Question

The mass and the diameter of a planet are three times the respective values for the Earth. The period of oscillation of a simple pendulum on the Earth is 2 seconds. The period of oscillation of the same pendulum on the planet would be?
A. $\dfrac{2}{{\sqrt 3 }}s$
B. $2\sqrt 3 s$
C. $\dfrac{{\sqrt 3 }}{2}s$
D. $\dfrac{3}{2}s$

Answer 1

In this question, we need to determine the time period of oscillations of the pendulum on the other planet, which is three times heavier and bigger than the Earth. For this, we will use the relation between the gravitational force, mass and the distance of the object along with the time period.
Formula used:
$g = \dfrac{{GM}}{{{R^2}}}$
$T \propto \sqrt {\dfrac{1}{g}}$

Complete step by step answer:
The ratio of the product of the gravitational constant and the square of the distance between the objects results in the force of gravity between the two objects. Mathematically, $g = \dfrac{{GM}}{{{R^2}}}$ where, G is the gravitational constant, M is the mass of the body, R is the distance between the objects, and g is the force of gravity between the objects.
Let the mass of the Earth be M, and the radius of the Earth be R.
Then, according to the question, the mass of the planet will be “3M”, and the radius of the planet will be “3R”.
Now, substituting the values of the mass and the radius of the Earth and the planet in the formula $g = \dfrac{{GM}}{{{R^2}}}$ to determine the value of the force of gravity on the Earth and the planet respectively.
${g_e} = \dfrac{{G{M_e}}}{{{R_e}^2}} - - - - (i)$
Similarly,
${g_p} = \dfrac{{G{M_p}}}{{{R_p}^2}} \\\ = \dfrac{{G\left( {3{M_e}} \right)}}{{{{\left( {3{R_e}} \right)}^2}}} \\\ = \dfrac{{G{M_e}}}{{3{R_e}^2}} - - - - (ii) \\\$
Dividing the equation (i) and (ii), we get
$\dfrac{{{g_e}}}{{{g_p}}} = \dfrac{{\left( {\dfrac{{G{M_e}}}{{{R_e}^2}}} \right)}}{{\left( {\dfrac{{G{M_e}}}{{3{R_e}^2}}} \right)}} \\\ = 3 - - - - (iii) \\\$
Now, the time period of the oscillation of the pendulum is inversely proportional to the force of gravity. Mathematically, $T \propto \sqrt {\dfrac{1}{g}}$.
So, we can write $\dfrac{{{T_e}}}{{{T_p}}} = \sqrt {\dfrac{{{g_p}}}{{{g_e}}}}$
Substituting the value from the equation (iii) in the above equation, we get
$\dfrac{{{T_e}}}{{{T_p}}} = \sqrt {\dfrac{{{g_p}}}{{{g_e}}}} \\\ \Rightarrow \dfrac{2}{{{T_p}}} = \sqrt {\dfrac{1}{3}} \\\ \Rightarrow {T_p} = 2\sqrt 3 \\\$
Hence, the time period of oscillations of the pendulum on the planet is $2\sqrt 3$ seconds.

So, the correct answer is “Option B”.

Note:
It is interesting to note here that the time period of the oscillations will be more where the force of gravity is less and vice versa.
Moreover, the force of gravity depends on the mass of the object.