The Student Room Group

M3 Simple Harmonic Motion, finding that time.

So basically Q4 iv) OCR MEI M3 January 2010.

Jan2010_Q4_SHM.png

All parts before part iv) were fine.

Now I am going to go through my new thoughts on the situation and hopefully some bodies may confirm, reject or adjust my thinking.

So to find the time taken, I started with finding the equation which describes the motion:
x=0.05cos(5.6t)x = 0.05cos(5.6t)
Simple enough.

Then to find the displacement when v=0.2ms1v = 0.2 ms^{-1}, I used the renowned formula:
v2=ω2(a2x2)v^{2} = \omega^{2} (a^{2} - x^{2})
Found that x=670x= \frac{\sqrt 6}{70}
Now inputting this newfound xx into the first formula, followed by rearrangement gave:
5.6t=0.79565.6t = 0.7956
Originally I would think to divide by 5.6 to get a straight forward answer for t, knowing that this is the time taken for the particle to reach a velocity of v=0.2ms1v = 0.2 ms^{-1} the first time.
However, now thoughts are beginning to rearrange themselves, new patterns are being seen and new found wisdom kicks in... From looking at the mark scheme.

So if I let 5.6t=q5.6t = q, and look at a normal cosine graph with period 2π2 \pi, I can find the point where I can see v=0.2ms1v = 0.2 ms^{-1} is going downwards.

Seeing that the particle is going downwards with v=0.2ms1v = 0.2 ms^{-1}, just past the point where q=τ2q = \frac{\tau}{2} or q=πq = \pi (depends on the way you swing), I have found that the particle travels downwards with v=0.2ms1v = 0.2ms^{-1} when q=π+0.7956q = \pi + 0.7956.

Using the substitution from earlier ( 5.6t=q5.6t = q ), I can now find the time by reverting back to the original cosine function.

Therefore,
t=π+0.79565.6t = \frac{\pi + 0.7956}{5.6}
t=0.703s(3S.F.)t = 0.703 s (3 S.F.)
I am also no longer a LaTeX virgin.

So, yay or nay?
Original post by Wunderbarr


So, yay or nay?


Yay.

You could have differentiated and got/used x˙=aωsinωt\dot{x}=-a\omega\sin \omega t

for a more direct method.
I've just done the question and a simpler method I found was to use x=asin(ωt) x = a\sin(\omega t) to find the time to travel up 670 \frac{ \sqrt{6} }{70} then add this to a quarter of the period. Let Y be the point this happens at. This gives us the time to get from A to Y (OR from Y to A) hence the time we need is the period - (time taken to go from Y to A).

I hope that made sense but it seems more straight forward to me personally in the way I work with these problems, I hope it's perhaps helped? :smile:
Reply 3
Original post by ghostwalker
x


Oh yes, I can see why that may be a more direct method.

Thanks, I will consider that way, as well as my own.

Original post by Foutre en L'air
x


Soz Oz, but it's not as straight forward for me >_< . But I think our ideas are quite similar.

Quick Reply

Latest