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    (Original post by DFranklin)
    Moonfield: Typo, sorry. You'll see it becomes a minus again in the next line...

    Alex: Yes, of course. The strip and pip undergo the same acceleration while they are in contact, and the strip's motion is given (i.e. the strip does SHM with amplitude 2cm and frequency 5 whatever the mass of the pip, even if this isn't exactly likely in the real world).
    Thanks for clarifying.
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    An interesting video I found while trying to revise SHM.
    https://www.youtube.com/watch?v=cUrWKYM97T8
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    (Original post by DFranklin)
    I'm not sure where this "pip leaves the strip when y = 0" is coming from, but it's not true. The pip leaves the strip when the deceleration of the strip is more than gravity can provide (the strip can only push on the pip to accelerate it - any deceleration comes from gravity).

    Following is more-or-less a full solution, but it's my only copy of the workings, so I don't have much choice but to post it.
    Spoiler:
    Show
    We know y = cos(10\pi t)/50, so \ddot{y} = -100\pi^2 cos(10\pi t)/ 50 = 2\pi^2 cos(10\pi t).

    We want \ddot{y} = -9.8 \implies cos(10\pi t) = \frac{-9.8}{2\pi^2} \approx -1/2. So 10\pi t = \arccos(-1/2) = 2\pi / 3

    So t = 1/15 = 0.0667 (slight discrepancy either because 9.8 isn't actually pi^2, just close to it, or because you were supposed to take g = 10).
    Spoiler:
    Show
    We want \ddot{y} = -9.8 \implies cos(10\pi t) = \frac{-9.8}{2\pi^2} \approx -1/2. So 10\pi t = \arccos(-1/2) = 2\pi / 3
    If you equate the acceleration (-2pi^2 cos10pi T) to -9.8, surely the negatives should cancel out?
 
 
 
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