The mass and diameter of a planet are twice those of earth. What will be the period of oscillation of a pendulum on this plenet. If is a second's pendulum on earth?

Options

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

Explanation

To solve the problem, we need to determine the period of oscillation of a pendulum on a planet that has twice the mass and diameter of Earth. We know that the period of a second's pendulum on Earth is 2 seconds.

### Step-by-Step Solution:

1. **Identify the known values**:
- The period of a second's pendulum on Earth, \( T_e = 2 \) seconds.
- Mass of the planet, \( M_p = 2 M_e \) (where \( M_e \) is the mass of Earth).
- Diameter of the planet, \( D_p = 2 D_e \) (where \( D_e \) is the diameter of Earth). Therefore, the radius of the planet is \( R_p = 2 R_e \) (where \( R_e \) is the radius of Earth).

2. **Use the formula for the period of a pendulum**:
The formula for the period of a simple 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.

3. **Determine the acceleration due to gravity on the planet**:
The acceleration due to gravity \( g \) is given by:
\[ g = \frac{GM}{R^2} \]
For Earth:
\[ g_e = \frac{G M_e}{R_e^2} \]
For the new planet:
\[ g_p = \frac{G M_p}{R_p^2} = \frac{G (2 M_e)}{(2 R_e)^2} = \frac{G (2 M_e)}{4 R_e^2} = \frac{1}{2} \frac{G M_e}{R_e^2} = \frac{1}{2} g_e \]

4. **Relate the periods of the pendulum on Earth and the planet**:
Since \( g_p = \frac{1}{2} g_e \), we can substitute this into the period formula for the pendulum on the planet:
\[ T_p = 2\pi \sqrt{\frac{L}{g_p}} = 2\pi \sqrt{\frac{L}{\frac{1}{2} g_e}} = 2\pi \sqrt{\frac{2L}{g_e}} = \sqrt{2} \cdot 2\pi \sqrt{\frac{L}{g_e}} = \sqrt{2} \cdot T_e \]
Since \( T_e = 2 \) seconds, we have:
\[ T_p = \sqrt{2} \cdot 2 \text{ seconds} = 2\sqrt{2} \text{ seconds} \]

5. **Final Result**:
The period of oscillation of a pendulum on this planet is \( 2\sqrt{2} \) seconds.