The acceleration due to gravity on the surface of the moon is `1.7ms^(-2)`. What is the time perioid of a simple pendulum on the surface of the moon, if its time period on the surface of earth is `3.5s ?` Take `g=9.8ms^(-2)` on the surface of the earth.

Explanation

To find the time period of a simple pendulum on the surface of the moon, we can use the relationship between the time period of a pendulum and the acceleration due to gravity. The formula for the time period \( T \) of a simple pendulum is given by:

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

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

### Step 1: Write the time period formula for Earth and Moon

For Earth:
\[ T_e = 2\pi \sqrt{\frac{L}{g_e}} \]
For Moon:
\[ T_m = 2\pi \sqrt{\frac{L}{g_m}} \]

Where:
- \( g_e = 9.8 \, \text{m/s}^2 \) (acceleration due to gravity on Earth),
- \( g_m = 1.7 \, \text{m/s}^2 \) (acceleration due to gravity on Moon).

### Step 2: Set up the ratio of the time periods

Since the length \( L \) of the pendulum is the same on both Earth and Moon, we can set up the ratio of the time periods:

\[ \frac{T_e}{T_m} = \sqrt{\frac{g_m}{g_e}} \]

### Step 3: Substitute the known values

We know:
- \( T_e = 3.5 \, \text{s} \)
- \( g_m = 1.7 \, \text{m/s}^2 \)
- \( g_e = 9.8 \, \text{m/s}^2 \)

Substituting these values into the ratio:

\[ \frac{3.5}{T_m} = \sqrt{\frac{1.7}{9.8}} \]

### Step 4: Solve for \( T_m \)

First, calculate \( \sqrt{\frac{1.7}{9.8}} \):

\[ \sqrt{\frac{1.7}{9.8}} \approx \sqrt{0.173469} \approx 0.416 \]

Now, substitute this back into the equation:

\[ \frac{3.5}{T_m} = 0.416 \]

Cross-multiplying gives:

\[ 3.5 = 0.416 \cdot T_m \]

Now, solve for \( T_m \):

\[ T_m = \frac{3.5}{0.416} \approx 8.4 \, \text{s} \]

### Final Answer

The time period of the simple pendulum on the surface of the moon is approximately \( 8.4 \, \text{s} \).

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