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Conservation of Net Mechanical Energy in SHM

  • Thread starter Habez
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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?
 
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  • #2
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Which is not 0... What am I doing wrong?
(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)
 
  • #3
2
0
(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)
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?
 
  • #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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