2022TrialShort answer

Q14 (a) (ii)

3 marks

Question

If the particle is moving at $\sqrt{15}\,\mathrm{m/s}$ when it is $2\,\mathrm m$ to the right of the centre of motion and at $2\sqrt3\,\mathrm{m/s}$ when it is $4\,\mathrm m$ to the right of the centre of motion, find the period and amplitude of the motion.

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Solution

(ii)

(\sqrt{15})^2=n^2(a^2-2^2)

15=n^2a^2-4n^2. \tag{1}

(2\sqrt3)^2=n^2(a^2-4^2)

12=n^2a^2-16n^2. \tag{2}

(1) − (2):

12n^2=3

n^2=\frac14

n=\frac12.

Period $=4\pi$ .

Sub. $n=\frac12$ in (1):

15=\frac14(a^2-4)

8=a.

Amplitude $=8$ .

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