
#1
Apr210, 10:53 AM

P: 3

1. The problem statement, all variables and given/known data
What is the phase constant (from 0 to 2π rad) for the harmonic oscillator with the velocity function v(t) given in Fig. 1530 if the position function x(t) has the form x = x_{m}cos(ωt + φ)? The vertical axis scale is set by vs = 7.50 cm/s. 2. Relevant equations x = x_{m}cos(ωt + φ) v=ωx_{m}sin(ωt + φ) v_{m}=ωx_{m} 3. The attempt at a solution From graph, v_{m}=9.375 cm/s v_{m}=9.375 cm/s = ωx_{m} x_{m}=9.375/ω At t=0, v(0)=7.5 cm/s=ωx_{m}sin(φ) φ=sin^{1}(7.5/ωx_{m}) 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. 



#2
Apr210, 11:15 AM

HW Helper
P: 4,442

Phase angle is always positive.
In this problem phase angle is in the fourth quadrant. 



#3
Apr210, 11:20 AM

HW Helper
Thanks
P: 9,818

It is correct, but try to give in positive angle with the same sine: piphi= 4.068 rad.
ehild 


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