Question

A bungee jumper undergoes simple harmonic motion with amplitude 3.5 m and frequency 0.135 Hz. Assume the bungee jumper follows the simple harmonic motion equation x=Acos(wt).

A. Determine the bungee jumper's velocity at 0.25 s
B. Determine the bungee jumper's velocity at 0.50 s
C. Determine the bungee jumper's velocity at 1.0 s

And this was the work shown:


displacement x=Acos(ωt)
A = 3.5 m
ω = 2π*f = 2π*0.135 rad/s = 0.848 rad/s

velocity v = dx/dt = -A*ω*sin(ωt)

v = -2.97*sin(0.848*t)
Note: calculator must be in RAD mode now!

A. v at t = 0.25 is -2.97*sin(0.212) = -0.625 m/s

B. v at t = 0.5 is -2.97*sin(0.424) = -1.22 m/s

C. v at t = 1.0s is -2.97*sin(0.848) = -2.23 m/s


***My only problem understanding this is where the -2.97 value comes from. How do I find this value using my calculator?*** Thanks

Answer 1

You have:

velocity v = dx/dt = -A*ω*sin(ωt)
so what is -A omega ?
A = 3.5
omega = 2 pi f = .848
so
- A omega = -2.968 or -2.97