# Why is the energy stored in a driven oscillator equal to the product of friction dissipated power $P_\text{fr}$ & decay time $\tau$ at resonance?

This is an excerpt from Waves by Frank S Crawford Jr.

[...] At steady-state the time-averaged power must equal the time-average of power dissipated by friction. The instantaneous frictional force is $-M\Gamma{\dot{x}_s^2}$ The instantaneous frictional power is the frictional force times the velocity. You can easily show that the time-averaged power loss to friction is given by $$P_\text{fr}= M\Gamma\omega^2\langle{{\dot{x}}_s^2}\rangle= \frac{1}{2} M\Gamma\omega^2[A_\text{ab}^2 + A_\text{el}^2] \tag{22}$$

[...] You can easily show that for steady-state oscillation the time-averaged stored energy $E$ is given by $$E = \dfrac{1}{2} m\left\langle {{\dot{x}}_s^2}\right\rangle + \dfrac{1}{2} m{\omega_0}^2 \left\langle x_s^2 \right\rangle = \frac{1}{2} M(\omega^2+\omega_0^2)\left(\frac{1}{2}A_\text{ab}^2 + \frac{1}{2}A_\text{el}^2\right)\tag{23}.$$

[...] Notice that for $\omega= \omega_0$ the stored energy $E$ given by Eq.$(23)$ is equal to the product of steady-state power dissipation [given by Eq.$(22)$] times the decay time $\tau$ for free oscillation. This can be qualitatively understandable: If we turned off the driving force, friction would cause the oscillator energy to "decay" exponentially with a mean decay time $\tau.$ When we drive the oscillator at its natural frequency, which is essentially $\omega_0$ for weak damping, the oscillation amplitude continues to build up until, at steady state, the power input is matched by the power loss due to friction. Since, friction dissipates most of the energy in a time $\tau,$ the steady-state stored energy is equal to that which has been supplied "recently" by the driving force, i.e., within a time $\tau.$ Thus we expect that at the equilibrium the stored energy will be approximately equal to the input power times $\tau,$ which is equal to the frictional power times $\tau.$

I'm having problem in conceiving the blocked line in the explanation.

What did he mean by recently? If friction dissipates the energy in $\tau$, how could there be any stored-energy then? I'm not getting that line. Could anyone, please explain this blocked line especially the word recently?

# Edit:

Friction dissipates the stored energy $E$ in time $\tau$ & the driving force compensates that by providing the energy dissipated by the friction. So, the stored energy is given by the friction dissipated power $P_\text{fr}$ times $\tau$ as this is equivalent to the stored energy $E$ dissipated by $\tau$ if the driving force were not present. Thanks to @Timaeus that I've understood that.

However, why doesn't the reasoning work for frequencies other than the resonant frequency?

The friction dissipated power for frequency $\omega\ne \omega_0$ is $P_\text{fr}=\frac{1}{2} M\Gamma\omega^2[A_\text{ab}^2 + A_\text{el}^2]$ . Now the energy dissipated in time $\tau$ is $\frac{1}{2} M\Gamma\omega^2[A_\text{ab}^2 + A_\text{el}^2]\cdot \tau= \frac{1}{2} M\omega^2[A_\text{ab}^2 + A_\text{el}^2]$. This is the energy dissipated by friction in $\tau$ & so the driving force must compensate that. However, it is not equal to the stored-energy $E$ which is given by Eq.$(23)$. Why does the reasoning that the energy dissipated by friction in time $\tau$ is not equal to the stored energy when the frequency is not resonant frequency that is $\omega\ne \omega_0?$ Why does the reasoning works only for resonant frequency?

• How sure are you that only $e^{-1}$ of the energy is dissipated during time $\tau$? Usually "decay time $\tau$" would mean that $e^{-1}$ of the energy will be left after time $\tau$, so $1-e^{-1}$ of it -- which is most! -- will have been dissipated by then. – Henning Makholm Nov 1 '15 at 22:00
• @Henning Makholm: Thanks, sir, I totally misinterpreted the definition. – user36790 Nov 2 '15 at 2:54
• @Henning Makholm: Also, thanks as you've answered my first question. – user36790 Nov 2 '15 at 3:06

Keep in mind that the bold text isn't a derivation, its a way to qualitatively understand, so it's going to be very very unconvincing. Instead of imaging that you are trying to compute a number imagine you are trying to estimate it by a factor of 10 or 100 or even a factor of 1000 or more. That's how unconvincing it will be.

So. At equilibrium amplitudes, power supplied equals power lost to friction. How much energy is lost to friction? Well, when no power is supplied it basically dissipates almost all of the available energy $E$ in time $\tau$. So friction saps away something like $E/\tau$ as a power lost to friction. So the power supplied is something like $E/\tau$ too.

The conclusion is that we expect the power supplied and the power lost to be somewhat equal to the energy stored divided by $\tau.$

But this is unconvincing because the power lost to friction changed and wasn't constant, if we waited longer then closer to 100% of the energy would be dissipated but the average power would be much smaller because we included more time of lower power loss.

The whole thing is incredibly wild estimates, not much different than dimensional analysis.

Most of the energy $E$ is lost by $\tau$ if the driving force is absent. Okay. But how is it related to the stored energy? Even if the driving force is present, it will have to supply $E/\tau$ so as to compensate the loss. How can it be stored, then?? it is just nullifying the dissipation; it is not stored?

The energy stored is defined to be $E$. Because the letter $E$ is the symbol I made up for the total stored energy. And $\tau$ is an almost arbitrary time related to how long it takes for the friction force to dissipate almost all of the energy $E$. Why is it that almost all of the energy is what is dissipated? Because we choose the zero of energy to be located at that place the friction approaches. Its not the absolute zero. There is more energy there, there is, force instance rest energy associated with every particle. There is chemical binding energy associated with each electron being in some atom/molecule. There is some energy associated with the thermal motion of the system. You could extract that energy with antimatter, chemical reactions with more reactive elements, and thermal contact with colder objects. But there isn't any more macroscopic kinetic energy or potential energy available besides $E$ that's all that is available. And that's the state the friction drives the system to. And it never gets perfectly there but it gets close in time $\tau$ and so in time $\tau$ almost $E$ is dissipated. Because of the definition of the two symbols.

Edit At resonance the amplitude can grow. As it grows the friction increases. When the amplitude grows it grows and grows and grows until the amplitude is so large that the friction dissipated is now exactly equal to the driving power. And that friction power is somewhat approximately equal to $E/\tau$ where $E$ is the total available energy stored.

At another frequency the amplitude doesn't grow so it doesn't get up to a nonzero steady state. In a complicated and real system there might be many modes that can be resonant and there might be shifting of modes over time.

• Most of the energy $E$ is lost by $\tau$ if the driving force is absent. Okay. But how is it related to the stored energy? Even if the driving force is present, it will have to supply $E/\tau$ so as to compensate the loss. How can it be stored, then?? it is just nullifying the dissipation; it is not stored? – user36790 Nov 2 '15 at 7:00
• Sir, I've edited my question. Could you please check that? – user36790 Nov 2 '15 at 18:05
• @user36790 Me too – Timaeus Nov 3 '15 at 4:33