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Math Help - Shm 2

  1. #1
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    Exclamation Shm 2

    A particle moves in SHM with centre O and passes through O with speed 10sqrt(3) cm/s. by integrating acceleration = - n^2 * x, calculate the speed when the particle is halfway between its mean position and a point of instantaneous rest.
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  2. #2
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    Quote Originally Posted by xwrathbringerx View Post
    A particle moves in SHM with centre O and passes through O with speed 10sqrt(3) cm/s. by integrating acceleration = - n^2 * x, calculate the speed when the particle is halfway between its mean position and a point of instantaneous rest.
    a = -n^2*x

    \frac{dv}{dt} = -n^2*x

    \frac{dv}{dx}*\frac{dx}{dt} = -n^2*x

    v*\frac{dv}{dx} = -n^2*x
    \int{vdv} = -n^2\int{xdx}

    Now find the integration. To find the constant of integration, put v = 0 when x = A, the amplitude.
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  3. #3
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    Hmmmm i dont get it, am i meant to get v^2 = 300-n^2 x^2 for the integration??? If so, how do i find n???

    2. what's 1/2 way between mean position + instantaneous rest?
    Last edited by xwrathbringerx; May 13th 2010 at 04:37 AM.
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  4. #4
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    Exclamation

    Also how exactly do i find the amplitude?
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  5. #5
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    Quote Originally Posted by xwrathbringerx View Post
    Also how exactly do i find the amplitude?
    When you do the integration, you get
    v^2/2 = - n^2*x^2/2 + C or
    v^2 = -n^2*x^2 + K.
    When x = 0, v = 10sqrt(3), v^2 = 300.

    v^2 = -n^2*(x^2) + 300.

    When x = A, the amplitude, v = 0. So

    0 = -n^2A^2 + 300.

    n^2*A^2 = 300.

    Velocity at A/2 is given by

    v^2 = -n^2*(A/2)^2 + 300

    Substitute the value for n^2A^2 and find v.
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