First slide

Applications of SHM

Question

A simple pendulum hanging from the ceiling of a stationary lift has a time period T₁. When the lift moves downward with constant velocity, the time period is T₂, then

Moderate

A

T₂ is infinity
B

$T_{2} > T_{1}$
C

$T_{2} < T_{1}$
D

$T_{2} = T_{1}$

Solution

$T \propto \frac{1}{\sqrt{g}}$ and g is same in both cases so time period remain same.

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