First slide

Applications of SHM

Question

The period of a simple pendulum measured inside a stationary lift is ‘T’. If the lift starts moving upwards with a acceleration g/3. What will be the time period.

Moderate

A

$\frac{T}{3}$
B

3T
C

$\frac{\sqrt{3} T}{2}$
D

$\sqrt{\frac{3}{2}} T$

Solution

$t i m e p e r i o d o f s i m p l e p e n d u l u m i s T=2π \sqrt{\frac{l}{g}} given acceleration of lift in which pendulum is oscillating is a= \frac{g}{3}$
$n e w t i m e p e r i o d o f s i m p l e p e n d u l u m i s T^{1} =2π \sqrt{\frac{l}{g+a}} =2π \sqrt{\frac{l}{g+g/3}} o n s u b s t i t u t i n g a = \frac{g}{3}, n o w d i v i d e t h e e q u a t i o n s o f T a n d T^{1} \frac{T^{1}}{T} = \sqrt{\frac{g}{g+a}} = \sqrt{\frac{g}{g+g/3}} = \sqrt{\frac{3g}{4g}} \Rightarrow \frac{T^{1}}{T} = \sqrt{\frac{3}{4}} \Rightarrow T^{1} = \sqrt{3} / 2 T$

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