1. Limited time only! Sign up for a free 30min personal tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

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]

    and

    [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?

    http://galileo.phys.virginia.edu/classes/152.mf1i.spring02/ComplexNumbersSHO.htm

    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.

    Argand_diagram.png
     
  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!
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook




Similar Discussions: Simple Harmonic Motion without damping
  1. Damped Harmonic Motion (Replies: 7)

Loading...