Question

The length of a seconds pendulum on the surface of the earth is 1m. Its length on the surface of the moon is:
$$\begin{array}{rl}& A.{\displaystyle \frac{1}{\sqrt{6}}}m\\ & B.{\displaystyle \frac{1}{6}}m\\ & C.{\displaystyle \frac{1}{36}}m\\ & D.36m\end{array}$$

Answer 1

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

Formula used: Time period of a simple pendulum $T=2\pi \sqrt{{\displaystyle \frac{L}{g}}}$
$g_m={\displaystyle \frac{1}{6}}\times g$
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{{\displaystyle \frac{L}{g}}}$,where $L$ is length of the simple pendulum and $g$ is gravity.
$g_m={\displaystyle \frac{1}{6}}\times g$ 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:
$2\pi \sqrt{{\displaystyle \frac{L_e}{g_e}}}=2\pi \sqrt{{\displaystyle \frac{L_m}{g_m}}}$

${\displaystyle \frac{L_e}{g_e}}={\displaystyle \frac{L_m}{g_m}}$
We know that $g_m={\displaystyle \frac{1}{6}}\times g_e$ and given that $L_e=1m$
${\displaystyle \frac{1}{g_e}}={\displaystyle \frac{L_m}{{\displaystyle \frac{1}{6}}\times g_e}}$
$L_m={\displaystyle \frac{1}{6}}m$
Hence the length of the seconds’ pendulum on the moon is B. ${\displaystyle \frac{1}{6}}m$

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 $-mg\sin \theta L=m{L}^2{\displaystyle \frac{d^2\theta }{d{t}^2}}$.
On simplification we get ${\displaystyle \frac{d^2\theta }{d{t}^2}}+{\displaystyle \frac{g}{L}}\theta =0$
On solving the above we get the solutions of the SHM as $\theta (t)={\theta }_o\cos (\omega t)$, where $\omega =\sqrt{{\displaystyle \frac{g}{L}}}$ is the frequency of the motion.
Then the time period $T$ is given by
$T={\displaystyle \frac{2\pi }{\omega }}=2\pi \sqrt{{\displaystyle \frac{L}{g}}}$

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 $g_m={\displaystyle \frac{1}{6}}\times g$