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:
\(T' =2\pi \sqrt{\frac L{g'}}=2\pi \sqrt {\frac {6L}g}\)
\(T'=\sqrt {6}T\)
\(f \propto \frac{1}{T}\)
\(f'=\sqrt\frac{1} {6}f\)
