Q: The mass and diameter of a planet are twice those of earth . What will be the period of oscillation of a pendulum on this planet if it is a seconds pendulum on the earth ?
(a) $\sqrt{2} sec$
(b) $2 \sqrt{2} sec$
(c) $\frac{1}{\sqrt{2}} sec$
(d) $\frac{1}{2 \sqrt{2}} sec$
Solution : Te = 2 sec (Having seconds pendulum)
$\displaystyle T = 2 \pi \sqrt{\frac{l}{g}}$
$\displaystyle g = \frac{G M}{R^2} $
$\displaystyle T = 2 \pi \sqrt{\frac{l}{\frac{G M}{R^2} }}$
$\displaystyle T = 2 \pi \sqrt{\frac{l R^2}{G M}}$
$\displaystyle \frac{T_p}{T_e} = \sqrt{\frac{R_p^2}{R_e^2} \times \frac{M_e}{M_p}}$
$\displaystyle \frac{T_p}{T_e} = \sqrt{\frac{4 R_e^2}{R_e^2} \times \frac{M_e}{2 M_e}}$
$\displaystyle \frac{T_p}{T_e} = \sqrt{2} $
$\displaystyle T_p = \sqrt{2} T_e$
$\displaystyle T_p = \sqrt{2} \times 2$
$\displaystyle T_p = 2 \sqrt{2} sec $
Correct option is (b)