Finding the ratio ω/ωo of an underdamped oscillator

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Homework Help Overview

The discussion revolves around finding the ratio of angular frequencies in an underdamped oscillator, specifically the ratio ω/ω0, given that the amplitude decreases to 1/e of its initial value after m complete oscillations. The problem involves understanding the equations governing underdamped motion and the implications of amplitude decay over time.

Discussion Character

  • Exploratory, Assumption checking, Mathematical reasoning

Approaches and Questions Raised

  • The original poster attempts to relate the amplitude decay to the equations of motion but expresses uncertainty about the next steps. Some participants question the assumptions regarding the constancy of angular frequency and the relationships between amplitude and time. Others suggest clarifying the amplitude equations and the implications of the period of oscillation.

Discussion Status

Participants are actively engaging with the problem, offering corrections and alternative perspectives on the amplitude relationships and the equations involved. There is a mix of interpretations regarding the behavior of angular frequency and the decay of amplitude, with some guidance provided on how to approach the problem further.

Contextual Notes

There are indications of potential misunderstandings regarding the equations and assumptions about the oscillator's behavior. The original poster's confusion about the changing angular frequency and the amplitude relationships is noted, along with the need for clarity on the definitions and parameters involved in the problem.

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Homework Statement



The amplitude of an underdamped oscillator decreases to 1/e of its initial value
after m complete oscillations. Find an approximate value for the ratio ω/ω0.

Homework Equations



x''+2βx'+ω02x = 0 where β=b/2m and ω0=√(k/m)

x(t) = Ae-βtcos(ω1t-δ) where ω1 has been defined as ω022

The Attempt at a Solution



The initial amplitude is equal to A0 = Ae-βt
and the final amplitude after m oscillations is equal to A0(1/e) = Af = Ae-(βt+1)

After this I honestly don't know where to go. I tried plugging in my Af into the underdamped motion equation and solving for ω but that didn't seem to make any sense. I'm assuming that ω0 will be equal to just √(k/m)?
Also, I thought that the frequency of an underdamped oscillator didn't change over time. So why would the angular frequency change?
If anyone could give me a push in the right direction that would be very helpful. I've been working on this for a quite while now.
 
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Normally, I think we'd have w^{2} = w_{0}^{2} - \beta ^{2}. I'm actually more accustomed to writing w^{2} = w_{0}^{2} - \frac{\gamma ^{2}}{4} , but regardless, you've goofed up your amplitude relationships.

After m periods, t = Tm where T is the period of oscillation in seconds. This also means the cosine term will have the same value as it does at t=0.

Thus, your equation should be A(t=Tm) = A_{0} e^{-1} = A_{0} e^{-\beta T m} \Rightarrow \beta T m = 1

See if you can go from there.
 
Also, what makes you think the angular frequency is changing? Even with your incorrect equation, I don't see how you would make that deduction.
 
HiggsBrozon said:

Homework Statement



The amplitude of an underdamped oscillator decreases to 1/e of its initial value
after m complete oscillations. Find an approximate value for the ratio ω1/ω0.

Homework Equations



x''+2βx'+ω02x = 0 where β=b/2m and ω0=√(k/m)

x(t) = Ae-βtcos(ω1t-δ) where ω12 has been defined as ω022.

First, without loss of generality, let δ = 0 since this is related to initial conditions & you weren't given any.

So the 1st (initial) amplitude is Aexp(-βt) with t=0.
And at the end of 1 cycle, which lasts t = ? seconds, what is the amplitude in terms of A, β and ω1? And so on 'till at the end of 5 cycles?

And what did the problem say the amplitude after 5 cycles was as a % of the first amplitude?
So how about an equation in β and ω1 plus the equation I corrected for you above (in red)?
 
Last edited:
My apologies on the late reply guys. It was a late night and I ended up finishing the problem the following morning. Thanks for your help!

For my equation of motion I used x(t) = Ae-βtcos(ω1t) = Ae-βmTcos(ω1mT)
where ω1 = √(ω022) (and yes my original formula was missing a square root.)
and T = 2pi/ω1

The cosine term is always equal to 1 so
x(t) = Ae-βmT = A/e
Therefore, βmT = 1
→ β = 1/(mT) = ω1/(2pim)

ω10 = √(ω022)/ω0 = √(1-β202) = 1 - β2/(2ω02) = 1 - 1/(8pi2m2)
 

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