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M3 - Simple Harmonic Motion

This is probably a really easy question. But alas the answer has eluded me thus far. Anyway, here is the question:-

Points O, A and B lie in that order on a straight line. A particle P is moving on the line with S.H.M period of 4s, amplitude 0.5m and centre O. OA is 0.1m and OB is 0.3m. When t = 0, P passes through B travelling in the direction OB. Calculate the time when P passes A

I tried using the following equation to find the first part of the motion, B to amplitude.

X = a Sin(wt + 0.3)
0.2 = 0.5 Sin(0.5Pi t + 0.3)

and for the second part of the motion, amplitude to A:-

0.4 = 0.5 Cos(wt)

However i get nowhere near the correct answer of 1.46s when i add the two times together.

Any help would be greatly appreciated,
Thanks,
Rob

Reply 1

dvs

Why do you have a 0.3 in x=asin(wt + 0.3)? You don't determine the phase angle like that. Suppose it's p, then:
x = a sin(wt + p)

When x=0.3, t=0. So we find p=0.64 rad. Therefore the equation of this motion is:
x = 0.5 sin(pi/2 t + 0.64)

Now when x=0.1, we have:
0.2 = sin(pi/2 t + 0.64)

Here we just use basic trig and find that:
0.201 = pi/2 t + 0.64
or
pi - 0.201 = pi/2 t + 0.64

From the second one we get the required answer.

Reply 2

dec0!

This is probably a really easy question. But alas the answer has eluded me thus far. Anyway, here is the question:-

LOL U SOUND LIKE MY MATHS TEACHER

Reply 3

robfinlay

I am your maths teacher!

Thank you for the reply. I take it that you take 0.201 from pi as it has gone through one period from the centre of oscillation? 2 X 0.5pi.

Is this correct?

Thank you.

Reply 4

dvs

I guess that's one way to think about it, but I was simply using the fact that sine is positive in the second quadrant.