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Compute the displacement and acceleration of a particle undergoing simple harmonic motion

by RoRi

A particle undergoing simple harmonic motion has initial displacement 1, initial velocity 2, and initial acceleration -12. Compute the displacement and acceleration when the velocity is \sqrt{8}.

First, since the initial displacement is 1 we have,

    \[ y = C \sin (kx + \alpha) \quad \implies \quad y(0) = C \sin \alpha = 1. \]

Next, since the initial velocity is 2 we have,

    \[ y' = kC \cos (kx + \alpha) \quad \implies \quad y'(0) = k C \cos \alpha = 2. \]

Finally, since the initial acceleration is -12 we have,

    \[ y'' = -k^2 C \sin (kx + \alpha) \quad \implies \quad y''(0) = -k^2 C \sin \alpha = -12. \]

Now, using the first equation we plug the value of C \sin \alpha into the third equation,

    \[ C \sin \alpha = 1 \quad \implies \quad -k^2 = -12 \quad \implies \quad k = 2 \sqrt{3}. \]

Then, from the second equation

    \[ k C \cos \alpha = 2 \quad \implies \quad 2 \sqrt{3} C \cos \alpha = 2 \quad \implies \quad C \cos \alpha = \frac{1}{\sqrt{3}}. \]

Since we now have values for C \sin \alpha and C \cos alpha we can compute \alpha,

    \[ \frac{C \sin \alpha}{C \cos \alpha} = \sqrt{3} \quad \implies \quad \tan \alpha = \sqrt{3} \quad \implies \quad \alpha = \frac{\pi}{3}. \]

We then compute the value of C,

    \[ C \sin \alpha = 1 \quad \implies \quad C \sin \left(\frac{\pi}{3}\right) = 1 \quad \implies \quad C = \frac{2}{\sqrt{3}}. \]

So now we have the complete equation of motion

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

Therefore, we can compute the quantities the problem requests,

    \begin{align*} y'(x_0) = \sqrt{8} && \implies && (2 \sqrt{3}) \left( \frac{2}{\sqrt{3}} \right) \cos \left( 2 \sqrt{3} x_0 + \frac{\pi}{3} \right) &= \sqrt{8} \\ && \implies && \cos \left( 2 \sqrt{3} x_0 + \frac{\pi}{3} \right) &= \frac{\sqrt{2}}{2} \\ && \implies && \left( \frac{1}{2 \sqrt{3}} \right) \left( \frac{\pi}{4} - \frac{\pi}{3} \right) &= x_0 \\ && \implies && x_0 &= -\frac{\pi}{24 \sqrt{3}}. \end{align*}

Finally, this gives us

    \begin{align*} y(x_0) &= \left( \frac{2}{\sqrt{3}} \right) \sin \left( 2 \sqrt{3} \cdot \frac{-\pi}{24 \sqrt{3}} + \frac{\pi}{3} \right) \\ &= \frac{2}{\sqrt{3}} \cdot \sin \left( \frac{\pi}{4} \right) \\ &= \frac{1}{3} \sqrt{6}, \end{align*}

and,

    \[ y''(x_0) = (-12)y(x_0) = -4 \sqrt{6}. \]

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