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Simple Harmonic Motion without damping

  1. Nov 4, 2012 #1
    So, simple harmonic motion without damping is described generally by
    [itex]x(t) = Acos(\omega*t +\delta)[/itex]

    Which is derived from the differential equation
    [itex]x''+\frac{k}{m}x = 0[/itex]

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


    [itex]tan\delta = \frac{c_1}{c_2}[/itex]

    With the differential equation, dealing with an initial condition is relatively easy, but it does not work as easily if using the generalized equation. Is there a way of making a relationship between [itex]A[/itex] and [itex]\delta[/itex]? I've worked with the equations a bit and can't find anything, but thought someone on here might know something different.
  2. jcsd
  3. Nov 4, 2012 #2
    I'm sorry, but you didn't define your variables [itex]c_1,c_2[/itex].
  4. Nov 4, 2012 #3
    Those are constants in the solution to the differential equation that are determined by the initial conditions.
  5. Nov 5, 2012 #4
    Not sure what you are trying to do.
    The solution expressed in terms of c1 and c2 is a general solution too.
    The one you wrote is just a different way to express the general solution.
    In both cases you have two independent parameters (c1,c2 or A, δ).
  6. Nov 5, 2012 #5
    In the problems that I have dealt with, I've been given the A value and had to calculate [itex]\delta[/itex]. I was trying to see if there is a simple way to relate it without solving every time, but the more I think about it, the less likely it seems.
  7. Nov 5, 2012 #6
    I don't get the question either. Is it the inverse tangent that is the problem? You actually have shown a relationship between A and δ. In general you can (so far as I know) not get exact values for arctan δ. It is a transcendental function.

    Someone please correct me if I am wrong. It's been years since I looked at this in depth.

    I asked about the constants of integration becaus I wanted to know how you arrived at them. It will not always be the case that you constants of integration work as you have presented.
  8. Nov 5, 2012 #7
    For a given A, δ may have all kind of values, depending on the initial conditions.
    Imagine that you can have a pendulum oscillating with the same amplitude but either starting from the maximum displacement (and zero v) or starting from equilibrium position but having initial velocity. A would be the same in both cases but δ will not.
    So you cannot find δ as a function of A only.
  9. Nov 5, 2012 #8
    Okay, that's what I was thinking but I saw that both A and [itex]\delta[/itex] were related to the constants, and thought there may be some way to relate the two.
  10. Nov 6, 2012 #9
    Are you familiar with the use of complex numbers to arive at the expressions you presented?


    I somewhat misspoke when I said you have a relationship between A and δ. A, as the other person clarified, is invariant for the ideal SHO in steady state. It's the radius of a circle described in the Gaussian/Argand plane. δ is the angle of rotation of the complex number expressed in polar coordinates.

  11. Nov 6, 2012 #10
    I was familiar with the use of complex numbers in solving the ODE, but not with relating to the rest of the concepts of SHO. That link gave me a bit more understanding. Thanks!
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