Question

A person bounces up and down on a trampoline, while always staying in contact with it. The motion is simple harmonic motion, and it takes 2.73 s to complete one cycle. The height of each bounce above the equilibrium position is 45.2 cm. Determine (a) the amplitude and (b) the angular frequency of the motion. (c) What is the maximum speed attained by the person?

Answer 1

he completed one cycle in 2.73 seconds.
frequency f =1/t =1/2.73 s =0.366 Hz
a)
amplitude A =45.2 cm =45.2*10^-2 m
b)
angular frequency =2pif =2.298 rad/s
c)
maximum speed v_max =wA =(2.298 rad/s)(45.2*10^-2 m) =1.0389 m/s

Answer 2

$\(T=2.73+s\)$

\(A=45.2 cm=.452 m\)

\(\omega=\frac{2\pi}{T}=\frac{2\pi}{2.73 s}=2.30 rad/s\)

\(v_{max}=A\omega=(.452 m)(2.30 rad/s)=1.04 m/s\)