A 24.00 kg block at rest on a horizontal frictionless air track is connected to the wall via a spring. The equilibrium position of the mass is defined to be at x=0. Somebody pushes the mass to the position x=0.3500 m, then lets it go. The mass undergoes simple harmonic motion with a period of 4.80000 s. What is the position of the mass 3.98 s after the mass is released?
I've calculated angular velocity using the equation:
w = 2pi*f
w = 2pi/period
w = 2pi/4.8s
I've tried using the following equation to calculate position:
x = Acos(wt)
x = (0.35m)*cos[(2pi/4.8s)(3.98s)]
x = 0.3486m
However this is not the correct answer. Could anybody find where I went wrong?
2 answers
horizontal frictionless air track... why are you using Pi?
T=2pai rootover 1 by g
1 answer