Peak of the amplitude resonance curve

[horserider37]

Hi everyone, I'm stuck on how to show the peak of the amplitude resonance curve is at w_d = w₀√(1-1/2Q^2), where Q = w₀/γ. My first instinct is to take a derivative of something and set = 0, but what eqn?Help?

 

[kuruman]

You are quoting a whole bunch of symbols. In what equation(s) do these symbols appear?

Is this a homework problem or part thereof?

 

[horserider37]

It's part of a textbook problem.

It's damped and driven harmonic oscillators - there's lots of equations but I think the relevant ones are x''+γx'+(w₀^2)x=F/m(cos(w_dt))
and
amplitude = A(w_d)=(F/m)/√(((w₀^2-w_d^2)^2)+((γw_d)^2))

but maybe there's others? I just confused in general. I don't know that I understand what the problem is asking.

 

[kuruman]

I will rewrite the amplitude equation in LaTeX (which you should learn how to use) and maybe you can see what's going on. $$A(\omega_d)=\frac{F/m}{\sqrt{(\omega_0^2-\omega_d)^2+(\gamma\omega_d)^2 }}.$$Can you just look at this expression and figure out for what value of $\omega_d$ it is maximized?

 

[vela]

Hi everyone, I'm stuck on how to show the peak of the amplitude resonance curve is at w_d = w₀√(1-1/2Q^2), where Q = w₀/γ. My first instinct is to take a derivative of something and set = 0, but what eqn?Help?

I'll assume that because you want to set a derivative equal to 0, you recognize that the word "peak" means maximum. What does "amplitude resonance curve" refer to? Answer that and you'll know what the problem is asking for.

 

[horserider37]

$$A(\omega_d)=\frac{F/m}{\sqrt{(\omega_0^2-\omega_d^2)^2+(\frac{\omega_0}{Q}\omega_d)^2 }}.$$

I think that's right. There was a missing square in your eqn.

No, I can't tell. Should I be able to? I feel like I'm missing something stupidly easy.

 

[horserider37]

When the denominator is smallest, right?. But how do I go from that to that formula?

 

[kuruman]

Going by "when the denominator is smallest" does not work in this case because you have two terms in the denominator with $\omega_d$. When $\omega_d$ increases, the first term in the denominator decreases while the second increases. Is there another, safer way to maximize the fraction as you vary $\omega_d$?

 

[horserider37]

I'm not sure what you mean by that. Like $$\omega_d<\omega_0$$ even as $\omega_d$ varies?

 

[sophiecentaur]

@horserider37 If you found a y on one side and an x on the other, would that take you into more familiar territory? dsomething by dsomethingelse is what you are after(?).

 

[Steve4Physics]

Hi @horserider37. I suspect you might benefit from a couple of hints.

To maximise $A(\omega_d)$ you need to find the value of $\omega_d$ which minimises the denominator in the RHS of your equation. For this purpose, the square root in the denominator can be ignored - do you see why?

So you need to find the value of $\omega_d$ which minimises
$(\omega_0^2-\omega_d^2)^2$$+(\frac{\omega_0}{Q}\omega_d)^2$

If we use some different symbols, the problem can be more familiarly expressed as finding the value(s) of x which minimise(s) $y(x) =(a-x^2)^2 + bx^2$.

How’s your calculus/algebra?!

 

[sophiecentaur]

How’s your calculus/algebra?!

Very difficult without it . However, a graph with chosen starting values with several points on it would give you a maximum value.

 

[horserider37]

Hi @horserider37. I suspect you might benefit from a couple of hints.

To maximise $A(\omega_d)$ you need to find the value of $\omega_d$ which minimises the denominator in the RHS of your equation. For this purpose, the square root in the denominator can be ignored - do you see why?

So you need to find the value of $\omega_d$ which minimises
$(\omega_0^2-\omega_d^2)^2$$+(\frac{\omega_0}{Q}\omega_d)^2$

If we use some different symbols, the problem can be more familiarly expressed as finding the value(s) of x which minimise(s) $y(x) =(a-x^2)^2 + bx^2$.

How’s your calculus/algebra?!

My algebra/calculus is horrible, hence the issue I think! I still don't understand how to find what minimizes that

 

[kuruman]

My algebra/calculus is horrible, hence the issue I think! I still don't understand how to find what minimizes that

Do you know at least how to take differentials? If so, when the denominator is at a minimum at a certain value of $\omega_d$ this means that $$d\left[\left(\omega_0^2-\omega_d^2\right)^2\right]+d\left[\left (\dfrac{\omega_0}{Q}\omega_d\right)^2\right]=0. $$

 

[Steve4Physics]

My algebra/calculus is horrible, hence the issue I think! I still don't understand how to find what minimizes that

As previously noted, the maths is equivalent to finding the value of x which minimises $y(x) =(a-x^2)^2 + bx^2$

Note that all occurences of $x$ are in the form $x^2$ so we can let $w = x^2$ and ask the simpler question: what value of $w$ minimises $y = (a-w)^2 + bw$?

Can you do that? You can do it with basic calculus or some algebra (completing the square). You will need to make an attempt and show your working - we are not going to do it for you!

 

[horserider37]

I got it! Thanks yall! I do *know* calculus, but I'm really bad with symbols and always overthink. I was severely complicating my derivatives lol, trying to make them much harder than they needed to be. Switching to a,b,c did the trick!

 

[sophiecentaur]

Switching to a,b,c did the trick!

Familiar territory.