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Find the equation of motion for 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.

Find the equation of motion of the particle if y = 3 and y' = 0 when x = 0 and if the period is \frac{1}{2}.


We have that the period is \frac{1}{2} this implies k = 4 \pi. So, the equation of motion is

    \[ y = C \sin (4 \pi x + \alpha), \qquad y' = 4 \pi C \cos (4 \pi x + \alpha). \]

Then using the given conditions,

    \[ y(0) = 3 \quad \implies \quad C \sin \alpha = 3, \]

and

    \[ y'(0) = 0 \quad \implies \quad 4 \pi C \cos \alpha = 0. \]

These two equations give us

    \[ \alpha = n \pi + \frac{\pi}{2} \quad \text{and} \quad C = 3. \]

Therefore,

    \[ y = 3 \sin (4 \pi x + \alpha) = 3 \sin \left( 4 \pi x + \frac{\pi}{2} \right) = 3 \cos (4 \pi x). \]

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