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Could someone please guide me through the following question. I honestly have no clue where to start. Thank you so much!

A particle moving in Simple Harmonic Motion starts from rest at a distance 10 metres to the right of its centre of oscillation O. The period of the motion is 2 seconds.

i. Find the speed of the particle when it is 4 metres from its starting point.

i. Find the time taken by the particle to first reach the point 4 metres from its starting point, in seconds correct to 2 decimal places.

BooScout
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1 Answers1

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The displacement from the center at any time $t$ is given by $$x(t) = 10\sin(\omega t) $$ where $$\omega =\frac{2\pi}{T} =\frac{2\pi}{2} =\pi \ \text{rad/s}$$ Note when $x=6$, $$6=10\sin(\pi t) \implies t=\frac{1}{\pi} \sin^{-1}\frac 35$$

and the velocity at $x=6$ is$$\frac{dx}{dt}\bigg|_{t=\frac{1}{\pi} \sin^{-1}\frac 35} =10\pi\cos\left(\sin^{-1}\frac 35\right)=8\pi \ \text{m/s}$$

Vishu
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  • Hi, I am just a bit confused that 10 turns out to be the amplitude but not the distance that the graph shifts to the right (my original equation was x=asin(wt)-10 because i thought 10 was the distance shifting to the right. Could you please tell me how to distinguish between these 2 cases? Thank you! – BooScout Jun 09 '20 at 20:25
  • @BooScout Your equation would shift the sinusoid $10$ units down, which means that the displacement is possibly always negative for small enough $a$, i.e. the particle is oscillating entirely towards the left of the center of oscillation. – Vishu Jun 09 '20 at 20:55
  • Could you also explain why x=6 not 4? Thank you! – BooScout Jun 09 '20 at 23:23
  • @BooScout Well, the particle being $4$ away from its extreme point is the same as it being $10-4=6$ away from the center. You’re welcome! – Vishu Jun 10 '20 at 10:08