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Simple Harmonic Motion proof

  1. Feb 18, 2015 #1
    1. The problem statement, all variables and given/known data
    Prove that:
    [tex]x=8sin2t+6cos2t[/tex] is undergoing S.H.M.
    (Not too sure about how to prove for solution.)
    2. Relevant equations
    Solution for S.H.M. [tex]x=asin(nt+α)[/tex] is [tex]\frac{d^{2}x}{dy^{2}}=-n^2x[/tex]


    3. The attempt at a solution
    [tex]r=\sqrt{8^{2}+6^{2}}=10\\α=tan^{-1}\frac{3}{4}\\∴x=10sin(2t+tan^{-1}\frac{3}{4})[/tex]
    Differentiating with respect to time: [tex]\frac{dx}{dt}=20cos(2t+tan^{-1}\frac{3}{4})\\\frac{d^{2}x}{dt^{2}}=-40sin(2t+tan^{-1}\frac{3}{4})[/tex]
     
  2. jcsd
  3. Feb 18, 2015 #2

    Nathanael

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    Homework Helper

    Right. Adding any two sinusoidal functions of the same frequency will result in another sinusoidal function, regardless of their amplitudes.

    If you solve the SHM differential equation, [itex]\frac{d^2x}{dt^2}=-kx[/itex] you will get [itex]x=C_1\sin(\sqrt{k}t)+C_2\cos(\sqrt{k}t)[/itex] and it because of the above fact that you can write the solution as [itex]x=C_3\sin(\sqrt{k}t+C_4)[/itex]
     
    Last edited: Feb 18, 2015
  4. Feb 19, 2015 #3

    rude man

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    You can just take this equation, compute x', then x'', and see that ω2 must = 4 by equating sine and cosine coefficients. Both yield the same answer ω2 = 4. Had the sine & cosine coeff. yielded differing ω then x(t) would not be shm.
     
  5. Apr 29, 2015 #4
    Take the second derivative of the given expression and express it in terms of x. The result would eliminate sin and cos and will prove your answer in form of a=-nx.
     
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