If a simple pendulum is taken on to the moon from the earth, then it

Options

A. Runs faster
B. Runs slower
C. Shows no change
D. Will not perform S.H.M.

Answer

To solve the problem of how a simple pendulum behaves when taken from Earth to the Moon, we need to analyze the relationship between the time period of the pendulum and the acceleration due to gravity on both celestial bodies.

### Step-by-Step Solution:

1. **Understand the Time Period of a Pendulum**:
The time period \( T \) of a simple pendulum is given by the formula:
\[ T = 2\pi \sqrt{\frac{L}{g}} \]
where \( L \) is the length of the pendulum and \( g \) is the acceleration due to gravity.

2. **Identify the Values of \( g \)**:
- On Earth, the acceleration due to gravity \( g_{Earth} \) is approximately \( 9.81 \, \text{m/s}^2 \).
- On the Moon, the acceleration due to gravity \( g_{Moon} \) is about \( 1.62 \, \text{m/s}^2 \).

3. **Compare the Time Periods**:
Since the length \( L \) of the pendulum remains constant when moving from Earth to the Moon, we can compare the time periods on both planets:
- Time period on Earth:
\[ T_{Earth} = 2\pi \sqrt{\frac{L}{g_{Earth}}} \]
- Time period on the Moon:
\[ T_{Moon} = 2\pi \sqrt{\frac{L}{g_{Moon}}} \]

4. **Analyze the Relationship**:
Since \( g_{Moon} < g_{Earth} \), it follows that:
\[ \sqrt{\frac{L}{g_{Moon}}} > \sqrt{\frac{L}{g_{Earth}}} \]
Therefore, \( T_{Moon} > T_{Earth} \). This means the time period of the pendulum on the Moon is greater than that on Earth.

5. **Conclusion**:
A longer time period indicates that the pendulum takes more time to complete one oscillation. Thus, the pendulum will run slower on the Moon compared to its oscillation on Earth.

### Final Answer:
The correct option is that the pendulum will **run slower** on the Moon.

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