RLC Circuits - Q Factor and Amplitude

[wjdgone]

Homework Statement

Imagine you have two RLC circuits you are trying to scan for resonances. They have identical resonant frequencies, but circuit 1 has a very high Q-factor
(Q1 >> 1), and circuit 2 has a very low Q-factor (Q2 < 1). Let's assume you are already
on resonance and looking at V(out) on the oscilloscope, and you change the frequency in either direction for both circuits. How will the amplitude response differ between circuits 1
and 2 as you move the driving frequency away from resonance?

Homework Equations

Q=R/(2*pi*f*L) - not sure if I need this in the first place

The Attempt at a Solution

I don't think I understand what the problem means by "as you move the driving frequency away from resonance." My best blind stab in the dark for this problem is that Q1 means that R>2*pi*f*L and Q2 means that R<2*pi*f*L, and I can relate the inductance to the change in amplitude such that if inductance decreases, amplitude increases? (I'm also not sure how to relate inductance to amplitude.)

 

[BvU]

not sure if I need this in the first place

You can use it if you know the definiton of $Q$ and understand what they mean with $\Delta \omega$

 

[wjdgone]

You can use it if you know the definiton of $Q$ and understand what they mean with $\Delta \omega$

I still don't understand how I can determine what amplitude does.

 

[Merlin3189]

Since you are not given any values at all, I'd take this as a qualitative question rather than quantitative.
What you need to understand is what Q represents and how it relates to resonance behviour.
I expect many PF readers will be best able to get this from the formulae, but I find the simple notion of what Q represents physically, shown vividly in graphs of amplitude vs frequency, is the easiest way to understand what will happen in the situation described - and no calculations needed!