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

  • #1
zenterix
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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

1722741471793.png


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

As a concrete example, consider a seismograph.

1722741516026.png


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?
 
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  • #2
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.##
 
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  • #3
zenterix said:
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,
 
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  • #4
haruspex said:
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.
 
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  • #5
zenterix said:
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.
 
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