Question

What would be the length of a seconds pendulum at a planet (where acceleration due to gravity is ${\displaystyle \frac{g}{4}}$) if its length on earth is $l$.
A. ${\displaystyle \frac{l}{2}}$
B. $2l$
C. ${\displaystyle \frac{l}{4}}$
D. $4l$

Answer 1

Hint:A seconds pendulum is a simple pendulum whose time period is exactly equal to 2 seconds. The time period of a simple pendulum is given as $T=2\pi \sqrt{{\displaystyle \frac{l}{g}}}$, where l is the length of the pendulum and g is acceleration due to gravity.

Formula used:
$T=2\pi \sqrt{{\displaystyle \frac{l}{g}}}$
Here, T is the time period, l is the length of the pendulum and g is acceleration due to gravity.

Complete step by step answer:
A seconds pendulum is a simple pendulum whose time period is exactly equal to 2 seconds. It takes one second to swing in one direction and another one second to swing in the opposite direction. This way it completes one full oscillation in two seconds.The time period of a simple pendulum is given as $T=2\pi \sqrt{{\displaystyle \frac{l}{g}}}$, where l is the length of the pendulum and g is acceleration due to gravity.

On earth the acceleration due to gravity is g and the length of the pendulum is said to be l. Also, we know that the time period of a seconds pendulum is 2 seconds. Therefore,
$T=2\pi \sqrt{{\displaystyle \frac{l}{g}}}=2s$ …. (i).
The time period of the seconds pendulum on the other must be 2 seconds also. It is given that the acceleration due to the gravity of this planet is ${\displaystyle \frac{g}{4}}$. Let its length be l’.
Then we write its time period as $T=2s=2\pi \sqrt{{\displaystyle \frac{l^{\prime }}{{\displaystyle \frac{g}{4}}}}}$
$2\pi \sqrt{{\displaystyle \frac{4l^{\prime }}{g}}}=2$ …. (ii).
Now, divide (ii) by (i).
${\displaystyle \frac{2\pi \sqrt{{\displaystyle \frac{4l^{\prime }}{g}}}}{2\pi \sqrt{{\displaystyle \frac{l}{g}}}=2}}={\displaystyle \frac{2}{2}}$
$\Rightarrow \sqrt{{\displaystyle \frac{4l^{\prime }}{l}}}=1$
$\Rightarrow {\displaystyle \frac{4l^{\prime }}{l}}=1$
$\therefore l^{\prime }={\displaystyle \frac{l}{4}}$.
This means that the required length of the seconds pendulum on the other planet is equal to ${\displaystyle \frac{l}{4}}$.

Hence, the correct option is C.

Note:From the formula for the time period of a simple pendulum, we understand that the time period of a pendulum only depends on its length and the acceleration due to gravity at its location. It is interesting to know that the time period of a pendulum is independent of its mass.