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WannabeNewton
#13
Feb12-13, 12:20 PM
C. Spirit
Sci Advisor
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Are you sure it is [itex]v_{max}^{2} + v^{2}[/itex] as opposed to [itex]v_{max}^{2} - v^{2}[/itex]? Otherwise you get that when [itex]v = v_{max}[/itex], [itex]\omega x = \pm \sqrt{2v_{max}^{2}}[/itex] but [itex]x(v = v_{max})[/itex] should be zero since the maximum speed will occur at the equilibrium position (I am assuming this is for simple harmonic motion and that [itex]x[/itex] is the displacement from equilibrium as usual).