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 C. Spirit Sci Advisor Thanks P: 5,636 Are you sure it is $v_{max}^{2} + v^{2}$ as opposed to $v_{max}^{2} - v^{2}$? Otherwise you get that when $v = v_{max}$, $\omega x = \pm \sqrt{2v_{max}^{2}}$ but $x(v = v_{max})$ should be zero since the maximum speed will occur at the equilibrium position (I am assuming this is for simple harmonic motion and that $x$ is the displacement from equilibrium as usual).