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 second pendulum on earth?
A. $\sqrt{2}$seconds
B. $2\sqrt{2}$ seconds
C. $\dfrac{1}{\,\sqrt{2}}$seconds
D. $\dfrac{1}{2\sqrt{2}}$seconds

Answer 1

In this question we are asked to calculate the time period of a pendulum under given conditions. We know that Time period of a pendulum is inversely proportional to the square root of the acceleration due to gravity of a planet therefore, we will be calculating the acceleration due to gravity for a given planet.

Formula used: $g=\dfrac{GM}{{{r}^{2}}}$
Where,
g is acceleration due to gravity
G is the universal gravitational constant
M is the mass of planet
R is the radius of planet
$T=2\pi \sqrt{\dfrac{l}{g}}$
Where,
L is the length of pendulum
T is the time period

Complete step by step answer:
Let us write the given conditions. It is said that the planet has radius and mass twice the size of earth. Therefore, let ${{M}_{p}}$be the mass of the planet and ${{M}_{e}}$ be the mass of earth. Also, let ${{r}_{p}}$ be the radius of the planet and ${{r}_{e}}$ be the radius of earth.
Therefore, from given condition we can say that
${{M}_{p}}=2{{M}_{e}}$ and ${{r}_{p}}=2{{r}_{e}}$ ……………. (1)
Now, we know that acceleration due to gravity is given by
$g=\dfrac{GM}{{{r}^{2}}}$
Therefore, acceleration due to gravity of earth will be given by
${{g}_{e}}=\dfrac{G{{M}_{e}}}{{{r}_{e}}^{2}}$ …………………. (2)
Similarly, acceleration due to gravity for the planet will be given by
${{g}_{p}}=\dfrac{G{{M}_{p}}}{{{r}_{p}}^{2}}$
Now, from (1) and (2) we can say that
${{g}_{p}}=\dfrac{G\times 2{{M}_{e}}}{4{{r}_{e}}^{2}}$
Therefore,
${{g}_{p}}=\dfrac{{{g}_{e}}}{2}$ ……………… (3)
Now, for the time period of pendulum we know that
$T=2\pi \sqrt{\dfrac{l}{g}}$
Therefore, the time period of pendulum on earth will be given by,
${{T}_{e}}=2\pi \sqrt{\dfrac{l}{{{g}_{e}}}}$
Similarly, time period of the pendulum on planet will be given by
${{T}_{p}}=2\pi \sqrt{\dfrac{l}{{{g}_{p}}}}$
From (3) we can say that
${{T}_{p}}=2\pi \sqrt{\dfrac{2l}{{{g}_{e}}}}$
Therefore,
${{T}_{p}}=\sqrt{2}{{T}_{e}}$ …………….. (3)
Now, we know that time period of a seconds pendulum is 2 seconds
Therefore,
${{T}_{e}}=2$ ………………… (4)
Therefore, from (3) and (4) we can say that
${{T}_{p}}=2\sqrt{2}$ seconds
Therefore, the time period of pendulum on planet of given conditions is $2\sqrt{2}$ seconds

So, the correct answer is “Option B”.

Note: A second pendulum is the type of pendulum whose time period of oscillation is precisely 2 seconds. The first second is for swing in positive direction and other for the return swing. Time period is the time taken by a particle to complete one whole vibration. In the case of a pendulum it is the time required for a pendulum to complete both swings.