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Find the velocity of a particle moving with simple harmonic motion

Assume a particle is moving with simple harmonic motion with its position governed by the equation

    \[ y = C \sin (kx+\alpha). \]

The velocity of the particle is defined to be the derivative y'. We define the frequency of the motion to be the reciprocal of the period.

Given that the amplitude C = 7 and the frequency is 10, find the velocity of the particle when y = 0.


Since the frequency is 10, we have

    \[ \frac{k}{2 \pi} = 10 \quad \implies \quad k = 20 \pi. \]

Then,

    \[ y = 0 \quad \implies \quad 7 \sin (20 \pi x + \alpha) = 0 \quad \implies quad x = -\frac{\alpha}{20 \pi} + n \pi. \]

Therefore,

    \[ y' = 140 \pi \cos (20 \pi x + \alpha) = \pm 140 \pi. \]

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