Question

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 earth?

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

Answer 1

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

Let,
The period of oscillation of a pendulum on Earth is
$T=2\pi \sqrt{\frac{l}{g}}$
where, $l$ is length of bob from rigid support and $g$ is the acceleration due to gravity on Earth.
And the period of oscillation of a pendulum on planet is
$T^{{}^{\prime }}=2\pi \sqrt{\frac{l}{g^{\prime }}}$
where, $l$ is length of bob from rigid support and $g^{{}^{\prime }}$ is the acceleration due to gravity on planet.
Hence,
$\frac{T^{{}^{\prime }}}{T}=\frac{2\pi \sqrt{\frac{l}{g^{\prime }}}}{2\pi \sqrt{\frac{l}{g}}}$
$\frac{T^{\prime }}{T}=\frac{g}{g^{{}^{\prime }}}$ .......................................$\left(1\right)$
Now,
$g=\frac{GM}{R^2}$ and $g^{{}^{\prime }}=\frac{G\left(2M\right)}{(2R{)}^2}$ ........................$($given$)$
where,$G$ is gravitational constant, $M$ is mass of Earth and $R$ is radius of Earth.
Therefore,
$\frac{g}{g^{{}^{\prime }}}=\frac{\frac{GM}{R^2}}{\frac{G\left(2M\right)}{(2R{)}^2}}$
$\frac{g}{g^{{}^{\prime }}}=\frac{GM}{R^2}\times \frac{(2R{)}^2}{G\left(2M\right)}$
$\frac{g}{g^{\prime }}=2$
$\Rightarrow \frac{T^{\prime }}{T}=\sqrt{2}$ ..............................................from $\left(1\right)$
$\therefore T^{{}^{\prime }}=T\sqrt{2}$
Since, it is the second's pendulum on Earth, it's time period is of 2s.
$\therefore T^{{}^{\prime }}=2\sqrt{2}second$