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Find the amplitude of a particle moving with simple harmonic motion

by RoRi

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 amplitude of the motion if the period is 2 \pi and the velocity is \pm v_0 when y = y''.

Since the period is 2 \pi we have k = 1. Since y' = \pm v_0 when y = y_0 we have

    \begin{align*} && y_0 &= C \sin (x_0 + \alpha) \quad \implies \quad x_0 = \arcsin \left( \frac{y_0}{c} \right) - \alpha \\ &&y' &= C \cos (x + \alpha) \\ \implies && y'(x_0) &= C \cos \left( \arcsin \left( \frac{y_0}{c} \right) \right) \\ &&&= C \cdot \frac{\sqrt{c^2-y_0^2}}{c} \\ &&&= \sqrt{c^2- y_0^2} \\ &&&= \pm v_0. \end{align*}

Therefore,

    \[ c^2 - y_0^2 = v_0^2 \quad \implies \quad c = \big( v_0^2 + y_0^2 big)^{\frac{1}{2}}. \]

2 comments

  1. tom says:

    Another way: y=Csin(kx + α) implies y=Bsin(kx)+Acos(kx), where A=f(0)=y0, B=f'(0)=+-v0. By definition C= sqr(A^2+B^2)=sqr(f0^2+v0^2).

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