Question

A pendulum on planet x, where the value of g is unknown, oscillates with a period of 2.0 s. what is the period of this pen dulum if: a. its mass is doubled? b. its length is doubled? c. its oscillation amplitude is doubled? note: you do not know the values of m, il, or g, so do not assume any specific values.

Answer 1

(a) If the mass (m) is doubled, the period will not change.

(b) when the length is double, the period becomes 2.83 s.

(c) If the oscillation amplitude is doubled, the period will not change.

How to calculate the period of the pendulum?

The period of a pendulum is given by the formula shown below.

T = 2π √L/g

where;

• L is the length
• g is acceleration due to gravity

(a) If the mass (m) is doubled, the period will not change because period of the pendulum does not depend on the mass of the pendulum.

(b) when the length is double, the period becomes;

T ∝ √L

T₁ / √L₁ = T₂ / √L₂

T₂ = (T₁ / √L₁ ) x √L₂

T₂ = (T₁ / √L₁ ) x √2L₁

T₂ = (T₁ / √L₁ ) x √2 x √L₁

T₂ = T₁ √2

T₂ = 2 s x √2

T₂ = 2.83 s

(c) If the oscillation amplitude is doubled, the period will not change because period of the pendulum does not depend on the amplitude of the pendulum.