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Damped harmonic oscillator

  1. Sep 17, 2012 #1
    1. The problem statement, all variables and given/known data
    The displacement amplitude of a lightly damped oscillator with m=0.250kg and k=6400N/m is observed to decrease by 15% in exactly five minutes
    a) Calculate the fraction (in%0 of the initial mechanical energy of the oscillator that has been converted to other forms of energy (such as thermal energy) in the five-minute interval.
    b)Calculate the Q value of this damped oscillator by first calculation omega initial and gama


    2. Relevant equations
    omega initial = sqrt(k/m)
    gama = b/m
    x = A initial exp^(gama t/2)
    TE = 1/2kA initial exp^(-gamat)
    A(t) = A inital exp^(bt/2m)
    T = 2pisqrt(m/k)


    3. The attempt at a solution
    I have calculated omega inital = 160rads and the period to be .04s. I know there is the relationship between the displacement amplitude and the total energy but I cannot seem to figure out how to relate them. Any help is greatly appreciated.
     
  2. jcsd
  3. Sep 18, 2012 #2
    At maximum displacement, what are the kinetic and potential energies of the oscillator?
     
  4. Sep 18, 2012 #3
    At maximum displacement, the kinetic energy should be zero. Should I be able to solve for the potential energy? I am so confused as to how to approach this. I know I have been given enough info but I feel like I don't have enough to find any additional values.
     
  5. Sep 18, 2012 #4
    So at maximum displacement the entire energy is the potential energy. What is the potential energy of a spring?
     
  6. Sep 18, 2012 #5
    1/2ka^2. How do i find this without being given a?
     
  7. Sep 18, 2012 #6
    What is a? What is amplitude?
     
  8. Sep 18, 2012 #7
    Ok, so I finally figured out I can find the % of TE in relation to the % of A lost. Which I can also use to find b which I can then use to find Q.
    A=Ainital exp^bt/2m where A/Ainital will equal .85 then the only unknown is b.
    And TE goes as A^2 so if A is 85% then TE is .85^2 or ~72%
     
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