First slide

Applications of SHM

Question

There is a simple pendulum hanging from the ceiling of a lift. When the lift is stand still, the time period of the pendulum is T. If the resultant acceleration becomes $g / 4,$ then the new time period of the pendulum is

Moderate

A

0.8 T
B

0.25 T
C

2 T
D

4 T

Solution

$T \propto \frac{1}{\sqrt{g}} \Rightarrow \frac{T_{2}}{T_{1}} = \sqrt{\frac{g_{1}}{g_{2}}} = \sqrt{(\frac{g}{g / 4})} \Rightarrow T_{2} = 2 T_{1} = 2 T$

Get Instant Solutions

When in doubt download our app. Now available Google Play Store- Doubts App

Recieve an sms with download link

Doubts App

OR

SCAN ME

Doubts App

Doubts App