Question

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 $2\text{Sec.}$ The Period of oscillation of the same pendulum on the planet would be:

• A. $\frac{3}{2}\text{sec}$
• B. $\frac{2}{\sqrt{3}}\text{sec}$
• C. $\frac{\sqrt{3}}{2}\text{sec}$
• D. $2\sqrt{3}\text{sec}$

Answer 1

The correct option is D $2\sqrt{3}\text{sec}$

Acceleration due to gravity is given by,
$g=\frac{GM}{R^2}$

$\Rightarrow \frac{g_p}{g_e}=\frac{M_p}{M_e}{\left(\frac{R_e}{R_p}\right)}^2$

$\frac{g_p}{g}=\frac{3M_e}{M_e}\times {\left(\frac{R_e}{3R_e}\right)}^2=\frac{1}{3}$

Also, $T\propto \frac{1}{\sqrt{g}}\Rightarrow \frac{T_p}{T_e}=\sqrt{\frac{g_e}{g_p}}=\sqrt{3}$

$\Rightarrow T_p=T_e\sqrt{3}=2\sqrt{3}\text{sec}$

Hence, $\left(D\right)$ is the correct answer.