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Math Help - Sinusodial distance formula

  1. #1
    Super Member wingless's Avatar
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    Sinusodial distance formula

    Well most of you know that the formula of distance in an oscillating (harmonic) movement is x = r.coswt. I know how to get at this formula using the circular movement. But I gotta reach it by using the acceleration.

    K.................................O...........obje ct............. L
    |------------------------|-----------o------------|
    ...................................|_________|
    ...........................................x

    x = distance from the center
    k = N/x (like the k of a spring)
    m = mass
    a = acceleration

    At any point, F = x*k
    and F = m*a so
    a = x*k/m

    But I cant figure out how to find the sinusodial distance formula from here.

    Thanks for all your help..
    Last edited by wingless; December 5th 2007 at 01:07 PM. Reason: typing errors
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  2. #2
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    Calculus needed

    First, you need to remember that the force, and thus acceleration, is in the opposite direction of the displacement x, so we have F=-k \cdot x and a=-\frac{k}{m}x.

    From here, we need calculus:

    Remember that a = \frac{dv}{dt} = \frac{d^{2}x}{dt^2}

    so we have \frac{d^{2}x}{dt^2} = -\frac{k}{m}x

    Solving this differential equation gives us the sinusoidial solution.

    (Hint: define \omega_0 = \sqrt{\frac{k}{m}})


    --Kevin C.
    Last edited by TwistedOne151; December 5th 2007 at 11:36 AM. Reason: fixed error in TeX
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  3. #3
    Super Member wingless's Avatar
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    Ok. That equation gave me,
    x = V_0 \text{Cos}\left(t\sqrt{\frac{k}{m}}\right) + x_0\text{Sin}\left(t\sqrt{\frac{k}{m}}\right)<br />

    As x_0 = 0 on the starting point, I cancel the sine function.

    So, x = V_0 \text{Cos}\left(t\sqrt{\frac{k}{m}}\right))<br />
    \omega_0 = \sqrt{\frac{k}{m}}
    V_0 = r\omega_0
    x = \omega r  Cos(\omega t)

    I think I did something wrong ..
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  4. #4
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    Constants are off

    From the differential equation \frac{d^{2}x}{dt^2} = -\frac{k}{m}x, which we use \omega \equiv \sqrt{\frac{k}{m}} to rewrite as \frac{d^{2}x}{dt^2} = -\omega^{2}x,
    you should get a general solution of the form x(t)=A\cos (\omega t) + B\sin (\omega t).

    Putting in t=0, and using \cos (0) = 1 and \sin (0) = 0, we get x_0 = x(0) = A \cdot 1 + B \cdot 0 = A.
    Similarly, v(t)=x'(t)=-A\omega \sin (\omega t) + B\omega \cos (\omega t), and v_0=v(0)=-A\omega \cdot 0 + B\omega \cdot 1 = B\omega, and B=\frac{v_0}{\omega}
    Thus you should have x(t)=x_0 \cos (\omega t) + \frac{v_0}{\omega} \sin (\omega t), not x(t)=v_0 \cos (\omega t) + x_0 \sin (\omega t).

    --Kevin C.
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  5. #5
    Super Member wingless's Avatar
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    Ok, I just thought it was an ordinary integration constant (like in linear motion). Thanks for helping out.
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