## Finding the ratio ω/ωo of an underdamped oscillator

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.
 Recognitions: Gold Member 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.
 Recognitions: Gold Member 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.

Recognitions:
Gold Member

## Finding the ratio ω/ωo of an underdamped oscillator

 Quote by HiggsBrozon 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 ω02-β2.
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)?
 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 = √(ω02-β2) (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) ω1/ω0 = √(ω02-β2)/ω0 = √(1-β2/ω02) = 1 - β2/(2ω02) = 1 - 1/(8pi2m2)

 Tags amplitude, mechanics, oscillator, underdamped