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Harmonic Motion Equation

  1. Nov 4, 2012 #1
    I'm trying to work out the differential equation for simple harmonic motion without damping,
    [itex]x''+\frac{k}{m}x = 0[/itex]

    I can solve it to
    [itex]x = c_1cos(\sqrt{\frac{k}{m}}) + c_2sin(\sqrt{\frac{k}{m}})[/itex]

    But the generalized solution is
    [itex]x = Acos(\omega*t + \delta)[/itex]

    [itex]A = \sqrt{c_1^2 + c_2^2}[/itex]

    I can understand the change of variables, but I don't really understand what happens to the sine term. Can anyone help me with this?
    Last edited: Nov 5, 2012
  2. jcsd
  3. Nov 5, 2012 #2


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    Try expanding the cos(wt+d) using cos(A+B) rule.
    (Btw, you left out the t's on the RHS of the cos-and-sine solution.)
  4. Nov 5, 2012 #3
    Let c1 = A cos(D) and c2= -A sin(D)
    Bring them back into x = c1 cos(K) + c2 sin(K) where K = sqrt(k/m)
    cos(D)cos(K)-sin(D)sin(K) = cos (K+D)
    x = A cos(K+D)
    and note that c1²+c2² = A²cos²(D)+A²sin²(D) = A²
    A = sqrt (c1²+c2²)
  5. Nov 5, 2012 #4
    So the cos(D) and -sin(D) represent the phase shift, and that's where the delta term comes from in the generalized solution?
  6. Nov 5, 2012 #5


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  7. Nov 6, 2012 #6
    delta = D = -arctan(c2/c1)
  8. Nov 22, 2012 #7
    I was also trouble with that.But now I have an idea.
    We know that mass-spring system or simple harmonic motion is one dimensional.
    From our coordinate system Y is vertical and X is horizontal.And also we know that mass-spring can be on vertical or horizontal.That's look what we found from differential equation :
    A = sin(wt+teta)

    So we are working for mass-spring system on vertical.We can add 90 degress to teta for change the dimension.So we get cos.
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