First slide

Applications of SHM

Question

A man measure the period of a simple pendulum inside a stationary lift and finds it to be T seconds. If the lift accelerates upwards with an accelerating g/4, then the period of the pendulum will be

Moderate

A

$T$
B

$\frac{T}{4}$
C

$\frac{2T}{\sqrt{5}}$
D

$2 T \sqrt{5}$

Solution

$i n s t a t o n a r y l i f t T=2π \sqrt{\frac{l}{g}} ---(1) I n u p w a r d m o t i o n o f l i f t T^{1} =2π \sqrt{\frac{l}{g + a}} ---(2) d i v i d i n g e q n (2) b y e q n (1) \frac{T^{1}}{T} = \sqrt{\frac{g}{g + a}} g i v e n a = \frac{g}{4} \frac{T^{1}}{T} = \sqrt{\frac{g}{g + \frac{g}{4}}} \frac{T^{1}}{T} = \sqrt{\frac{4}{5}} = \frac{2}{\sqrt{5}} T^{1} = \frac{2 T}{\sqrt{5}}$ -

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