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Formula for the harmonic oscillator is f"(x)+W^2 * X(t)=0

  1. Nov 19, 2003 #1
    can anyboy show to me why the formula for the harmonic oscillator is f"(x)+W^2 * X(t)=0. Please I spent a whole afternoon trying to figure it out and I just wasted my time.
  2. jcsd
  3. Nov 19, 2003 #2
    Do you mean x''(t), not f''(x)?

    I'm not sure where you're starting from. Are you starting from a harmonic oscillator as something whose motion is defined by,

    [tex]x(t) = A \sin(\omega t+\phi)[/tex]

    ? If so, if you just calculate its second time derivative [tex]\ddot{x}(t)[/tex], you can see immediately that it satisfies the equation,

    [tex]\ddot{x}(t) + \omega^2 x(t) = 0[/tex]
  4. Nov 20, 2003 #3
    Hopefully i can get this right.

    Using the example of a mass and spring oscillator. By Newton's £rd Law the forces on the spring are equal. The forces on the spring are the force on the mass (F1=ma) and the resistive force on the spring
    (F2+-kx, negative because it is resitive and opposite to F1=ma)



    By definition a= x"(t)

    So mx"(t)+ kx=0
    Dividing by m gives x"(t)+(k/m)x=0

    The standard equation for the period of oscillation (T) of a mass spring
    pendulum is

    T = (2(Pi))sqrt(m/K)

    omega= w = (2(Pi))/T = 2(pi)/((2(Pi))sqrt(m/K))

    The 2(pi) should cancel and leave you with w = sqrt(K/m)
    Squaring both sides gives w^2=(K/m)

    Substitute this into the above equation and get

    x"(t) + (w^2)x = 0

    I hope thats understandable as I never tried writing such an equation in pure text before. Hope it works and helps.
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