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M3 Simple Harmonic Motion, finding that time.

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Why bother with a post grad? Are they even worth it? Have your say! 26-10-2016
    • Thread Starter

    So basically Q4 iv) OCR MEI M3 January 2010.

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    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)
    Simple enough.

    Then to find the displacement when v = 0.2 ms^{-1}, I used the renowned formula:
    v^{2} = \omega^{2} (a^{2} - x^{2})
    Found that x= \frac{\sqrt 6}{70}
    Now inputting this newfound x into the first formula, followed by rearrangement gave:
    5.6t = 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 = 0.2 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 = q, and look at a normal cosine graph with period 2 \pi, I can find the point where I can see v = 0.2 ms^{-1} is going downwards.

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

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

    t = \frac{\pi + 0.7956}{5.6}
    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?

    You could have differentiated and got/used \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 = a\sin(\omega t) to find the time to travel up  \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?
    • Thread Starter

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