Question

The mass of 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?

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

Answer 1

The correct option is B $2\sqrt{3}s$

$\because g=\frac{GM}{R^2}$
$\frac{g_p}{g_o}=\frac{M_o}{M_o}{\left(\frac{R_o}{R_p}\right)}^2=3{\left(\frac{1}{3}\right)}^2=\frac{1}{3}$
Also $T\propto \frac{1}{\sqrt{g}}$
$\Rightarrow \frac{T_p}{T_o}=\sqrt{\frac{g_o}{g_p}}=\sqrt{3}$
$\Rightarrow T_p=2\sqrt{3}s$.