2020HSCShort answer

Q13 (a)

3 marks

Question

A particle is undergoing simple harmonic motion with period $\dfrac{\pi}{3}$ . The central point of motion of the particle is at $x=\sqrt3$ . When $t=0$ the particle has its maximum displacement of $2\sqrt3$ from the central point of motion. Find an equation for the displacement $x$ of the particle in terms of $t$ .

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Solution

Criteria	Marks
Provides correct solution	3
Correctly finds any two of amplitude, $n$ , function with consistent phase	2
Correctly finds one of amplitude, $n$ , function with consistent phase	1

Sample answer:

The equation has the form

x(t)=A\cos(nt)+\sqrt3

A=2\sqrt3,\qquad \frac{2\pi}{n}=\frac\pi3,\qquad n=6

\therefore x(t)=2\sqrt3\cos(6t)+\sqrt3

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Syllabus alignment

Mechanics

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

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