A pendulum having a period of oscillation of 2 seconds is taken on a planet where g is four times that on the earth. The period of the pendulum would be

Options

A. 2s
B. 1 s
C. 4 s
D. `2sqrt(2)` s

Answer

To solve the problem, we need to determine the period of a pendulum on a planet where the acceleration due to gravity is four times that of Earth. 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-by-Step Solution:

1. **Identify the given data:**
- The period of the pendulum on Earth, \( T_e = 2 \) seconds.
- The acceleration due to gravity on Earth, \( g_e = 9.8 \, \text{m/s}^2 \).
- The acceleration due to gravity on the new planet, \( g_p = 4g_e \).

2. **Calculate the effective gravity on the planet:**
\[ g_p = 4g_e = 4 \times 9.8 = 39.2 \, \text{m/s}^2 \]

3. **Use the period formula for the pendulum on Earth:**
\[ T_e = 2\pi \sqrt{\frac{L}{g_e}} \]
Rearranging this to find \( L \):
\[ L = \frac{T_e^2 g_e}{4\pi^2} \]
Substituting \( T_e = 2 \) seconds and \( g_e = 9.8 \, \text{m/s}^2 \):
\[ L = \frac{(2)^2 \times 9.8}{4\pi^2} = \frac{4 \times 9.8}{4\pi^2} = \frac{9.8}{\pi^2} \]

4. **Calculate the period of the pendulum on the planet:**
Using the same formula but substituting \( g_p \):
\[ T_p = 2\pi \sqrt{\frac{L}{g_p}} = 2\pi \sqrt{\frac{\frac{9.8}{\pi^2}}{39.2}} \]
Simplifying this:
\[ T_p = 2\pi \sqrt{\frac{9.8}{39.2\pi^2}} = 2\pi \sqrt{\frac{1}{4}} = 2\pi \times \frac{1}{2} = \pi \]

5. **Convert \( \pi \) to seconds:**
Since \( \pi \) is approximately 3.14, we can conclude:
\[ T_p \approx 1 \, \text{second} \]

### Final Answer:
The period of the pendulum on the planet would be **1 second**.

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