2021HSCShort answer

Q13 (d) (ii)

2 marks

Question

For the simple harmonic motion described in part (i), find the first value of $t$ for which $x=0$ , giving the answer correct to 2 decimal places.

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Solution

Criteria	Marks
Provides correct solution	2
Finds the displacement function, or equivalent merit	1

Sample answer:

Consider $x=a\cos(nt+\alpha)+c$ , where $a=5$ , $n=2$ , $c=3$ .

$x=5\cos(2t+\alpha)+3$
$5.5=5\cos\alpha+3$
$\frac{2.5}{5}=\cos\alpha$
$\alpha=\cos^{-1}\left(\frac{2.5}{5}\right)=1.04719\ldots\text{ radians}$

$0=5\cos(2t+1.04719\ldots)+3$
$\cos(2t+1.04719\ldots)=-\frac35$
$2t+1.04719\ldots=2.214\ldots\text{ radians}$
$t=0.583\ldots$

$\therefore\text{First value of }t\text{ when }x=0\text{ is }t=0.58\text{ seconds (2 decimal places).}$

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Mechanics

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

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