First slide

Applications of SHM

Question

The frequency of oscillation of a simple pendulum suspended in a satellite that revolve around the earth is

Easy

A

1 Hz
B

2 Hz
C

zero
D

infinity

Solution

frequency = zero
$T = 2 π \sqrt{\frac{l}{g}}$
in an artificial satellite g=0
frequency = $\frac{1}{t i m e p e r i o d} = \frac{1}{T}$
if g=0 in above equation, T = $\infty$
then frequency = $\frac{1}{\infty} = 0$

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