The mass and diameter of a planet are twice those of earth. The period of oscillation of pendulum on this planet will be (if it is a second’s pendulum on earth)

Options

A. `1//sqrt(2)` second
B. `2//sqrt(2)` second
C. `2sqrt(2)` seconds
D. `(1//2)` seconds

Explanation

To solve the problem, we need to determine the period of oscillation of a pendulum on a planet where the mass and diameter are both twice that of Earth. We will use the formula for the period of a simple pendulum and the relationship between gravitational acceleration and mass and radius of the planet.

### Step-by-Step Solution:

1. **Identify the parameters:**
- Let the mass of Earth be \( m_e \) and the radius of Earth be \( r_e \).
- The mass of the new planet \( m_p = 2m_e \) and the radius of the new planet \( r_p = 2r_e \).

2. **Calculate gravitational acceleration on the new planet:**
- The formula for gravitational acceleration \( g \) at the surface of a planet is given by:
\[ g_p = \frac{G m_p}{r_p^2} \]
- Substituting the values for \( m_p \) and \( r_p \):
\[ g_p = \frac{G (2m_e)}{(2r_e)^2} = \frac{G (2m_e)}{4r_e^2} = \frac{G m_e}{2r_e^2} \]
- We know that the gravitational acceleration on Earth is:
\[ g_e = \frac{G m_e}{r_e^2} \]
- Therefore, we can express \( g_p \) in terms of \( g_e \):
\[ g_p = \frac{g_e}{2} \]

3. **Use the formula for the period of a pendulum:**
- The period \( T \) of a simple pendulum is given by:
\[ T = 2\pi \sqrt{\frac{L}{g}} \]
- For a second's pendulum on Earth, the period \( T_e = 2 \) seconds, which means \( L \) (the length of the pendulum) is such that:
\[ 2 = 2\pi \sqrt{\frac{L}{g_e}} \]

4. **Calculate the period on the new planet:**
- The period on the new planet \( T_p \) will be:
\[ T_p = 2\pi \sqrt{\frac{L}{g_p}} = 2\pi \sqrt{\frac{L}{\frac{g_e}{2}}} = 2\pi \sqrt{\frac{2L}{g_e}} = \sqrt{2} \times 2\pi \sqrt{\frac{L}{g_e}} = \sqrt{2} \times T_e \]
- Substituting \( T_e = 2 \) seconds:
\[ T_p = \sqrt{2} \times 2 = 2\sqrt{2} \text{ seconds} \]

5. **Conclusion:**
- The period of oscillation of the pendulum on the new planet is \( 2\sqrt{2} \) seconds.

### Final Answer:
The period of oscillation of the pendulum on the new planet is \( 2\sqrt{2} \) seconds.