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Conservation of Net Mechanical Energy in SHM

  1. May 12, 2016 #1
    MENTOR Note: Thread moved here from Classical Physics hence no template

    I have a question set that I need to be able to answer before my exam next month, I know how to answer all of them except this one. I get the feeling I'm being an idiot.

    Show that the simple harmonic motion solution of the simple pendulum in the form $$\theta (t) = A\cos ({\omega _0}t)$$ (constant A) conserves net mechanical energy E = K + U.

    I have the equation for E as [tex]E = \frac{1}{2}(m{v^2} + mgl{\theta ^2})[/tex]
    I want to show its derivative is equal to 0 obviously. After I substitute in [tex]\theta \left( t \right)[/tex] and differentiate, I get [tex]\frac{{dE}}{{dt}} = \frac{{{A^2}mgl{\omega _0}\sin (2{\omega _0}t)}}{2}[/tex]

    Which is not 0... What am I doing wrong?
     
    Last edited by a moderator: May 13, 2016
  2. jcsd
  3. May 12, 2016 #2
    (m.v.v )' is not 0, v changes with time
    here is derivation---
    te= ke + pe
    = (i.w.w/2) + (c.theta.theta/2)
    =(i.w0.w0.a.a.(sin(w0.t)^2). 0.5) + (c.a.a.(cos(w0.t)^2)/2) (since w=theta' , so w= -w0.A.sin(w0.t)
    =(c.a.a.(sin(w0.t)^2)/2) +(c.a.a.(cos(w0.t)^2)/2) (since w0=(c/i)^0.5 , so c=i.w0.w0)
    =ca.a/2 (which is a const)
     
  4. May 12, 2016 #3
    I'm sorry, I can't see what you're saying as I am struggling to see what you have wrote. $${{dE} \over {dt}}$$ has to be equal to 0 so that there is no change in the total energy, i.e. energy is conserved - right?
     
  5. May 13, 2016 #4

    vanhees71

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    I guess in your notation
    $$v=\dot{\theta}=-A \omega_0 \sin(\omega_0 t).$$
    Plug this into the formula for the energy and remember what's ##\omega_0## in terms of the other parameters in the problem.
     
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