2023TrialShort answer

Q14 (d)

4 marks

Question

A particle is moving in simple harmonic motion with centre around the origin, starting at $x=m$ , where $m>0$ . The displacement equation is given by $x=a\cos(nt+\alpha)$ . After 1 second, the particle is at $x=r$ , where $r>m$ and after another second, it returns to $x=r$ .

Show that

\cos n=\frac{r+m}{2r}

Try the question before revealing the answer.

Solution

\begin{aligned} t&=0,\quad x=m\\ &m=a\cos\alpha\\[2pt] t&=1,\quad x=r\\ &r=a\cos(n+\alpha)\\[2pt] t&=2,\quad x=r\\ &r=a\cos(2n+\alpha) \end{aligned}

Method 1:

\begin{aligned} r+m&=a\cos(2n+\alpha)+a\cos\alpha\\ &=a\bigl(\cos(2n+\alpha)+\cos\alpha\bigr)\\ &=2a\cos(n+\alpha)\cos n\\ &=2r\cos n \end{aligned}

\cos n=\frac{r+m}{2r}

Method 2:

\begin{aligned} r&=a\cos(n+\alpha)\\ &=a\cos n\cos\alpha-a\sin n\sin\alpha\\ &=m\cos n-a\sin n\sin\alpha \end{aligned}

a\sin n\sin\alpha=m\cos n-r

\begin{aligned} r&=a\cos(2n+\alpha)\\ &=a(\cos 2n\cos\alpha-\sin 2n\sin\alpha)\\ &=m(2\cos^2n-1)-2a\sin n\cos n\sin\alpha\\ &=m(2\cos^2n-1)-2\cos n(m\cos n-r)\\ &=-m+2r\cos n \end{aligned}

r+m=2r\cos n

\cos n=\frac{r+m}{2r}

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Mechanics

Simple harmonic motionDerive and interpret displacement, velocity and acceleration models for simple harmonic motion.

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