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

Hint: 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.