Difference between maximizing amplitude by choosing omega_0 given omega_d versus choosing omega_d given omega_0.

[zenterix]

Homework Statement

Consider the expression

$$A=\frac{f}{\sqrt{(\omega_0^2-\omega_d^2)^2+\gamma^2\omega_d^2}}\tag{1}$$

Relevant Equations

For context, $A$ represents the amplitude of a solution to the equation

$$\ddot{y}+\gamma\dot{y}+\omega_0^2y=f\cos{\omega_d t}\tag{2}$$

If $\omega_0$ is given (by the nature of the physical system under consideration, for example the spring constant and mass of a simple pendulum) then $A$ can be thought of as a function of the driving angular frequency $\omega_d$.

We can differentiate (1) and find the $\omega_d$ that maximizes amplitude.

We find

$\omega_{d,max}=\sqrt{\omega_0^2-\frac{\gamma^2}{2}}\tag{3}$

Pictorially

What if we consider $\omega_d$ as fixed and consider $A$ a function of $\omega_0$?

As a concrete example, consider a seismograph.

The floor (the Earth) vibrates at a certain angular frequency that we can measure but not control. This drives the mass on the spring.

We want to choose $\omega_0$ such that the amplitude of the mass is the largest.

We have

$$A(\omega_0)=\frac{f}{\sqrt{(\omega_0^2-\omega_d^2)^2+\gamma^2\omega_d^2}}\tag{4}$$

and if we differentiate and equate to zero we find the solutions

$$\omega_0=0\tag{5}$$

$$\omega_0=\pm \omega_d\tag{6}$$

This is not what I expected.

After all, given $\omega_0$ we find that amplitude is maximized at an $\omega_d$ slightly smaller than $\omega_0$.

Given $\omega_d$, I would have expected that amplitude is maximized at a slightly larger $\omega_0$.

What am I missing here?

 

[kuruman]

You don't need to take derivatives. Just look at the ratio $$A=\frac{f}{\sqrt{(\omega_0^2-\omega_d^2)^2+\gamma^2\omega_d^2}}.$$It is maximum when its denominator is minimum. This clearly happens when $\omega_0=\omega_d.$

 

[haruspex]

What am I missing here?

What you are missing is that there is not really any basis to that reasoning.
It might help to consider other examples. Minimising $z=(x-y)^2+y^2$ wrt x gives $x=y$, but minimising wrt y gives $y=x/2$. Try sketching that as a surface,

 

[zenterix]

What you are missing is that there is not really any basis to that reasoning.

True, now that I think of it the question in the OP is quite silly.

$A$ is a function of both $\omega_d$ and $\omega_0$ and the graph is a surface.
Fixing one of the variables makes $A$ a function of the other variable. The two functions obtained this way do not have to be the same, and indeed in the case of $A(\omega_0, \omega_d)$ are not the same.

Glad I got that cleared up.

I asked the question at the end of the day. Sometimes you just need some sleep and a fresh new day.

That being said, this entire question arose from a problem solving video from the course "Vibrations and Waves.

At the very end, the lecturer discusses exactly what I asked in the OP. You can see it at around the time 1:08:30 of the video.

I think he made a small mess of his explanation and it turned out to be misleading. Indeed, the lecturer basically plots $A$ as a function of $\omega_d$ and proceeds to say that the $\omega_d$ that maximizes $A$ is slightly smaller than $\omega_0$. But, the context of the explanation was the seismograph in which we are not tuning $\omega_d$ but rather $\omega_0$.

At this point, what I understand is that if you are tuning $\omega_0$ then to maximize $A$ you must set it equal to $\omega_d$, the vibration angular frequency of the Earth in that location.

 

[hutchphd]

Glad I got that cleared up.

Yes. Also a more common example would be the tuner of any simple analog radio reciever. The knob typically changes a capacitance which tunes the resonant frequency to match that of the carrier for the incoming signal, maximizing the response. A very important and ubiquitous idea.