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This is another one of those questions that I feel like I am almost there, but not quite, and it's the math that gets me. But here goes:

For a driven damped harmonic oscillator, show that the full width at half maximum of the response function $| R(\omega)|^2$ is $\gamma$. Where $\gamma $ is the damping factor. So we start off with: $$\ddot x + \gamma \dot x + \omega_0x = \frac{F_{ext}(t)}{m} = f(t)$$

And I move to solve this. I know that the solution will be $x_0 e^{i\omega t}$ and that I should get the following: $$-\omega^2 x_0 e^{i\omega t} + (i\omega)x_0 e^{i\omega t}\gamma+ \omega_0^2x_0e^{i\omega t} = \frac{F_0 e^{i\omega t}}{m}$$ which turns into $$x_0 (\omega_0^2 -\omega^2 + (i\omega) \gamma ) = \frac{F_0}{m}= f_0$$ which then becomes $$x_0 = \frac{f_0}{\omega_0^2 -\omega^2 + (i\omega) \gamma }$$ The denominator should give me the response function $R(\omega)$. So I take the absolute value and square it and I get: $$|R(\omega^2)| = \frac{f_0}{\omega_0^2 -\omega^2 t + (i\omega) \gamma } \frac{f_0}{\omega_0^2 -\omega^2 t - (i\omega) \gamma } = \frac{f_0}{(\omega_0^2 -\omega^2)^2 + \gamma^2 \omega^2 }$$

So far so good. I want to find the maximum of this function and that means I want to know where the derivative of the denominator is zero. Taking that derivative: $$\frac{d}{d\omega}(\omega_0^2 -\omega^2)^2 + \gamma^2 \omega^2) = 2((\omega_0^2 -\omega^2)(2\omega) + 2\gamma^2 \omega = 0$$ and since it = 0 we can divide thru by $2\omega$ and we are left with $$-2(\omega_0^2 - \omega^2) + \gamma^2 = 0$$ and solving for $\omega$: $\omega = \sqrt{\omega_0^2 - \frac{\gamma^2}{2}} $. We will replace $\gamma$ with $2\beta$, leaving us $\omega = \sqrt{\omega_0^2 - 2 \beta^2} $

we know where the maximum is, but now we want to find the half-maximum. Plugging my $\omega$ back into my $|R(\omega)|^2$ expression, I have $$\frac{1}{(\omega_0^2 - \omega_0^2 + 2 \beta^2)^2 + 4\beta^4 (\omega_0^2 - 2\beta^2)}=\frac{1}{(2 \beta^2)^2 + 4\beta^4 (\omega_0^2 - 2\beta^2)}=\frac{1}{4 \beta^4 + 4\beta^2 \omega_0^2 - 8\beta^6}$$

which shows me the maximum of $|R(\omega)|^2$. At 1/2 that is the half maximum and I want to know what $\omega$ is at that point. So going back to my original equation I posit: $$|R(\omega^2)| = \frac{1}{(\omega_0^2 -\omega^2)^2 + 4\beta^2 \omega^2 }=\frac{1}{8 \beta^4 + 8\beta^2 \omega_0^2 - 16\beta^6}$$

But at this point I feel I have lost the plot. Going through this the whole thing struck me as more complicated than it needs to be. My text says showing this should be "an easy exercise." So I am turning to people here to see where I messed up.

Best to you all, and thanks.

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I believe that some approximations are in order here. This seems like the kind of question where we assume damping is relatively small. Note that the response function isn't even symmetric about $\omega_0$. There may be some fancy way to define FWHM here, but if we just take the limit where $\gamma$ is small, and the peak value occurs at about $\omega_0$, then it is simple:

$\omega = \omega_0 + \delta\omega$

Expand $(\omega_0^2 - \omega^2)$ to first order in $\delta\omega$, and we get $-2\omega_0\delta\omega $. Then:

$(2\omega_0\delta\omega)^2 = \gamma^2(2\omega_0^2-(\omega_0 + \delta\omega)^2)$

$\omega_0 >> \delta\omega$, so $\omega_0^2$ dominates the RHS.

The full width half max is then twice $\delta\omega$. A simple calculation shows this is $\gamma$.

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I've only read the first part of your question and recognize it as nearly a problem in Thornton|Marion 5th ed. {3-19} The solutions manual is freely or nominally priced available on the internet.

You want the particular solution. T|M use $A \cos(\omega +t)$ for the RHS explicitly and use the substitution X sub $p(t) = D \cos (\omega t + \delta)$ to obtain: $$ D = \frac{A}{ \sqrt{\left(\omega_0^2 - \omega^2\right)^2 + 4\omega^2\beta^2}} $$ $\omega$ is the driving frequency and $\omega_0$ is the natural frequency of the osc. e.g. sqrt. (k/m or g/l)


sorry if you want the answer for a general driver this isn't it, and I believe not easily found (if at all) i.e. must use Green's fcns. or Dirac delta methods. See "The Pendulum / a case study in physics" Baker and Blackburn in addition to T|M. Many intermediate physics mechanics texts also show these methods.

bc unfortunately, maths. declined

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    $\begingroup$ Welcome to math.SE! For some basic information about writing math at this site see e.g. here, here, here and here. Please try to avoid linguistic abbreviations (like fcns. or bc), as many of our users are not native English speakers, and this may serve to confuse. $\endgroup$ – Cameron Buie Oct 20 '13 at 16:42
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The problem is not so much the calculation, as that is all about quadratic functions, but to select suitable auxiliary variables so that the calculation gets a readable and thus verifiable organization.

I'll interpret the task, as did @Aurey, that the task asks for the lowest order terms of the answer assuming $0<γ=2β\ll ω_0$.


Set $x=ω^2−ω_0^2$, then the denominator is a quadratic form in $x$, \begin{align} (ω_0^2−ω^2)^2+4β^2ω^2 &=x^2+4β^2x+4β^2ω_0^2 \\ &=(x+2β^2)^2+4β^2(ω_0^2-β^2) \\ &=D(x) \end{align} which shows the minimum value and thus the maximum of the frequency response at $x_{\min}=-2β^2$. Now we want to find the positions $x_\pm$ where the parabola has twice the minimum value, so that the fraction has half the maximum value. \begin{align} &&D(x_\pm)&=2·D(x_\min) \\ \iff&&(x_\pm+2β^2)^2+4β^2(ω_0^2-β^2)&=2·4β^2(ω_0^2-β^2) \\ \iff&& (x_\pm+2β^2)^2&=4β^2(ω_0^2-β^2) \\ \iff&&x_\pm&=-2β^2\pm2β\sqrt{ω_0^2-β^2} \\ &&&=\pm2βω_0-2β^2+O(β^3) \end{align} or $$ ω_\pm^2=ω_0^2\pm2βω_0-2β^2+O(β^3)=(ω_0\pm β)^2-3β^2+O(β^3) $$ of which the positive square roots are, at the same order in $β$, $$ ω_\pm=ω_0\pmβ-\frac{3β^2}{2ω_0}+O(β^3)=ω_0\pm\frac{γ}2-\frac{3γ^2}{8ω_0}+O(γ^3) $$ Thus the distance between these points is not influenced by the asymmetry, as both roots are shifted, in the lowest terms, by the same amount. The searched for width is $$ ω_+-ω_-=γ+O(γ^3). $$

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