2021HSCShort answer

Q13 (d) (i)

2 marks

Question

An object is moving in simple harmonic motion along the $x$ -axis. Its acceleration is $\ddot x=-4(x-3)$ , where $x$ is its displacement from the origin in metres. Initially, it is $5.5$ metres to the right of the origin and moving towards the origin. Its speed is $8\,\mathrm{m\,s^{-1}}$ as it passes through the origin. Between which two values of $x$ is the particle oscillating?

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Solution

Criteria	Marks
Provides correct solution	2
States the amplitude of the motion, or equivalent merit	1

Sample answer:

$\ddot x=-4(x-3)$
$n^2=4,\qquad\text{centre is }x=3,\qquad n=2$
$\text{period}=\frac{2\pi}{2}=\pi$

$v^2=n^2\left(a^2-(x-c)^2\right)$

When $v=8$ , $x=0$ :

$64=4\left(a^2-(0-3)^2\right)$
$16=a^2-9,\qquad a^2=25,\qquad a=5$

$\therefore\text{The particle oscillates between }x=3+5=8\text{ and }x=3-5=-2.$

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Mechanics

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

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