Question

The simple pendulum acts as second's pendulum on earth. Its time on a planet, whose mass and diameter are twice that of earth is

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

Answer 1

Answer: (B)

$2\sqrt 2 s$

Answer 2

Second's pendulum is that simpie pendulum whose time period of vibration is two second. The bob of such pendulum while oscillating passes through the mean position after every one second. Now, Time period of simple penduium is given by $\, \, \, \, \, \, \, \, \, \, \, T=2\pi \sqrt{\bigg(\frac{l}{g}\bigg)}$ or $\, \, \, \, \, \, \, \, T \propto \frac{1}{\sqrt g} \, \, \, \, \, \, \, \, \, \, \, \,$ ... (i) but $\, \, \, \, \, \, \, \, \, g=\frac{GM}{R^2} \, \, \, \, \, \, \, \, \, \, \, \,$(on earth) and $\, \, \, \, \, \, \, \, \, \, g'=\frac{G(2m)}{4R^2} \, \, \, \, \, \, \, \, \, \, \, \,$(on planet) $\, \, \, \, \, \, \, \, \, \, =\frac{1}{2}\frac{GM}{R^2} =\frac{g}{2}$ E (i) gives $\, \, \, \, \, \, \, \frac{T'}{T} =\frac{\sqrt g}{\sqrt g'} =\sqrt 2$ or $\, \, \, \, \, \, \, T' =\sqrt 2 T$ $\, \, \, \, \, \, \, \, \, \, \, \, \, T' =\sqrt 2T$ $\, \, \, \, \, \, \, \, \, \, \, \, \, =\sqrt 2 \times 2 \, \, \, \, \, \, \, \, \, \, \, (T =2 s)$ or $\, \, \, \, \, \, \, \, \, \, \, \, = 2\sqrt 2 s$