(adsbygoogle = window.adsbygoogle || []).push({}); 1. The problem statement, all variables and given/known data

In a damped harmonic oscillator, show that [itex]|R(\omega)|^2 = \gamma[/itex] at full width half maximum.

2. My attempt at a solution

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 [itex]| R(\omega)|^2[/itex] is [itex]\gamma[/itex]. Where [itex]\gamma [/itex] 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 [itex]x_0 e^{i\omega t}[/itex] 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 [itex]2\omega[/itex] and we are left with $$-2(\omega_0^2 - \omega^2) + \gamma^2 = 0$$ and solving for [itex]\omega[/itex]: [itex]\omega = \sqrt{\omega_0^2 - \frac{\gamma^2}{2}} [/itex]. We will replace [itex]\gamma[/itex] with [itex]2\beta[/itex], leaving us [itex]\omega = \sqrt{\omega_0^2 - 2 \beta^2} [/itex]

we know where the maximum is, but now we want to find the half-maximum. Plugging my [itex]\omega[/itex] back into my [itex]|R(\omega)|^2[/itex] 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 [itex]|R(\omega)|^2[/itex]. At 1/2 that is the half maximum and I want to know what [itex]\omega[/itex] 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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# Homework Help: Deriving a value for Full Width Half Maximum

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