2022TrialShort answer

Q14 (b) (i)

3 marks

Question

A particle is moving so that $x=2\sin3t-2\sqrt3\cos3t$ . By expressing the motion in the form $A\sin(3t-\alpha)$ , find the period and amplitude.

Try the question before revealing the answer.

Solution

(b)(i)

2\sin3t-2\sqrt3\cos3t=A\sin(3t-\alpha)

2\sin3t-2\sqrt3\cos3t =A\sin3t\cos\alpha-A\cos3t\sin\alpha.

Equating parts:

2=A\cos\alpha \tag{1}

2\sqrt3=A\sin\alpha. \tag{2}

(1)^2+(2)^2=A^2(\cos^2\alpha+\sin^2\alpha)=2^2+(2\sqrt3)^2

A=4.

Sub. $A=4$ in (1):

2=4\cos\alpha\Rightarrow\cos\alpha=\frac12.

Sub. $A=4$ in (2):

2\sqrt3=4\sin\alpha\Rightarrow\sin\alpha=\frac{\sqrt3}{2}.

\alpha=\frac\pi3.

x=4\sin\left(3t-\frac\pi3\right),\quad \text{Period}=\frac{2\pi}{3},\quad \text{Amplitude}=4.

Supplied marking guidance · human reviewed

Compare this with your working, not only your final sentence.

Syllabus alignment

Mechanics

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

Spot a problem? Send feedback