# The value of acceleration due to gravity at the moon is g/6, where g is the value of acceleration due to gravity at earth. The value of frequency of o

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The value of acceleration due to gravity at the moon is g/6, where g is the value of acceleration due to gravity at earth. The value of frequency of oscillation of simple pendulum on moon as compared to earth will be

1. $\sqrt6$ times
2. 6 times
3. 1/6 times
4. $\sqrt\frac{1} {6}$ times

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Correct Answer - Option 4 : $\sqrt\frac{1} {6}$ times

Concept:

The time period of a simple pendulum is

$T =2\pi \sqrt{\frac Lg}$

The frequency of oscillation (f) is inversely proportional to the time period

$f \propto \frac{1}{T}$

where L = Length of the pendulum, g = acceleration due to gravity

Calculation:

Given:

The acceleration due to gravity at the moon is $g'=\frac g6$

The time period of the pendulum at the moon surface is:

$T' =2\pi \sqrt{\frac L{g'}}=2\pi \sqrt {\frac {6L}g}$

$T'=\sqrt {6}T$

$f \propto \frac{1}{T}$

$f'=\sqrt\frac{1} {6}f$