Question

11. A pendulum is oscillating with its length 120 cm on Earth. (a) Find out its time period. (b) If pendulum is taken to Mars (g = 3.7 m/s^2), what would be its time period? (C) If mass of the pendulum is doubled, what will happen to its time period?

Answer 1

a) Let the period of the pendulum on the earth is T_e.

Given that the length of the earth L=120 cm=1.20 m.

The period of the pendulum on the earth

  $T_{e}=2\pi \sqrt{ \frac{L}{g}}$

  $or\: \: \: \: \: \: T_{e}=2\pi \sqrt{ \frac{1.20}{9.8}}$

$\therefore \: \: \: \: \: \: T_{e}=2.198656852\approx 2.2\: \: \: \: s$

$\therefore$ The period of the pendulum on the earth is 2.198656852 s or 2.2 s approx.

b) Let the period of the pendulum on Mars is T_m.

The period of the pendulum on Mars is

$T_{m}=2\pi \sqrt{ \frac{L}{g}}$

$or\: \: \: \: \: \: T_{m}=2\pi \sqrt{ \frac{1.20}{3.7}}$

$\therefore \: \: \: \: \: \: T_{m}=3.578241344\approx 3.58\: \: \: \: s$   

  

$\therefore$ The period on Mars is 3.578241344 s or 3.58 s.

c) If mass of the pendulum is doubled, then the time period remains the same.

Explanation: The period of the pendulum does not depend on the mass of the pendulum. The period of a pendulum depends on the length of the pendulum and the gravitational acceleration.