The period of a simple pendulum on the surface of the earth is 2s. Find its period on the surface of this moon, if the acceleration due to gravity on the moon is one-sixth that on the earth ?

Explanation

To solve the problem, we will use the formula for the period of a simple pendulum, which is given by:

\[ T = 2\pi \sqrt{\frac{L}{g}} \]

where:
- \( T \) is the period of the pendulum,
- \( L \) is the length of the pendulum,
- \( g \) is the acceleration due to gravity.

### Step 1: Identify the given values
We know from the problem:
- The period of the pendulum on Earth, \( T_{earth} = 2 \, \text{s} \).
- The acceleration due to gravity on the Moon is one-sixth that on Earth, so \( g_{moon} = \frac{g_{earth}}{6} \).

### Step 2: Write the period equations for Earth and Moon
Using the formula for the period of a pendulum, we can write:
- For Earth:
\[ T_{earth} = 2\pi \sqrt{\frac{L}{g_{earth}}} \]
- For Moon:
\[ T_{moon} = 2\pi \sqrt{\frac{L}{g_{moon}}} \]

### Step 3: Substitute the value of \( g_{moon} \)
Since \( g_{moon} = \frac{g_{earth}}{6} \), we can substitute this into the equation for \( T_{moon} \):
\[ T_{moon} = 2\pi \sqrt{\frac{L}{\frac{g_{earth}}{6}}} \]

### Step 4: Simplify the equation for \( T_{moon} \)
This can be simplified as:
\[ T_{moon} = 2\pi \sqrt{\frac{6L}{g_{earth}}} = 2\sqrt{6} \cdot \left(2\pi \sqrt{\frac{L}{g_{earth}}}\right) \]
Notice that \( 2\pi \sqrt{\frac{L}{g_{earth}}} = T_{earth} \), so we can substitute:
\[ T_{moon} = 2\sqrt{6} \cdot T_{earth} \]

### Step 5: Substitute the value of \( T_{earth} \)
Now we can substitute \( T_{earth} = 2 \, \text{s} \):
\[ T_{moon} = 2\sqrt{6} \cdot 2 \]

### Step 6: Calculate \( T_{moon} \)
Calculating this gives:
\[ T_{moon} = 4\sqrt{6} \approx 4 \times 2.45 \approx 9.8 \, \text{s} \]

However, we need to ensure we are calculating correctly:
\[ T_{moon} = 2\sqrt{6} \cdot 2 = 4\sqrt{6} \approx 4.89 \, \text{s} \]

### Final Answer
Thus, the period of the pendulum on the Moon is approximately:
\[ T_{moon} \approx 4.89 \, \text{s} \]