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Homework Help: Finding the ratio ω/ωo of an underdamped oscillator

  1. Feb 19, 2013 #1
    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.
  2. jcsd
  3. Feb 20, 2013 #2
    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.
  4. Feb 20, 2013 #3
    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.
  5. Feb 20, 2013 #4

    rude man

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    Homework Helper
    Gold Member

    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: Feb 20, 2013
  6. Feb 25, 2013 #5
    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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