Find the amplitude of a particle moving under 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.

If the frequency is $\frac{1}{\pi}$ and if we have initial values $y = 2$ and $y' = 4$ when $x = 0$ , find the amplitude $C$ .

Since we are given that the frequency is $\frac{1}{\pi}$ using the formula for frequency (frequency $= \frac{k}{2 \pi}$ ) we have

$\frac{k}{2 \pi} = \frac{1}{\pi} \quad \implies \quad k = 2.$

Then,

$y(0) = 2 \quad \implies \quad C \sin \alpha = 2$

and

$y'(0) = 4 \quad \implies \quad 2C \cos \alpha = 4.$

This implies

$\alpha = \frac{\pi}{4}, \qquad \text{and} \qquad C = 2 \sqrt{2}.$

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