Physics Question Need Help?

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 1.90s to complete one cycle. The height of each bounce above the equilibrium is 45 cm. Determine the amplitude and the angular frequency of the motion. What is the maximum speed attained by the person?

Please tell the equation used and describe what was done in detail.....Thanks

Answer 1

y(t)=A sin(2 pi f t)

A=0.45 m
f= 1/T= 1/1.9=.53Hz
w=2pif=3.3 rad/sec
dy/dt=v(t)= A w cos(w t)
then
Vmax occurs at cos(wt) =1 or
w t= 0, pi, 2pi, 3pi...npi
t= 0, pi/w, 3pi/w...
Vmax= Aw=.45 x 3.3=1.5 m/s

Answer 2

If we understand "the height of each bounce above the equilibrium is 45 cm.", as the height above the point of equilibrium position when the person stays without motion on this trampoline, then 45 cm is the amplitude of the motion, and the person bounces up 45 cm and down 45 cm from this point of equilibrium.

Period of motion is 1.90 sec., so frequency is V=1/T = 1 / 1.9 = 0.526315789 Hz.
Angular frequency is W=2PiV = 6.28*0.526 = 3.303 Radian/sec.
Equation of motion will be
H = Ho + A sin Wt,
where A is amplitude (0.45 m), t is time in seconds, W is angular frequency.
Maximal speed will be as maximum of derivative of this equation in t:
Vspeed = AWcosWt, and it attains maximum when cosWT becomes 1 or -1.

Vspeed max = AW = 0.45*3.303 (m/s) = 1.486 m/s
That is the answer

Answer 3

A. amplitude = .45m
B. Just use rads=(2*3.14) / T . So 6.28/1.9s = 3.305rads
C. Vmax = .45 x 3.305 = 1.487m/s