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If the velocity time graph of a SHM is the derivative of the Distance time graph, and the kinetic energy of the mass in the SHM is maximum when the displacement is 0, how can the maximum velocity be Amplitude* angular velocity?

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    $\begingroup$ Take the derivative of position equation and you see that the maximum of velocity is amplitude of position times the frequency which could be considered as the angular velocity. $\endgroup$ – Amin R. Dec 14 '15 at 14:07
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This relationship comes from Hooke's Law and the maximum amount of energy that can be stored in a spring. From Hooke's law, $$F=-kx$$ we know that the frequency of oscillation is given by $$\omega = \sqrt{\frac{k}{m}}$$ We also know that the maximum amount of energy that can be stored in a spring is when the spring is stretched fully to its amplitude, and is given by $$U = \frac{1}{2}kA^2$$ where $A$ is the amplitude. When the displacement is zero, the velocity of the mass is maximum, and all of the springs potential energy is converted into kinetic energy, which leaves us with $$\frac{1}{2}kA^2 = \frac{1}{2}mv_{max}^2$$

and thus

$$v_{max}^2 = A^2\frac{k}{m} = A^2\omega^2$$

and therefore

$$v_{max} = A\omega$$

As Amin R. points out, this result is the same as the result of taking the derivative of the position function:

$$v(t) = \frac{d}{dt}A\sin(\omega t + \phi) = A\omega \cos(\omega t+\phi)$$

and $v(t)$ is maximum when $\cos(\omega t + \phi) = 1$

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