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, given that the pendulum is a second's pendulum on Earth (which has a period of 2 seconds).

### Step-by-Step Solution:

1. **Understanding the Pendulum Period Formula**:
The period \( T \) of a simple pendulum is given by the formula:
\[ T = 2\pi \sqrt{\frac{L}{g}} \]
where \( L \) is the length of the pendulum and \( g \) is the acceleration due to gravity.

2. **Acceleration due to Gravity**:
The acceleration due to gravity \( g \) on a planet is given by:
\[ g = \frac{GM}{R^2} \]
where \( G \) is the gravitational constant, \( M \) is the mass of the planet, and \( R \) is the radius of the planet.

3. **Given Conditions**:
- The mass of the planet \( M_p = 2M_e \) (twice the mass of Earth).
- The diameter of the planet is twice that of Earth, which means the radius \( R_p = 2R_e \).

4. **Calculating \( g_p \)**:
Substituting the values into the formula for \( g \):
\[ g_p = \frac{G(2M_e)}{(2R_e)^2} = \frac{2GM_e}{4R_e^2} = \frac{1}{2} \frac{GM_e}{R_e^2} = \frac{1}{2} g_e \]
Thus, the acceleration due to gravity on the planet is half that of Earth.

5. **Finding the Period on the New Planet**:
Since the pendulum is a second's pendulum on Earth, we know:
\[ T_e = 2 \text{ seconds} \]
The period of the pendulum on the new planet can be expressed as:
\[ T_p = 2\pi \sqrt{\frac{L}{g_p}} \]
Since \( g_p = \frac{1}{2} g_e \), we can substitute this into the equation:
\[ T_p = 2\pi \sqrt{\frac{L}{\frac{1}{2} g_e}} = 2\pi \sqrt{\frac{2L}{g_e}} = \sqrt{2} \cdot 2\pi \sqrt{\frac{L}{g_e}} = \sqrt{2} \cdot T_e \]

6. **Calculating \( T_p \)**:
Since \( T_e = 2 \) seconds:
\[ T_p = \sqrt{2} \cdot 2 \text{ seconds} = 2\sqrt{2} \text{ seconds} \]

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

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