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