The mass and the diameter of a planet are three times the respective values for the Earth. The period of oscillation of a simple pendulum on the Earth is 2s. The period of oscillation of the same pendulum on the planet would be:

Engineering

Physics

Gravitational Acceleration

Question

$\frac{3}{2} s$

$\frac{2}{\sqrt{3}} s$

$\frac{\sqrt{3}}{2} s$

$2 \sqrt{3} s$

$∵ g = \frac{GM}{R^{2}}$

$\frac{g_{p}}{g_{e}} = \frac{M_{p}}{M_{e}} {(\frac{R_{e}}{R_{p}})}^{2} = 3 {(\frac{1}{3})}^{2} = \frac{1}{3}$

$Also, T \propto \frac{1}{\sqrt{g}}$

$\Rightarrow \frac{T_{p}}{T_{e}} = \sqrt{\frac{g_{e}}{g_{p}}} = \sqrt{3}$

$\Rightarrow T_{p} = 2 \sqrt{3} s$

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