A 25.0 kg child on a swing kicks upward on the downswing, changing the distance from the pivot point to her center 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.

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***

Answered by Everest2017 | 2024-06-10

The difference in resonant frequency of the swing before and after the child kicks is approximately 0.008 Hz. The initial frequency is calculated at a distance of 2.40 m, and the new frequency at 2.28 m shows an increase of 0.008 Hz. This is determined using the resonant frequency formula relating frequency to the length of the pendulum and gravitational acceleration.
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Answered by Everest2017 | 2024-12-19