Finding Phase Constant for Harmonic Oscillator

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SUMMARY

The phase constant for the harmonic oscillator with the given velocity function is determined to be approximately 4.068 radians. The position function is defined as x = xmcos(ωt + φ), while the velocity function is v = -ωxmsin(ωt + φ). The maximum velocity (vm) was calculated as 9.375 cm/s, leading to the conclusion that φ = sin-1(7.5/-9.375) results in a negative angle, which can be converted to a positive angle in the fourth quadrant.

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Homework Statement


What is the phase constant (from 0 to 2π rad) for the harmonic oscillator with the velocity function v(t) given in Fig. 15-30 if the position function x(t) has the form x = xmcos(ωt + φ)? The vertical axis scale is set by vs = 7.50 cm/s.
[PLAIN]http://img227.imageshack.us/img227/4729/qu1512.gif

Homework Equations


x = xmcos(ωt + φ)
v=-ωxmsin(ωt + φ)
vm=ωxm

The Attempt at a Solution



From graph, vm=9.375 cm/s

vm=9.375 cm/s = ωxm

xm=9.375/ω

At t=0, v(0)=7.5 cm/s=-ωxmsin(φ)

φ=sin-1(7.5/-ωxm)

φ=sin-1(7.5/-ω*9.375/ω)

φ=sin-1(7.5/-9.375)= -.927 rad

I still got it wrong and not sure where I messed up. Only thing that I can think of is that I incorrectly assumed t=0 is 7.5 cm/s and if that's the case then I don't know where to begin on this problem.
 
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Phase angle is always positive.
In this problem phase angle is in the fourth quadrant.
 
It is correct, but try to give in positive angle with the same sine: pi-phi= 4.068 rad.

ehild
 
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