The length of a second's pendulum on the surface of Earth in 1m. What will be the length of a second's pendulum on the moon?

Options

A. `(1)/(6)`m
B. 1m
C. `(1)/(36)m`
D. 36 m

Explanation

To find the length of a second's pendulum on the Moon, we can follow these steps:

### Step 1: Understand the time period of a second's pendulum
The time period \( T \) of a second's pendulum is 2 seconds. The formula for the time period of a pendulum is given by:
\[ T = 2\pi \sqrt{\frac{L}{g}} \]
where \( L \) is the length of the pendulum and \( g \) is the acceleration due to gravity.

### Step 2: Calculate the acceleration due to gravity on Earth
Given that the length of the pendulum on Earth \( L = 1 \, \text{m} \) and \( g \) on Earth is approximately \( 9.81 \, \text{m/s}^2 \), we can rearrange the formula to find \( g \):
\[ T = 2 \, \text{s} = 2\pi \sqrt{\frac{1}{g}} \]
Squaring both sides gives:
\[ 4 = 4\pi^2 \frac{1}{g} \]
Rearranging this, we find:
\[ g = \pi^2 \, \text{m/s}^2 \]

### Step 3: Determine the acceleration due to gravity on the Moon
The acceleration due to gravity on the Moon \( g_{\text{moon}} \) is approximately \( \frac{1}{6} \) of that on Earth:
\[ g_{\text{moon}} = \frac{g}{6} = \frac{\pi^2}{6} \, \text{m/s}^2 \]

### Step 4: Set up the equation for the time period on the Moon
Using the same formula for the time period on the Moon:
\[ T = 2\pi \sqrt{\frac{L_m}{g_{\text{moon}}}} \]
Substituting \( T = 2 \, \text{s} \) and \( g_{\text{moon}} = \frac{\pi^2}{6} \):
\[ 2 = 2\pi \sqrt{\frac{L_m}{\frac{\pi^2}{6}}} \]

### Step 5: Simplify and solve for \( L_m \)
Squaring both sides:
\[ 4 = 4\pi^2 \frac{L_m}{\frac{\pi^2}{6}} \]
This simplifies to:
\[ 4 = 24 L_m \]
Thus:
\[ L_m = \frac{4}{24} = \frac{1}{6} \, \text{m} \]

### Conclusion
The length of a second's pendulum on the Moon is \( \frac{1}{6} \, \text{m} \).

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