Question

The length of a second pendulum on earth is $l$. The length of a seconds pendulum on a planet the mass and radius of which are twice those of the earth will be:
A.$l$
B. $\dfrac{l}{\sqrt 2}$
C. $\dfrac{l}{2}$
D. $\sqrt{2l}$

Answer 1

We know that the time period of the pendulum depends on the length of the pendulum and the acceleration due to gravity acting on it. Here, since the dimensions of the planet is changing, we can say that the acceleration due to gravity and hence the time period of the pendulum in the new planet varies.
Formula used:
$g=\dfrac{GM}{R^{2}}$
$t=2\pi\sqrt{\dfrac{l}{g}}$

Complete step by step answer:
We know that the time period $t$ of the pendulum is directly proportional to the square root of the length $l$ of the pendulum and is inversely proportional to the square roof of the acceleration due to gravity $g$.
$t\propto \sqrt{l}, t\propto\sqrt{\dfrac{1}{g}}$
$\implies t=2\pi\sqrt{\dfrac{l}{g}}$
Then, let $l$ be the length of the pendulum on earth due to acceleration due to gravity $g$ Also, let the mass of earth be $m$ and radius be $r$. And let $L$ be the length of the pendulum on the other planet, with mass $2\;m$ and radius $2\;r$, then the acceleration due to gravity $g\prime$ is given as
$g\prime=\dfrac{G\times 2m}{(2r)^{2}}$
Similarly, $g=\dfrac{G\times m}{r^{2}}$, where $G$ is the universal gravitational constant.
Since, the time period of a seconds pendulum is $2\;s$ which is a constant, irrespective of the place where it is measured, we can equate the time period of the two cases, then we have
$2\pi\sqrt{\dfrac{l}{g}}=2\pi\sqrt{\dfrac{L}{g\prime}}$
$2\pi\sqrt{\dfrac{l}{\dfrac{G\times m}{r^{2}}}}=2\pi\sqrt{\dfrac{L}{\dfrac{G\times 2m}{(2r)^{2}}}}$
$\dfrac{l}{L}=\dfrac{\dfrac{G\times m}{r^{2}}}{\dfrac{G\times 2m}{(2r)^{2}}}$
$\dfrac{l}{L}=2$
$\implies L=\dfrac{l}{2}$

So, the correct answer is “Option C”.

Note: We know that a pendulum is a small bob, which swings freely by a lightweight string. We also know that the pendulum undergoes simple harmonic motion which is also called the to and fro motion. A seconds pendulum is a pendulum whose time period is $2\;s$, that is it takes exactly $1\;s$ in one direction and takes another $1\;s$ to come back to its mean position.