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Find amplitude and frequency given positions and speeds

  1. Aug 11, 2013 #1

    I'm working my way through Classical Mechanics by J.R. Taylor. I'm stumped by the one-star ("easiest") Problem 5.11: "You are told that, at known positions x1 and x2, an oscillating mass m has speeds v1 and v2. What are the amplitude and angular frequency of the oscillations?"

    2. Relevant equations

    Taylor gives the following equations for simple harmonic motion:

    5.5 x(t) = C[itex]_{1}[/itex]e[itex]^{iωt}[/itex] + C[itex]_{2}[/itex]e[itex]^{iωt}[/itex]
    5.6 x(t) = B[itex]_{1}[/itex]cos(ωt) + B[itex]_{2}[/itex]sin(ωt)
    5.11 x(t) = Acos(ωt - [itex]\delta[/itex])
    5.14 x(t) = Re[Aei(ωt - [itex]\delta[/itex])]

    Taylor gives relationships among the constants A, B's, and C's.

    3. The attempt at a solution

    Solutions are given at the back of the book:

    A = sqrt( (x[itex]_{2}[/itex][itex]^{2}[/itex]v[itex]_{1}[/itex][itex]^{2}[/itex] - x[itex]_{1}[/itex][itex]^{2}[/itex]v[itex]_{2}[/itex][itex]^{2}[/itex]) / (v[itex]_{1}[/itex][itex]^{2}[/itex] - v[itex]_{2}[/itex][itex]^{2}[/itex]))


    Equations 5.6 and 5.11 immediately give positions at times ωt[itex]_{1}[/itex] and ωt[itex]_{2}[/itex]; differentiating those equations gives speeds. It's easy to get the amplitude A in terms of ωt, e.g.

    A = x[itex]_{1}[/itex]/cos(ωt[itex]_{1}[/itex]).

    But I can't figure out how to get A in terms of the x's and v's. I don't see how to separate ω from the t's, or how to get ω out of the trig functions.
  2. jcsd
  3. Aug 11, 2013 #2


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    Staff: Mentor

    Those formulas all assume that the oscillation is around x=0. Do we know that? Let's assume we do.
    We need the additional assumption that the mass is in a harmonic potential.

    Can you write down the potential?
    Based on this potential and the velocities, what can you tell about the energies at x1 and x2? What about energy conservation?
    With the known energy and shape of the potential, you can calculate everything else.
  4. Aug 12, 2013 #3
    Oh. Potential energy. Of course. That works--thanks!

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