A clock `S` is based on oscillations of a spring and clock `P` is based on pendulum motion, both clocks run at the same rate on Earth. On a planet having the same mass, but twice the radius that of the earth

Options

A. S will run faster than P
B. P will run faster than S
C. Both run at same rate
D. Both run at same rate but different than earth

Answer

To solve the problem, we need to analyze the time periods of both clocks based on their respective mechanisms and how they change when moving to a new planet with the same mass but twice the radius of Earth.

### Step-by-Step Solution:

1. **Identify the Time Period of the Spring Clock (S)**:
The time period \( T_s \) of a spring oscillator is given by the formula:
\[ T_s = 2\pi \sqrt{\frac{m}{k}} \]
where \( m \) is the mass and \( k \) is the spring constant.

2. **Identify the Time Period of the Pendulum Clock (P)**:
The time period \( T_p \) of a simple pendulum is given by:
\[ T_p = 2\pi \sqrt{\frac{l}{g}} \]
where \( l \) is the length of the pendulum and \( g \) is the acceleration due to gravity.

3. **Determine the Acceleration Due to Gravity on the New Planet**:
The acceleration due to gravity \( g \) on a planet is given by:
\[ g = \frac{GM}{R^2} \]
where \( G \) is the gravitational constant, \( M \) is the mass of the planet, and \( R \) is its radius. For the new planet with the same mass as Earth but twice the radius, we have:
\[ g' = \frac{GM}{(2R)^2} = \frac{GM}{4R^2} = \frac{g}{4} \]
Hence, the new gravity \( g' \) is half of that on Earth.

4. **Calculate the New Time Period for the Pendulum Clock**:
Substituting \( g' \) into the time period formula for the pendulum:
\[ T_p' = 2\pi \sqrt{\frac{l}{g'}} = 2\pi \sqrt{\frac{l}{\frac{g}{4}}} = 2\pi \sqrt{\frac{4l}{g}} = 2\sqrt{2}\pi \sqrt{\frac{l}{g}} = \sqrt{2} T_p \]
This shows that the time period of the pendulum clock increases by a factor of \( \sqrt{2} \).

5. **Determine the Frequency of Each Clock**:
The frequency \( f \) is the reciprocal of the time period \( T \):
\[ f_s = \frac{1}{T_s} \quad \text{and} \quad f_p = \frac{1}{T_p} \]
For the new pendulum clock:
\[ f_p' = \frac{1}{T_p'} = \frac{1}{\sqrt{2} T_p} = \frac{f_p}{\sqrt{2}} \]
The frequency of the spring clock remains unchanged as it is independent of gravity:
\[ f_s' = f_s \]

6. **Comparison of Frequencies**:
Since \( f_p' = \frac{f_p}{\sqrt{2}} \) and \( f_s' = f_s \), we need to compare the frequencies:
- On Earth, both clocks run at the same rate, so \( f_s = f_p \).
- On the new planet, \( f_p' < f_s' \).

7. **Conclusion**:
Therefore, the spring clock \( S \) will run faster than the pendulum clock \( P \) on the new planet.

### Final Answer:
The spring clock \( S \) will run faster than the pendulum clock \( P \) on the new planet.