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 seconds pendulum on earth)

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

Answer 1

Answer: (B)

$2 \sqrt{2} s$

Answer 2

Gravity, $g=\frac{G M}{R^{2}}$ $\therefore \frac{g_{\text {earth }}}{g_{\text {planet }}}=\frac{M_{e}}{M_{P}} \times \frac{R_{P}^{2}}{R_{e}^{2}}$ $\Rightarrow \frac{g_{e}}{g_{p}} =\frac{2}{1}$ $\Rightarrow T \propto \frac{1}{\sqrt{g}}$ $\Rightarrow \frac{T_{e}}{T_{p}}=\sqrt{\frac{g_{p}}{g_{e}}}$ Also, $\frac{2}{T_{p}} =\sqrt{\frac{1}{2}}$ $\Rightarrow T_{p} =2 \sqrt{2} s$ $\therefore \frac{g_{\text {earth }}}{g_{\text {planet }}}=\frac{M_{e}}{M_{P}} \times \frac{R_{P}^{2}}{R_{e}^{2}}$ $\Rightarrow \frac{g_{e}}{g_{p}}=\frac{2}{1}$ Also, $T \propto \frac{1}{\sqrt{g}}$ $\Rightarrow \frac{T_{e}}{T_{p}}=\sqrt{\frac{g_{p}}{g_{e}}}$ $\Rightarrow \frac{2}{T_{p}}=\sqrt{\frac{1}{2}}$ $T_{p}=2 \sqrt{2} s$