I have a question that I can reason physically but mathematically I am not sure if my approach is correct.

The amplitude of the oscillator is: $$A(\omega) = \frac{QF_{0}}{m}(\frac{1}{\omega_{0} \omega})\sqrt{R(\omega)}$$

with $$Q=\frac{\omega_0}{l}$$ and $$R(\omega) = \frac{(l \omega)^2}{(\omega_0 ^2 - \omega^2)^2 + (l \omega)^2}$$

So I have to prove what happens as a) $\omega \rightarrow 0$ and b) $\omega \rightarrow \infty$

So I know that for case a the amplitude will return to the natural amplitude of $\frac{F_0}{k}$ so if I do this mathematically I get $A(0) = \frac{F_0}{k}$ Which is what I want

For case b, I also know that the amplitude should go towards 0, is it sufficient to show that as the $\omega \rightarrow \infty$, the $\frac{1}{\omega_0 \omega}$ goes to zero, therefore the amplitude goes to zero.

Also for the record, this is not a homework question, it's just me trying to fully understand how to prove statements I have come across.


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  • $\begingroup$ Generally, when you take a limit of a product, you can't analyze separately factors. It is because of indetermined forms. So, the answer to b) question is no, you can't. $\endgroup$ – Reversal Apr 18 '16 at 11:34
  • $\begingroup$ Okay, so in that case the mathematics behind case a are wrong as well $\endgroup$ – user Apr 18 '16 at 11:36
  • $\begingroup$ Moreover, you have dimensional problems in the definition of $A$ and $R$. $\endgroup$ – Reversal Apr 18 '16 at 11:36
  • $\begingroup$ The definition of $A$ and $R$ were taken from books. Basically my reasoning behind the answers I expect to get are for part a, that no driving force => system goes back to natural state and part b very high driving force => system cannot respond fast enough and results in 0 net amplitude. $\endgroup$ – user Apr 18 '16 at 11:46
  • $\begingroup$ Sorry, I forgot a square term in R, part a now works correctly $\endgroup$ – user Apr 18 '16 at 11:53

I first thought, that you have a $\frac 0 0$ or $\frac\infty\infty$ in both cases, but it's wrong.
In (a) you get $R = l/\omega$ and in (b) too you can just naively insert $\infty$ for $\omega$ and get $R = \frac{(l\omega)^2}{\omega^4}$, and thus an $\frac 1{\infty^2}$ as the result.

But actually I think the physical meaning is more interesting and I'm not sure that you understand the results, you reasoning in the comment sounds not quite right.
To have a formula for $A(\omega)$ only makes sense if you mean the motion after a long time - the damping will then have destroyed any initial information of the motion, and you just have an oscillation with the driving frequency.
So $\omega\rightarrow 0$ does not mean "no driving force", it means the force is so slow, that the system is always in the equilibrium position (which is shifted by the force). So $\omega=0$ does not mean a zero displacement, it could just as well be a constant displacement of $F/k$ or any in between. That's why the question is about $\omega\rightarrow 0$ and not about $\omega=0$.

The reasoning in the case of fast oscillations could be:
Here as opposed to the slow case, where the phase shift is negligible, it will be maximal so as to translate maximal energy. The power translated from the driving force will be proportional to $Fv = F\omega x$. The power consumed by friction will depend on the velocity squared since the force is also proportional, so it will be $\gamma\omega^2x^2$. Since those must be equal, you see that $\omega x$ has to remain finite, i.e. $x\rightarrow 0$

  • 1
    $\begingroup$ No need for l'Hopital - terms simply cancel, depending on whether $\omega >> \omega_0$ or $<<$. $\endgroup$ – Floris Apr 18 '16 at 12:39
  • 1
    $\begingroup$ oh no... I was blinded by the comments to the question :) $\endgroup$ – Ilja Apr 18 '16 at 12:57
  • $\begingroup$ Happens to the best of us. $\endgroup$ – Floris Apr 18 '16 at 13:00
  • $\begingroup$ Thanks for taking the time to go into the reasoning, helped a lot $\endgroup$ – user Apr 18 '16 at 14:01

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