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 $2 \, s$ , 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,
$g=\frac{G M}{R^{2}}$
$\Rightarrow \frac{g_{p}}{g_{e}}=\frac{M_{p}}{M_{e}}\times \left(\frac{R_{e}}{R_{p}}\right)^{2}=3\times \frac{1}{9}=\frac{1}{3}$
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 $=2s$
$\therefore T_{p}=2\sqrt{3}s$