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 can follow these steps:

### Step-by-Step Solution:

1. **Understand the Formula for 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 \( g \) is approximately \( 9.81 \, \text{m/s}^2 \). On the Moon, the acceleration due to gravity \( g' \) is about \( \frac{1}{6} \) of that on Earth:
\[ g' = \frac{g}{6} = \frac{9.81}{6} \approx 1.635 \, \text{m/s}^2 \]

3. **Calculate the Time Period on Earth**:
Let the time period of the pendulum on Earth be \( T_E \):
\[ T_E = 2\pi \sqrt{\frac{L}{g}} \]

4. **Calculate the Time Period on the Moon**:
Let the time period of the pendulum on the Moon be \( T_M \):
\[ T_M = 2\pi \sqrt{\frac{L}{g'}} = 2\pi \sqrt{\frac{L}{\frac{g}{6}}} = 2\pi \sqrt{\frac{6L}{g}} \]

5. **Relate the Time Periods**:
Now, we can express \( T_M \) in terms of \( T_E \):
\[ T_M = 2\pi \sqrt{6} \sqrt{\frac{L}{g}} = \sqrt{6} \cdot T_E \]

6. **Conclusion**:
Since \( \sqrt{6} \) is greater than 1, we find that:
\[ T_M > T_E \]
This means that the time period of the pendulum on the Moon is greater than that on Earth, indicating that the pendulum will take longer to complete one oscillation. Therefore, the clock (or pendulum) will run slower on the Moon.

### Final Answer:
The simple pendulum taken to the Moon will run slower than it does on Earth.

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