Question

A 25.0 kg child on a swing kicks upward on the downswing thus changing the distance from the pivot point to her centre of gravity from 2.40 m to 2.28 m. What is the difference in the resonant frequency of her swing before the kick and afterwards? Answer to three significant digits.

Answer 1

The time period of the swing is given by T = 2π √ (L / g)
The natural or resonant frequency is n = 1/2π √ (g / L)

L = distance of the center of gravity of child from the pivot.
g = acceleration due to gravity

1 √9.81
So n1 = --------------- * ------- = 0.3217 times per second
2 * 3.14 √2.40

1 √9.81
So n2 = --------------- * ------- = 0.3301 times per second
2 * 3.14 √2.28

So the increase in the resonant frequency is : 0.0084 times per second
= 0.008 / second