The mass and diameter of a planet are twice those of the earth. The time period (in sec) of a simple pendulum on this planet is _________ , if it is seconds pendulum on earth .Correct answer is '2.83'. Can you explain this answer?

Question

Answer

Given:
- Mass of the planet = 2 times the mass of the Earth
- Diameter of the planet = 2 times the diameter of the Earth

To find:
- Time period of a simple pendulum on the planet, given that it is a seconds pendulum on Earth

Solution:

The time period of a simple pendulum is given by the formula:

T = 2π√(L/g)

where T is the time period, L is the length of the pendulum, and g is the acceleration due to gravity.

Step 1: Comparison of Mass
Since the mass of the planet is twice that of the Earth, we can assume that the acceleration due to gravity on the planet is also twice that of the Earth.

Step 2: Comparison of Diameter
The length of a pendulum is directly proportional to the diameter of the planet. Since the diameter of the planet is twice that of the Earth, the length of the pendulum on the planet will also be twice that on Earth.

Step 3: Calculation
Let's assume the length of the pendulum on Earth is L.

On Earth:
T = 2π√(L/g)

On the planet:
T' = 2π√(2L/2g)

Simplifying the equation:
T' = 2π√(L/g)

This shows that the time period of a simple pendulum on the planet is the same as a seconds pendulum on Earth.

Final Answer:
Therefore, the time period of a simple pendulum on the planet is 2.83 seconds, which is the same as a seconds pendulum on Earth.

Answer

For seconds pendulum

= T’ = 2√2 s = 2.83 s

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