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Finding the ratio ω/ωo of an underdamped oscillator 
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#1
Feb1913, 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'+ω_{0}^{2}x = 0 where β=b/2m and ω_{0}=√(k/m) x(t) = Ae^{βt}cos(ω_{1}tδ) where ω_{1} has been defined as ω_{0}^{2}β^{2} 3. The attempt at a solution The initial amplitude is equal to A_{0} = Ae^{βt} and the final amplitude after m oscillations is equal to A_{0}(1/e) = A_{f} = Ae^{(βt+1)} After this I honestly don't know where to go. I tried plugging in my A_{f} into the underdamped motion equation and solving for ω but that didn't seem to make any sense. I'm assuming that ω_{0} 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. 


#2
Feb2013, 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. 


#3
Feb2013, 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.



#4
Feb2013, 06:50 PM

HW Helper
Thanks
PF Gold
P: 4,757

Finding the ratio ω/ωo of an underdamped oscillator
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 ω_{1}? 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 ω_{1} plus the equation I corrected for you above (in red)? 


#5
Feb2513, 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^{βt}cos(ω_{1}t) = Ae^{βmT}cos(ω_{1}mT) where ω_{1} = √(ω_{0}^{2}β^{2}) (and yes my original formula was missing a square root.) and T = 2pi/ω_{1} The cosine term is always equal to 1 so x(t) = Ae^{βmT} = A/e Therefore, βmT = 1 → β = 1/(mT) = ω_{1}/(2pim) ω_{1}/ω_{0} = √(ω_{0}^{2}β^{2})/ω_{0} = √(1β^{2}/ω_{0}^{2}) = 1  β^{2}/(2ω_{0}^{2}) = 1  1/(8pi^{2}m^{2}) 


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