Question

The length of a seconds pendulum on the surface of the earth is 1m. Its length on the surface of the moon is:

& A.~\dfrac{1}{\sqrt{6}}m~ \\\ & B.\dfrac{1}{6}m \\\ & C.\dfrac{1}{36}m \\\ & D.36m \\\ \end{aligned}$$

Answer 1

- Hint: The time period of a second’s pendulum is $T=$ 2 seconds, the time period $T=2\pi \sqrt{\dfrac{L}{g}}$ where $L$ is length of the simple pendulum and $g$ is gravity. The gravity on the moon is about $\dfrac{1}{6}$ times the gravity of the earth.

Formula used: Time period of a simple pendulum $T=2\pi \sqrt{\dfrac{L}{g}}$
$g_{m}=\dfrac{1}{6}\times g$
Where, $g_{m}$ is the gravity on the moon and $g$ is the gravity on the earth.

Complete step-by-step solution
The time period of a second’s pendulum is $T=$2 seconds, the time period$T=2\pi \sqrt{\dfrac{L}{g}}$,where $L$ is length of the simple pendulum and $g$ is gravity.
$g_{m}=\dfrac{1}{6}\times g$ where,$g_{m}$is the gravity on the moon and $g$ is the gravity on the earth.
We know that the time period of a second’s pendulum is $T=$2 seconds, on both moon and earth.
We can compare the time period of both the systems as:
$2\pi \sqrt{\dfrac{L_{e}}{g_{e}}}=2\pi \sqrt {\dfrac{L_{m}}{g_{m}}}$

$\dfrac{L_{e}}{g_{e}}=\dfrac{L_{m}}{g_{m}}$
We know that $g_{m}=\dfrac{1}{6}\times g_{e}$ and given that $L_{e}=1m$
$\dfrac{1}{g_{e}}=\dfrac{L_{m}}{\dfrac{1}{6}\times g_{e} }$
$L_{m}=\dfrac{1}{6}m$
Hence the length of the seconds’ pendulum on the moon is B. $\dfrac{1}{6}m$

Additional Information:
A simple pendulum is a system which consists of a point-mass bob m hanging from a massless string of length l, from a fixed point. When the bob is displaced or given a small push, the pendulum undergoes an periodic to and fro motion, due to gravity it tries to restore to the a position in equilibrium. After a few oscillations, the energy of the system is lost and it comes to rest.
From Newton’s second law, we can write the equation of motion of the pendulum as $-mg \sin\theta L=mL^{2}\dfrac{d^{2}\theta}{dt^{2}}$.
On simplification we get $\dfrac{d^{2}\theta}{dt^{2}}+\dfrac{g}{L}\theta=0$
On solving the above we get the solutions of the SHM as $\theta(t)=\theta_{o}\cos(\omega t)$, where $\omega = \sqrt {\dfrac{g}{L}}$ is the frequency of the motion.
Then the time period $T$ is given by
$T=\dfrac{2\pi}{\omega}=2\pi \sqrt{\dfrac{L}{g}}$

Note: This might seem like a complex question but it can be solved easily, if the concept of SHM and the formulas are known. This question is asked frequently. Also remember that the time period of a second’s pendulum is $T=$ 2 seconds, and $g_{m}=\dfrac{1}{6}\times g$