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M3 - Shm watch

    • Thread Starter

    Hi, can anyone help me with the attached qu please? It continues on another sheet hence why there are 2 attachments.


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    At equilibirum:
    (mg/l)(x-l) = (2mg/2l)(5l-x-2l), where x is the distance of the particle from A

    When the particle (P) is at a distance x from O:
    Extension in AP = 2l+x-l = l+x
    Extension in BP = 3l-x-2l = l-x

    Newton's second law gives:
    T(BP) - T(AP) = ma
    (mg/l)(l-x) - (2mg/2l)(l+x) = ma
    (g/l)(-2x) = -(2g/l)x = a
    So it's SHM with w²=(2g/l).

    Period = T = 2pi/w = 2pi/sqrt[2g/l] = 2pi.sqrt[l/2g] s
    Amplitude = A = (midpoint of AB) - (AO) = 2.5l - 2l = 0.5l m -- I'm not sure about this!

    This is all fine if both strings remain taut. If one of the strings became slack, then the system won't perform complete oscillations anymore. Since the particle was initially projected towards O, and since O is 2l from A, then the particle was projected towards A (since it was initially at the midpoint of A and B - i.e. 2.5l from each). So it makes no sense if the string attached to B became slack since there is no possible way that string would return to its natural length; however, the string attached to A can return to its natural length IF the particle is given enough intial speed to make it reach that point.

    We already know that the system begins SHM regardless of the magnitude of u. So we can use the equation v²=w²(A²-x²).
    If the amplitude of the motion is l with center O, then the string will return to its natural length and become slack. So:
    A=l, w²=(2g/l) and x=0.5l when v=u
    That is:
    u² = (2g/l)(l²-0.25l²) = (2g/l)(3l²/4) = (3/2)gl

    So that's the maximum speed you can apply to the particle and have it still perform complete oscillations -- i.e.:
    u² = (3/2)gl is max
    => u² < (3/2)gl
    => 2u² < 3gl as required.
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Updated: March 2, 2005

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