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

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

where
$l=$ length of pendulum
$g=$ acceleration due to gravity.
$R=$ Radius of earth
- Various cases
A. If $l\ll R$, then $\frac{1}{l}\gg \frac{1}{R}$ so $T=2\pi \sqrt{\frac{l}{g}}$
B. If $l\gg R($ or $l\to \mathrm{\infty })$ 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 $l=R$ so $T=2\pi \sqrt{\frac{R}{2g}}\cong 1$ hour