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

$3 T$
C

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

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

Solution

$T=2π \sqrt{\frac{l}{g}} given a= \frac{g}{3} T^{1} =2π \sqrt{\frac{l}{g+a}} \frac{T^{1}}{T} = \sqrt{\frac{g}{g+a}} = \sqrt{\frac{g}{g+g/3}} = \sqrt{\frac{3g}{4g}} = \sqrt{\frac{3}{4}} T^{1} = \frac{\sqrt{3} T}{2}$

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