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Given the oscillator described by: $$m\ddot{x}+\gamma \dot{x}+kx=F_0\cos(\omega t)$$ And supposing the system is at it's stable state, I wish to calculate the following:
1) The system's energy at any time.
2a) The power of the driving force and the power of the dampening force.
2b) Show the power of the dampening force is equal to the power of the driving force.

Now, I began with $E_\text{tot} = E_k+U = m\dfrac{\dot x ^2}{2}+k\dfrac{x^2}{2}$ and got that the total energy isn't constant. Which seems odd to me since after a very long time (stable state) i would expect the energy to be constant. Is my first intuition correct? I got an energy that is non-constant.

Then I calculated the power of the forces with the formula: $P_f=f\cdot \dot x$ but didn't get equality - which once again is strange (because the question specifically asked to show equality).
But, again, this is odd. If the powers are equal (and ofcourse opposite in sign) then where does the power of the returning force $kx$ go to ?

Edit: For the first problem i wrote the stable solution as $x(t)=A\cos(\omega t+\phi)$ ($A$ and $\phi$ are defined as in the solution below). $\dot x = -A\omega\sin(\omega t+\phi)$. Then calculating the energy: $$E= \dfrac{m}2\big(-A\omega\sin(\omega t+\phi)\big)^2+\dfrac{k}2\big(A\cos(\omega t+\phi)\big)^2=\dfrac{A}2\big(m\omega ^2\sin^2(\omega t+\phi)+k\cos^2(\omega t+\phi)\big)$$

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You have $$m\ddot x+\gamma \dot x + k x= F(t)$$ Calculate the variation of energy between time $a$ and time $b$: $$\int_a^b (m\ddot x+\gamma \dot x + kx)\dot x\, \text{d}t=\int_a^b F(t)\dot x\text{d}t$$ You can easily integrate two of the terms on the left: $$ \left[\frac{1}{2}m\dot x^2 + \frac{1}{2}kx^2\right]_a^b=\int_a^b F(t)\dot x\text{d}t-\int_a^b \gamma\dot x^2\,\text{d}t$$ You have four terms in total:

  • the variation of kinetic energy
  • the variation of potential energy
  • energy brought by the external force between $a$ and $b$
  • dissipated energy by viscosity between $a$ and $b$ (the sign is constant because the velocity is squared)

Now, in your particular case, you are looking for steady state so $x(t)=A\cos(\omega t+\varphi)$ and $F(t)=F\cos(\omega t)$. $A$ and $\varphi$ are given in @VictorPira's answer (I didn't check). So you can calculate the two remaining integrals and you get: $$ \left[\frac{1}{2}m\dot x^2 + \frac{1}{2}kx^2\right]_a^b=A\omega F\left[\frac{-\cos(2\omega t +\varphi)-2\omega t\sin(\varphi)}{4\omega}\right]_a^b+\gamma A^2\omega^2\left[\frac{\sin(2(t\omega+\varphi)-2(t\omega+\varphi)}{4\omega}\right]_a^b$$

Since you're considering the stable state, everything is periodic, of period $T=2\pi/\omega$. You might thus want to choose $b=a+2\pi\omega$. Obviously, because $x$ and thus $\dot x$ are periodic, the left-hand side will be 0. So you see that the energy dissipated by the damper over one period is equal to the energy brought by the external force. This does not mean that each term is constant of course, but their variation over one period is.

If think with all this you should be able to answer your questions.

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This is homework-and-exercises question so I will not answer it completely, I just give hints.

Work with the solution of the equation in stable state. It should be:

$$ x = A\cos (\omega t + \phi) $$

where

$$ A = \frac{F_0}{m}\sqrt{\frac{1}{(k/m-\omega^2)+\gamma^2\omega^2}} $$ $$ \tan \phi = \frac{-\gamma \omega}{(k/m-\omega^2)} $$

Make sure you are using right trigonometric operations, use $k/m=\omega_0^2$ for brevity, calculate $\dot{x}$ and use your equation for energy. It will be tim independent. Don't forget: $\cos^2 \alpha + \sin^2 \alpha = 1$.

The same with powers.

Better now?

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  • $\begingroup$ I got $E_{tot}=\dfrac{mA}{2}[\omega ^2\sin^2(\omega t+\phi)+\omega_0^2\cos^2(\omega t+\phi)]$ Which is time dependent (since the pythagorean theorem cannot be used) $\endgroup$ – Ranc Oct 21 '15 at 12:38
  • $\begingroup$ @Ranc - I suggest that you add your working to the question. Your intuition is correct; your math must be wrong. In the circumstances I think a little "check my work" is in order. $\endgroup$ – Floris Oct 21 '15 at 12:45
  • $\begingroup$ @Floris OK, I added my first calculations. I sure hope my differation technique is alright. $\endgroup$ – Ranc Oct 21 '15 at 17:36

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