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The mass and diameter of a planet are twice those of earth. The period of oscillation of a pendulum on this planet will be? (If the pendulum is a second’s pendulum on earth)
A) $\dfrac{1}{{\sqrt 2 }}s$
B) $2\sqrt 2 s$
C) $2s$
D) $\dfrac{1}{2}s$

Last updated date: 13th Jun 2024
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Hint:Acceleration due to gravity is dependent on mass and radius of a planet. We will use its definition to calculate the acceleration due to gravity for the planet which has double mass and diameter. Now we will use the acceleration due to gravity on the planet to calculate the period of oscillation for the pendulum.

Complete step by step solution:
In the simple harmonic motion of a pendulum, the time period of a single oscillation is given by the following expression.
$T = 2\pi \sqrt {\dfrac{L}{g}} $ ……….(1)
Where $L$ is the length of the pendulum
$g$ is the acceleration due to gravity.
From the above equation it is clear that the time period of a pendulum is independent of mass of the pendulum.
We will look at the above equation later.
Acceleration due to gravity is defined as the acceleration at which the planet is pulling another object or in simple terms, the acceleration generated from the force of gravity.
It is given by the formula,
$g = \dfrac{{GM}}{{{r^2}}}$ ………. (2)
Where, $G$ is the gravitational constant
$M$ is the mass of the planet
$r$ is the radius of the planet
For earth it is written as ${g_e} = \dfrac{{G{M_e}}}{{{r_e}^2}}$ where the subscripted terms represent the respective terms for earth.
The planet of our concern has a double mass and diameter from earth.
Mathematically we can write this as
${M_p} = 2{M_e}$
${r_p} = 2{r_e}$
Here the subscript ‘p’ means the term represents the values for the planet of concern.
So the value of acceleration due to gravity for this planet is,
${g_p} = \dfrac{{G{M_p}}}{{{r_p}^2}}$
Substituting the values we get,
$ \Rightarrow {g_p} = \dfrac{{G \times 2{M_e}}}{{{{(2{r_e})}^2}}}$
Or we can also write,
$ \Rightarrow {g_p} = \dfrac{1}{2}{g_e}$ …… (3)
Now using this value of acceleration due to gravity we can estimate the value of the time period of a second’s pendulum.
On earth the time period of the pendulum is given by,
${T_e} = 2\pi \sqrt {\dfrac{L}{{{g_e}}}} $ ………. (4)
As the pendulums in each planet are similar, we can say that the length of the pendulums will also be equal.
So the time period of second’s pendulum on the planet will be,
${T_p} = 2\pi \sqrt {\dfrac{L}{{{g_p}}}} $
Substituting the value of ${g_p}$ in the above equation, we get,
${T_p} = 2\pi \sqrt {\dfrac{{2L}}{{{g_e}}}} $
$ \Rightarrow {T_p} = \sqrt 2 \times 2\pi \sqrt {\dfrac{L}{{{g_e}}}} $
From equation (4),
$ \Rightarrow {T_p} = \sqrt 2 \times {T_e}$
Time period of a second’s pendulum on earth is $2s$, so the time period of the same on the planet would be,
$ \Rightarrow {T_p} = \sqrt 2 \times 2 = 2\sqrt 2 s$

So the correct answer is option (B).

Note:These types of questions are very common in many examinations and a student requires to understand how each definition in mechanics works. Another thing to keep in mind, sometimes the examiner might ask whether the time period depends on the following factors or not and chances are there will be an option mentioning ‘mass’. From the very formula of time-period for oscillation in SHM we know that $T$ is independent of mass of the pendulum or its velocity.