Consider a linear harmonic oscillator subject to a periodic force:
$$ \ddot x + 2 \beta \dot x + \omega _0 ^2x = f_0\cos \omega t$$
The solution tends to:
$$A \cos (\omega t – \delta)$$
where:
$$A^2=\dfrac{f _0 ^2}{(\omega _0 ^2-\omega ^2)^2+4\beta ^2\omega ^2}$$
The maximum of this function is reached when:
$$\omega ^2 = \omega_0^2-2\beta ^2$$
which is:
$$A^2 _{max}=\dfrac{f_0 ^2}{4\beta ^2\omega_0 ^2-4\beta^4}\approx \dfrac{f_0 ^2}{4 \beta ^2\omega_0 ^2}$$
if $\beta$ is small compared to $\omega _0$.
I need to find the FWHM of $A^2$. I put:
$$\dfrac {A_{max}^2}{2}=\dfrac {f_0 ^2}{8 \beta ^2\omega_0 ^2}=^!\dfrac{f _0 ^2}{(\omega _0 ^2-\omega ^2)^2+4\beta ^2\omega ^2}$$
$$\iff (\omega _0 ^2-\omega ^2)^2+4\beta ^2\omega ^2=8\beta^2\omega_0^2$$
My book gives the hint "Is it ok to consider $\omega + \omega _0 \approx 2\omega _0$ or $\omega \approx \omega _0$, it is not ok to consider $\omega – \omega _0 \approx 0$".
With this hint it's not hard to find the solution:
$$( \omega – \omega _0)^2(\omega + \omega_0)^2+4\beta^2\omega^2\approx 4\omega _0 ^2[(\omega-\omega_0)^2+\beta^2]$$
$$\therefore (\omega-\omega_0)^2\approx\beta^2\iff\omega=\omega_0 \pm\beta $$
solved. Now my question is: it's pretty clear that the approximation $\omega -\omega _0 \approx 0$ wouldn't work. It is also clear to me that it is not contradictory to assume $\omega \approx \omega _0$ , for we use this approximation for the other term $\beta ^2 \omega ^2 $. Is there a more precise way to see this (i.e. that the approximation $\Delta \omega \approx 0 $ is not correct)??
Best Answer
Consider the positive quantity $X = (\omega - \omega_0)^2 (\omega + \omega_0)^2$. Let us make the approximation $\omega \approx \omega_0$:
The second approximation may or may not be accurate, depending on $\omega$ and $\omega_0$. The first one, however, is always disastrous, since it obliterates completely the quantity you try to estimate.