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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)??

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2 Answers

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Consider the positive quantity $X = (\omega - \omega_0)^2 (\omega + \omega_0)^2$. Let us make the approximation $\omega \approx \omega_0$:

  • In the first factor: we get $X_1 = 0$ and the relative error $\epsilon _X = |\frac{X_1-X}{X}| = 100 \%$.
  • In the second factor: we get $X_2 = 4 \omega _0^2 (\omega - \omega_0)^2$ and $\epsilon _X = | \frac{X_2-X}{X}| = |[2 \omega_0/(\omega + \omega_0)]^2 -1|$.

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.

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Short answer (rough idea): I think the basic idea here is that, while $\omega$ is close to $\omega_0$ (which is expected if you have weak damping), the sum of the two is a "large" quantity that you can freely approximate, but the difference $\omega-\omega_0$ is a small number. Actually, it is the very small number that you are trying to solve for, so clearly you cannot set it to zero!

Long answer (general techniques): In this case you can actually solve for $\omega$ exactly (solve the quadratic equation for $\omega^2$, then take another square root) and do a series expansion to see how it depends on $\beta$, but there is a more general set of techniques.

The way to rigorously establish results like this is asymptotic analysis. It is well worth studying, even if only to learn some basics (I'm far from an expert on it). The basic idea is to study how different quantities scale relative to each other. For example $\Delta\omega \approx 0$ is not a precise statement. What you really mean is $\Delta\omega$ is small, but small relative to what? You can establish this by studying scaling and limits. It turns out the correct statement is that $\Delta\omega \sim \beta$ as $\beta\to0$.

It is helpful to choose some reference scale and go to dimensionless variables relative to that scale. In this case $\omega_0$ is a natural choice (you could make a different one). Let's define $ \epsilon = \beta/\omega_0 $, and $\omega = \omega_0 + \Delta\omega = (1 + \delta) \omega_0$. In these variables your main equation becomes:

$$ \delta^4+4 \delta^3+4\epsilon^2\delta^2+4\delta^2 +8 \epsilon^2\delta-4 \epsilon^2=0 .$$

We can solve this equation easily when $\epsilon=0$. The solutions are $\delta=0,-2$. The $-2$ solution gives us negative frequencies which are unphysical, so we focus on the solutions in the neighbourhood of $0$. Since $\delta\to0$ as $\epsilon\to0$, in the limit $\delta\sim\epsilon^a$ for some positive power $a$.

Now we can apply the method of dominant balance. The idea is to satisfy the equation by balancing the largest terms against each other. There are naively 6 choose 2 = 15 ways of doing this, but not all of them are consistent. Proceed systematically. (You'll see this gets a bit dull. The advantage is the procedure is systematic. You can take shortcuts if you have intuition, but I'm showing the brute, dumb way.)

First try to balance the first two terms. This requires $\delta^4 \sim \delta^3$ which requires $4 a = 3 a$ in the limit $\epsilon\to0$. The only solution is $a=0$, but this violates the condition that all the other terms go away in the limit (in particular the $\delta^2$ term sticks around and can't be neglected). So this doesn't work.

Try the first and third term. This gives $4a=2+2a$, so $a=1$ and $\delta\sim\epsilon$ as $\epsilon\to0$. But now the $\epsilon^2$ terms dominate the limit. Scratch this.

The next interesting case is the first and fifth term. $4a=2+a$, $a=2/3$, $\delta\sim\epsilon^{\frac{2}{3}}$ as $\epsilon\to0$. Again other terms dominate.

The only case that works is fourth and last term. $2a=2$, $a=1$... this is the same scaling as before, but now we have the actual dominant terms. We can consistently drop all the other terms in the limit $\epsilon\to0$. Dropping all the other terms we get

$$ 4\delta^2 - 4\epsilon^2 = 0 \implies \delta = \pm \epsilon. $$

There are two solutions, as there should be, corresponding to the two solutions $\omega = \omega_0 \pm \beta + \cdots$ of the original equation.

Now if you want you can use this information to find a more accurate solution. Set $\delta = \pm\epsilon (1 + f)$ where $f$ is some thing that vanishes as $\epsilon\to0$. This is just the dominant behaviour, which we just worked out, times some general thing that doesn't modify the dominant behaviour in the limit $\epsilon\to0$. Now we substitute this in the equation, simplify it quite a bit (noting that in combinations like $1+\epsilon$ you can always drop the $\epsilon$ in the limit) and do dominant balance again to find the leading behaviour for $f$.

Lather. Rinse. Repeat. (Until thoroughly bored or your paycheck arrives.)

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thank you for your answers (both two!). I don't understand why do you say that $\delta \approx \epsilon ^a $, i see that since $\delta $ is defined as $\Delta \omega / \omega _0$, $\delta \approx \epsilon \iff \Delta \omega \approx \beta \omega ^{1-a}$ . Why? –  pppqqq Apr 1 '13 at 16:21
    
BTW, i tried a little different method, suggested by the introduction of $\delta,\epsilon$ variables: if you don't expand the $(1+\delta)^n$ terms you get the equation $$x^2-2(2\epsilon ^2 -1)x + 1-8\epsilon ^2=0$$ where $x=(1+\delta)^2$. Solving for $x$ you get $$x=1-2\epsilon ^2 \pm 2\epsilon \sqrt{\epsilon ^2 +1}$$. From this and with a Taylor expansion you get to the same answer: $$\omega = \omega _0 \pm \beta + O(\beta^2)$$. –  pppqqq Apr 1 '13 at 16:29
    
@Kazz8 I'm a little confused about how you got your $\beta \omega^{1-a}$ expression? I guess I didn't explain that the idea of $f\sim g$ is to express a scaling behaviour. It is a precise notion that the limit of the ratio $f/g$ approaches a constant in a certain limit. This is different from the fuzzier notion of $f \approx g$. I'd go into more detail, but sorry, I've got to go now. In the meantime here is a decent looking set of notes on asymptotic analysis that might help get you started. –  Michael Brown Apr 1 '13 at 17:00
    
sorry, it was $\beta ^a \omega ^{1-a}$. From: $$\delta=\Delta \omega / \omega , \qquad \epsilon = \beta / \omega $$. Thanks for the notes , i'll try to give a look. –  pppqqq Apr 2 '13 at 14:09
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