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Homework Help: Damping I made an assumption I am not sure is right

  1. Jan 31, 2008 #1
    I now I have several threads but I need help with them all and thanks to who ever gives their time.

    1. The problem statement, all variables and given/known data
    A damped harmonic oscillator consists of a block (m=2.00kg), a spring (k=10.0N/m), and a damping force (F=-bv). Initially, it oscillates with and amplitude of 25.0cm; because of damping, the amplitude falls to three-fourths of this initial value at the completion of 4 oscillations.
    a. what is the value of b?
    b. how much energy has been lost during these four oscillations?

    2. Relevant equations
    X(t) = Ae^(-bt/2m)
    t = 4T
    T = 2π(m/k)^(1/2)
    E(t) = .5(k)(A^2)(e)^(-bt/m)

    3. The attempt at a solution

    a. Can I assume damping does not affect the period of each oscillation.
    t = 4T
    T = 2π(m/k)^(1/2)
    t = 4[2π(m/k)^(1/2)]
    E(t) = .5(k)(A^2)(e)^(-bt/m)
    (3/4)A = Ae^(-b(2π)(m/k)^(1/2)/2m) A are divided out
    (3/4) = e^(-b(2π)(m/k)^(1/2)/2m)
    In (3/4) = In e^(-b(2π)(m/k)^(1/2)/2m)
    b = [(2m)In (3/4)]/[-b(2π)(m/k)^(1/2)]
    = [(2x2.00kg)In (3/4)]/[-(2π)(2.00kg/10.N/m)^(1/2)]
    = .102kg/s or 102g/s

    b. Can I assume, once again, damping does not affect the period of each oscillation.
    t = 4T
    T = 2π(m/k)^(1/2)
    t = 4[2π(m/k)^(1/2)]
    E(t) = .5(k)(A^2)(e)^(-bt/m)
    = .5(k)(A^2)(e)^(-b2π)(m/k)^(1/2)/m)
    = .5(10.0n/m)(.250^2)(e)^(-.102kg/s2π)(2.00kg/10.0N/m)^(1/2)/2.00kg)
    = .176KJ

    Is my assumption wrong for this question?
    Last edited: Feb 1, 2008
  2. jcsd
  3. Jan 31, 2008 #2
    Your assumption is fine, but why are you using energy? Furthermore you want

    [tex]x(t) = A e^{-bt/2m} cos(\omega t + \delta)[/tex]

    where delta is the phase shift, which I think you can assume zero because there is no initial velocity, and [itex]\omega = \sqrt{(\omega_n^2 - (b/2m)^2)}[/itex].

    You want to use an energy equation for b) but I don't see why you would in a).
  4. Feb 1, 2008 #3
    I am using energy for b

    where would I use: where delta is the phase shift, which I think you can assume zero because there is no initial velocity, and [itex]\omega = \sqrt{(\omega_n^2 - (b/2m)^2)}[/itex]
  5. Feb 1, 2008 #4
    Oh, well you had an energy equation in your answer for a), so I didn't quite know what you were doing there.

    Really what you want to do is look at x(0) because that will be the greatest amplitude and x(4T) because that is when the amplitude is 3/4 of what it originally was. It looks like you did most of that, you just don't have your Cos term in there. It looks like the second part is correct for answer to the first part though.
  6. Feb 2, 2008 #5
    ok, If I add [itex]cos(\omega t + \delta)[/itex], [itex]\omega = 2 \pi / T[/itex] and [itex] \delta = 0 [/itex]it will be right

    or is [itex] \omega = \sqrt{(\omega_n^2 - (b/2m)^2)}[/itex] but what is [itex]\omega_n[/itex]
  7. Feb 3, 2008 #6
    Sorry, I forgot to define the terms.

    [tex]\omega_n = \frac{2 \pi}{T}[/tex]

    called omega_n because it is the "natural frequency," the frequency you would get from just a plain harmonic oscillator. Yes, if you put the cos term in there you should be good. Your previous way and this should actually produce similar answers.
  8. Feb 4, 2008 #7
    thank I will try
  9. Feb 6, 2008 #8
    a. t = 4T
    T = 2π(m/k)^(1/2)
    t = 4[2π(m/k)^(1/2)]
    X(t) = Ae^(-bt/2m) cos(w't + theta), b is small so w' = w, and theta is 0
    (3/4)A = Ae^(-bt/2m) cos(wt) A cancel
    .75 = e^(-bt/2m) cos(wt), w=(k/m)^(1/2)
    .75/[cos[(k/m)^(1/2)(t)]] = e^(-bt/2m), where t = 4[2π(m/k)^(1/2)]
    In .75/[cos[(k/m)^(1/2)(4[2π(m/k)^(1/2)])]] = (-b[4[2π(m/k)^(1/2)]]/2m)
    -[2m In .75/[cos[(k/m)^(1/2)(4[2π(m/k)^(1/2)])]] ]/ [4[2π(m/k)^(1/2)]]=b
    b=.808kg/s or 808g/s

    b. t = 4T
    T = 2π(m/k)^(1/2)
    t = 4[2π(m/k)^(1/2)]
    E(t) = .5(k)(A^2)(e)^(-bt/m)
    = .5(k)(A^2)(e)^(-b2π)(m/k)^(1/2)/m)
    = .5(10.0n/m)(.250^2)(e)^(-.808kg/s2π)(2.00kg/10.0N/m)^(1/2)/2.00kg)
    = 177kJ
  10. Feb 6, 2008 #9
    How is that, do I need to change something?
    Can I assume b is same from the energy formula only working for a small b, allowing me to say W' is about W
  11. Feb 7, 2008 #10
  12. Feb 7, 2008 #11
    Sorry, I put a subscript on [itex]\omega[/itex] that should not have been there. I confused myself on accident.

    [tex]\omega = 2 \pi/T[/tex]
    [tex]\omega_n = \sqrt{k/m}[/tex]

    When I said the two ways should produce similar answers, they should be the exact same. So if the phase, delta, is zero then you will have:

    [tex]Ae^{(stuff)}cos(\omega t)[/tex]

    and when you have 4T you get 8π, so the cosine term goes away. Now you said it went away because b was small, but when you look at it, you can't really make that assumption. b must be smaller by an order or two of magnitude before you can say that.

    Sorry I led you astray and gave you the wrong equation, but technically the way you did the first time was wrong because you weren't really using x(t), and I didn't see you reason through why the cosine would be gone. The reason I screwed up is because I knew the cosine term would go away, and I wanted you to just work through why it would.
  13. Feb 7, 2008 #12
    so why did the cos not go away?
    I used [tex]\omega_n = \sqrt{k/m}[/tex] and t=4T, but t = 4[2π(m/k)^(1/2)], so how do you get 8π. Should it not be 8(pi)w

    My first answer is right then?
  14. Feb 8, 2008 #13
    Since the angular frequency is 2π/T and if t is 4T then you would get 2π*4=8π.

    Yes, your procedure for getting b in the first problem is okay, now that the cosine part is sorted out, but I think you made a typo:

    b = [(2m)In (3/4)]/[-b(2π)(m/k)^(1/2)]
    = [(2x2.00kg)In (3/4)]/[-(2π)(2.00kg/10.N/m)^(1/2)]
    = .102kg/s or 102g/s

    you should have (if I did the mental math right)
    [tex]ln(3/4) = (-b(2 \pi)(m/k)^{(1/2)}/2m)[/tex]

    [tex]- ln(3/4)\sqrt{km} / \pi = b[/tex]
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