First slide

Applications of SHM

Question

The time period of a simple pendulum inside a stationary lift is $\sqrt{5}$ sec. What will be the time period when the lift moves upward with an acceleration g/4?

Moderate

A

$\sqrt{5}$ seconds
B

$2 \sqrt{5}$ seconds
C

$(2+ \sqrt{5})$ seconds
D

2 seconds

Solution

In upward accelerating lift, $g_{e f f} = g + a$
$T=2π \sqrt{\frac{l}{g}} {and T}^{1} = 2π \sqrt{\frac{l}{(g+g/4)}} \Rightarrow \frac{T^{1}}{T} = \frac{2}{\sqrt{5}} given T= \sqrt{5} \Rightarrow T^{1} = \frac{2}{\sqrt{5}} T= \frac{2}{\sqrt{5}} (\sqrt{5}) \Rightarrow T^{1} =2 seconds$

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