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For a damped, driven oscillator, show that the average kinetic energy is the same at a frequency of a given number of octaves* above the kinetic energy resonance as at a frequency of the same number of octaves below resonance.

*An octave is a frequency interval in which the highest frequency is just twice the lowest frequency.

We start with: $\ddot{x} + 2 \beta \, \dot{x} + \omega_0^2 x = A \cos \omega t$. If $t \gg \frac{1}{\beta}$, then $x_h (t) \to 0$ (the homogenous solution vanishes), so $x(t) \approx x_p(t) = D \cos(\omega t - \delta)$. We also have that: $$T_\text{avg} = \frac{1}{2} m v_\text{avg}^2 = \frac{1}{2} m \dot{x}_\text{avg}^2$$ Then, $$\dot{x}(t) \approx \dot{x_p}(t) = - \omega D \sin(\omega t - \delta)$$ Then how do I calculate $\dot{x}_{avg}$?

This is where I am stuck, and do not know how to proceed from here. I know that the average value of a function is defined to be $f_\text{avg} := \frac{1}{2L} \int_{-L}^{L} f(x)\,\mathrm{d}x$. So I know I would need to integrate $\dot{x}(t)$ over some period and it would be something like $$\dot{x}_\text{avg} \approx \frac{1}{2L} \int_{-L}^{L} \dot{x}(t) \, \mathrm{d}t$$ But what period should I use?

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closed as off-topic by John Rennie, Jon Custer, Gert, user36790, Bill N Sep 23 '16 at 2:08

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  • $\begingroup$ Duplicate? The method to use is explained here physics.stackexchange.com/q/60722. Kinetic energy resonance is velocity (current) resonance as opposed to displacement (charge) charge resonance physics.stackexchange.com/q/279729 $\endgroup$ – Farcher Sep 20 '16 at 14:46
  • $\begingroup$ This is asking too many separate questions. Javier, could you narrow down the post to just ask one thing? $\endgroup$ – David Z Sep 20 '16 at 14:52
  • $\begingroup$ In particular, I would like to know how to calculate the average value of $\dot{x}(t)$. That would be sufficient for my purposes. But I really do think that answering what the kinetic energy resonance is, is not only relevant but also helpful because that is what the questions asks yet it is not defined. $\endgroup$ – Javier Sep 20 '16 at 14:54
  • $\begingroup$ Please take my question off hold @DavidZ $\endgroup$ – Javier Sep 20 '16 at 15:01
  • $\begingroup$ Well, it'd be better to tell me why. Normally I'd ignore someone who just asks for their question to be taken off hold without giving a justification. But in this case, I went ahead and made the required final edit for you. You can ask about the kinetic energy resonance in a separate question. $\endgroup$ – David Z Sep 20 '16 at 15:10
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At resonance with a sinusoidal driver the driven system also undergoes sinusoidal oscillations.
The frequency of the driven system is equal to the frequency of the driver.
You are asked to find the average of the kinetic energy $\frac 1 2 m v^2_o \sin^2 \omega t$.

If you remember your alternating current theory (rms amd all that) the the average of the function $\sin^2 \omega t$ over a period is $\frac 1 2$.

$$\displaystyle \left <\sin^2 \omega t\right > = \dfrac{\displaystyle \int_0^{\frac{2 \pi}{\omega}}\sin^2 \omega t\;dt}{\left( \dfrac{2 \pi}{\omega}\right )}= \dfrac 1 2$$

So the average kinetic energy is the peak kinetic energy divided by 2.

You made an error by trying to find the average speed and then squaring that average to find the average kinetic energy. That does not work because the average speed is not equal to the square root of the average of the speed squared. You may have met this ideal in the kinetic theory of gases.

So it does not matter how the resonance graph is plotted peak kinetic energy of average kinetic energy against time you will still get the same shape only the energy axis will change in scale.
This is also the same as velocity resonance which is current resonance in an electrical circuit.

In the link that I have given you the system is an electrical one.
The equations which are dealt with are not fundamentally different from the ones you are using.

You can change from one to the other if you remember $L \leftrightarrow m, C \leftrightarrow \frac 1 k,R \leftrightarrow r, q \leftrightarrow x $ and $i \leftrightarrow v$ where I have usad the conventional symbols for electrical and mechanical system elements.

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You already have that $$\dot{x}(t) \approx \dot{x_p}(t) = - \omega D \sin(\omega t - \delta),$$

and you say you want to calculate $\dot{x}_{\textrm{avg}}$, which you define as the average velocity. It should be clear that since the system is oscillating in place, the average velocity must be zero, but let's see how to get this result from the formula you give.

The formula you give is $$\dot{x}_\text{avg} \approx \frac{1}{2L} \int_{-L}^{L} \dot{x}(t) \, \mathrm{d}t.$$ You were unsure of what $L$ to integrate over. In this case it makes the most sense to take the limit $L \to \infty$, so this system will oscillate forever. Thus we get $$\dot{x}_\text{avg} = \lim_{L\to \infty} \frac{1}{2L} \int_{-L}^{L} \dot{x}(t) \, \mathrm{d}t.$$

Now recall that the integral of the velocity is the change in position, so that we have $$\dot{x}_\text{avg} = \lim_{L\to \infty} \frac{x(L)-x(-L)}{2L}.$$ And now notice additionally, that since the motion is bounded to a region of size $2D$, we have $|x(L)-x(-L)|<2D$, and so the numerator of the limit above stays bounded while the denominator increases without bound, so $\dot{x}_\textrm{avg}=0$.

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