Question

A clock S is based on oscillation of a spring and a clock P is based on pendulum motion. Both clocks run at the same rate on earth. On a planet having the same density as earth, twice the radius –
A) S will run faster than P
B) P will run faster than S
C) They will both run at the same rates as on the earth
D) They will both run at the equal rates, but not the same as on the earth.

Answer 1

We need to understand the relation between the planet’s dimensions and the physical parameters which can change with the change in these dimensions. We can use this idea on the acceleration due to gravity to solve this problem very easily.

Complete answer:
We know that the spring oscillations and the pendulum oscillations possess different properties which governs their time periods of oscillation. The time period of oscillation of the spring is given by the dependence of mass of the spring (not weight) and the spring constant as –
${{T}_{s}}=2\pi \sqrt{\dfrac{m}{k}}$
The time period of oscillation of a pendulum is given by the length of the pendulum and the acceleration due to gravity as –
${{T}_{p}}=2\pi \sqrt{\dfrac{l}{g}}$
Now, we understand that the time period of the pendulum is dependent on the acceleration due to the gravitational force on a particular planet. If we consider ‘g’ as the acceleration due to gravity which is given as –
$g=\dfrac{GM}{{{R}^{2}}}$
The acceleration due to gravity on a planet which has the radius twice as earth and same mass as that of earth can be given as –

& g'=\dfrac{GM}{{{(2R)}^{2}}} \\\ & \Rightarrow g'=\dfrac{GM}{4{{R}^{2}}} \\\ & \therefore g'=\dfrac{g}{4} \\\ \end{aligned}$$ Now, we can calculate the time period of the pendulum oscillation on this new planet as – $$\begin{aligned} & {{T}_{p}}'=2\pi \sqrt{\dfrac{l}{g'}} \\\ & \Rightarrow {{T}_{p}}'=2\pi \sqrt{\dfrac{4l}{g}} \\\ & \therefore {{T}_{p}}'=2{{T}_{p}} \\\ \end{aligned}$$ We understand that the time period of the pendulum clock is twice from the initial condition. The clock P moves half as fast as when it is earth, whereas the clock S runs at the same speed as on earth. Therefore, the S will run faster on the new planet. **The correct answer is option A.** **Note:** The time period of oscillation and the rate of running the clock shouldn’t be confused. They are inversely proportional as the faster the clock runs the shorter the time period of oscillation as we can clearly see in the case of pendulum in this example.