I am asked to deduce by how much does damping change the angular frequency of a harmonic system. I immediately thought to use the equation $\omega$ = $\sqrt{\omega_0^2 - \frac{\gamma^2}{4}}$. However, my lecturer told us to use this identity, $\sqrt{1+y}$ = 1 + $\frac{y}{2} - \frac{y^2}{8} + \frac{y^3}{16}$. I tried to factor out $\omega_0^2$ from the first equation to get the form $\sqrt{1+y}$ she would like, but this does not make sense to me as it seems to complicate a simple equation. Can anyone help me figure out if I am going down the right path or if there's a whole other concept I'm missing to which I can apply her identity?
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$\begingroup$ pls show you effort to factor out $\omega_0$. $\endgroup$– JEBCommented Dec 7, 2020 at 21:06
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$\begingroup$ @JEB $\omega = \sqrt{\omega_0^2 (1- \frac{\gamma^2}{4\omega_0^2})} = \omega_0 \sqrt{1- \frac{\gamma^2}{4\omega_0^2}}$ $\endgroup$– D RamCommented Dec 7, 2020 at 21:10
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When you factor out the $\omega_o^2$ you get $\omega$ = $\omega_o \sqrt {1 - (\gamma ^2/\omega_o^2)/4}$. I cannot remember from my first year waves lecture but I think $\gamma ^2/\omega_o^2$ equals to some other important variable. Then you can use the Maclaurin series to simplify it out. Once you manage to do that, then get rid of all the ordered terms greater than and equal to 2. This should leave you with a simplified answer.
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$\begingroup$ $\frac{\gamma^2}{\omega_0^2} = \frac{1}{Q^2}$ where Q is the quality factor. However, why would I want to eliminate the ordered terms greater than and equal to 2? $\endgroup$– D RamCommented Dec 7, 2020 at 21:19
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$\begingroup$ my result is $\frac{\omega}{\omega_0} = \sqrt{1-\frac{1}{4Q^2}}$. if i let y = $-\frac{1}{4Q^2}$, then i can use the series. But my question still stands why do I eliminate the ordered terms greater than two ? $\endgroup$– D RamCommented Dec 7, 2020 at 21:21
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$\begingroup$ Suppose quality factor $Q$ = 2. Then $\gamma ^2/\omega_o^2$ = $1/4$. The second order term of this is $1/16$ which is small compared to $1/4$ and so on, hence, you can choose to ignore it. This is a tool that is used very often in Physics and is very helpful for simplification etc. $\endgroup$ Commented Dec 7, 2020 at 21:24
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$\begingroup$ Note, this only works if $\mid y \mid < 1 $ $\endgroup$ Commented Dec 7, 2020 at 21:27
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$\begingroup$ okay so following these steps i got $\frac{\omega}{\omega_0}$ = $\sqrt{1+\frac{-1}{4Q^2}}$ = $1-\frac{1}{8Q^2}-\frac{1}{128Q^4}$+... And if i only choose the first two terms, I get $\frac{\omega}{\omega_0} \approx 1-\frac{1}{8Q^2}$. In the context of my questions, this yields $\frac{\omega}{\omega_0} \approx 1$. Which makes $\omega \approx \omega_0$. This would lead me to conclude that damping did not change the angular frequency by any significant amount? $\endgroup$– D RamCommented Dec 7, 2020 at 21:37
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So you have:
$$ \omega = \omega_0\sqrt{1-\xi^2} $$
with:
$$ \xi = \frac c {2\sqrt{mk}} \equiv \frac 2 Q $$
where $c, k, m$ are the damping constant, spring constant, and inertia constant (aka: mass).
As pointed out, the radical is negative for $\xi > 1 $ ($Q<\frac 1 2$), this means you have an over damped oscillator that never oscillates.
The critical oscillator ($Q=\frac 1 2$) asymptotically approaches equilibrium. For any larger value, the oscillator is underdamped (see wikipedia figure).
For small $\xi$, or large quality factor $Q$, you can expand the radical:
$$ \sqrt{1-\frac 1 {4Q^2}} \approx 1 - \frac 1 {8Q^2} - \frac 1 {128Q^4}$$
You can drop the 2nd and higher term because the formula is a correction to the frequency, and at $Q=1/2$, it reads:
$$ 1 - \frac 1 2 - \frac 1 8 $$
but you don't even have a frequency..it's not oscillating, so a 12.5% error is not a concern.
The exact resonant frequency is of interest particularly when the oscillator is driven by a sinusoidal force with frequency $\omega$. In that case, the amplitude of the response is (as shown in the (wikipedia) figure):
$$ Z = \sqrt{(2\omega_0\xi)^2 + \frac{(\omega_0^2-\omega^2)^2}{\omega^2}} $$
and it is here that you can justify the dropping of higher order terms. The width of the resonances is much greater than the 2nd order correction to the frequency.
figure:
Moreover, in the real world oscillator where the frequency matters, $Q$ is big, $10^4$ to $10^6$ is common in quartz oscillators (so the 2nd order correction is $10^{-18}$ to $10^{-26}$).
This may also come up in mechanical systems, say positive aerodynamic feedback on an airplane control surface, or a bridge responding to wind. (This is a critical application: the resonance can destroy the structure). If the $Q$ factor is low, the system response to forcing is not very strong, and exists over a wide range of frequencies. If there is a high-Q resonance, then you need to now the frequency, but because it is high-Q, the 2nd order correction is negligible.
