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 secor,d's pendulum on earth)

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

Answer 1

Answer: (B)

$2\sqrt{2}s$

Answer 2

Gravity, g=\frac{GM}{R^2}\hspace15mm (G is constant) $\therefore\, \, \, \, \, \frac{g_{earth}}{g_{planet}}=\frac{M_e}{M_p}\times\frac{R^2_p}{R^2_e}\Rightarrow\frac{g_e}{g_p}=\frac{2}{1}$ Also, $\, \, \, \, \, \, \, \, \, \, T ??frac{1}{\sqrt{g}}\Rightarrow\frac{T_e}{T_p}=\sqrt{\frac{g_p}{g_e}}$ $\Rightarrow \, \, \, \, \, \, \, \, \, \, \, \, \, \frac{2}{T_p}=\sqrt{\frac{1}{2}}$ $\, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, T_p=2\sqrt{2}s$