A pendulum with a period of 1 s on Earth, where the acceleration due to gravity is g, is taken to another planet, where its period is 2s. The acceleration due to gravity on the other planet is most nearly

1. ag = g

2. ag = g/4

3. ag = 4 g

4. ag = g/2

5. ag = 2 g

g ' = ¼ g correct answer is 2

Explanation:

The expression for the angular velocity of a simple pendulum is

w = √ g / L

The angular velocity and the period are related

w = 2π / T

T = 2π √L / g

Let's look for the length of the pendulum with the data on Earth

L = T² g / 4π²

L = 1² g / 4π²

L = 2.533 10⁻² 9.8

L = 0.2482 m

Now let's look for the gravity of the planet

g' = 4π² L / T'²

g' = 4π² 0.2483 / 2²

g' = 2.45 m / s²

The relationship between this value and the earth's gravity is

g ' / g = 2.45 / 9.80

g ' = 0.25 g

g ' = ¼ g

correct anwers is 2