Question

The mass and diameter of a planet are twice those of earth. The period of oscillation of pendulum on this planet will be (If it is a second's pendulum on earth)

• A. $\frac{1}{\sqrt{2}}$ sec
• B. $2\sqrt{2}$sec
• C. 2 sec
• D. $\frac{1}{2}$sec

Answer 1

Answer: (B)

$2\sqrt{2}$sec

Answer 2

As we know $g = \frac{GM}{R^{2}}$

$\Rightarrow$ $\frac{g_{\text{earth}}}{g_{\text{planet}}} = \frac{M_{e}}{M_{p}} \times \frac{R_{\rho}^{2}}{R_{e}^{2}} \Rightarrow \frac{g_{e}}{g_{p}} = \frac{2}{1}$

Also$T \propto \frac{1}{\sqrt{g}} \Rightarrow \frac{T_{e}}{T_{p}} = \sqrt{\frac{g_{p}}{g_{e}}} \Rightarrow \frac{2}{T_{p}} = \sqrt{\frac{1}{2}}$

$\Rightarrow$ $T_{p} = 2\sqrt{2}\sec$.