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.

See Answers (1)

Accepted Answer

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

Related Question in Physics