First slide

Applications of SHM

Question

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

Moderate

A

T
B

$\frac{T}{4}$
C

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

$2 T \sqrt{5}$

Solution

In stationary lift $T = 2 π \sqrt{\frac{l}{g}}$
In upward moving lift $T^{'} = 2 π \sqrt{\frac{l}{(g + a)}}$
( $a =$ Acceleration of lift)
$\Rightarrow \frac{T^{'}}{T} = \sqrt{\frac{g}{g + a}} = \sqrt{\frac{g}{(g + \frac{g}{4})}} = \sqrt{\frac{4}{5}}$ $\Rightarrow T^{'} = \frac{2 T}{\sqrt{5}}$

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