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. $2s$
• D. $\frac{1}{\sqrt{2}}s$

Answer 1

The correct option is B $2\sqrt{2}s$

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