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 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 \[ n = 1/2π √ (g / L) \]

So, variable L would be the distance of the center of gravity of child from the pivot.

Variable G would be, acceleration because of gravity

Ok so, n1 ⬇

___1_____* _______√9.81___ Which = 0.3217 times/second

2 * 3.14 √2.40


Ok so, n2 ⬇

_____1_____ * _____√9.81______ Which = 0.3301 times per second
2 * 3.14 √2.28


The increase or grow in the resonant frequency = ?