The mass and the diameter of a planet are three times the respective values for the Earth. The period of…
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 $2 \mathrm{~s}$. The period of oscillation of the same pendulum on the planet would be:
$\frac{\sqrt{3}}{2} \mathrm{~s}$
$\frac{2}{\sqrt{3}} \mathrm{~s}$
$\frac{3}{2} s$
$2 \sqrt{3}$ s
Solution
Acceleration due to gravity $\mathrm{g}=\frac{\mathrm{GM}}{\mathrm{R}^{2}}$
$\frac{g_{p}}{g_{e}}=\frac{M_{p}}{M_{e}}\left(\frac{R_{e}}{R_{p}}\right)^{2}=3\left(\frac{1}{3}\right)^{2}=\frac{1}{3}$
Also $\mathrm{T} \propto \frac{1}{\sqrt{\mathrm{g}}} \Rightarrow \frac{\mathrm{T}_{\mathrm{p}}}{\mathrm{T}_{\mathrm{e}}}=\sqrt{\frac{\mathrm{g}_{\mathrm{e}}}{\mathrm{g}_{\mathrm{p}}}}=\sqrt{3}$
$\Rightarrow \mathrm{T}_{\mathrm{p}}=2 \sqrt{3} \mathrm{~s}$