Finding the ratio ω/ωo of an underdamped oscillator


by HiggsBrozon
Tags: amplitude, mechanics, oscillator, underdamped
HiggsBrozon
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#1
Feb19-13, 08:07 PM
P: 7
1. The problem statement, all variables and given/known data

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.


2. Relevant 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

3. 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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bossman27
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#2
Feb20-13, 04:16 AM
P: 204
Normally, I think we'd have [itex] w^{2} = w_{0}^{2} - \beta ^{2} [/itex]. I'm actually more accustomed to writing [itex] w^{2} = w_{0}^{2} - \frac{\gamma ^{2}}{4} [/itex] , but regardless, you've goofed up your amplitude relationships.

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

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

See if you can go from there.
bossman27
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#3
Feb20-13, 04:22 AM
P: 204
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.

rude man
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#4
Feb20-13, 06:50 PM
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Finding the ratio ω/ωo of an underdamped oscillator


Quote Quote by HiggsBrozon View Post
1. The problem statement, all variables and given/known data

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.

2. Relevant 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)?
HiggsBrozon
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#5
Feb25-13, 05:32 PM
P: 7
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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