Question

The mass and the diameter of a planet are three times of the respective values for the earth. If the time period of oscillation of a simple pendulum on earth is $2s$ , then the time period of oscillation of the same pendulum on the surface of the given planet would be

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

Answer 1

Answer: (D)

We know that,
$g=\frac{GM}{R^2}$
$\Rightarrow \frac{g_p}{g_e}=\frac{M_p}{M_e}\times {\left(\frac{R_e}{R_p}\right)}^2=3\times \frac{1}{9}=\frac{1}{3}$
Also,
$T=2\pi \sqrt{\frac{1}{g}}$
$\Rightarrow \frac{T_p}{T_e}=\sqrt{\frac{g_e}{g_p}}=\sqrt{3}$
$\Rightarrow T_p=\sqrt{3}{T}_e$

Time period at earth for seconds pendulum $=2s$
$\therefore T_p=2\sqrt{3}s$