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Oscillation Problem w/Damping

  1. Apr 8, 2009 #1
    1. The problem statement, all variables and given/known data
    The suspension system of a 2200 kg automobile "sags" 14 cm when the chassis is placed on it. Also, the oscillation amplitude decreases by 55% each cycle. Estimate the values of (a) the spring constant k and (b) the damping constant b for the spring and shock absorber system of one wheel, assuming each wheel supports 550 kg.

    2. Relevant equations
    2vim3cg.jpg


    3. The attempt at a solution
    I found k by making mg=-kx
    k=38500 N/m
    part B i found it to be 876 kg/s.
    However, this answer is wrong.
    I'm not sure what's wrong here but the formula I used works with everyone elses example.
    e ^ (-bt / 2m) = 55/100
    I found T to be 0.751 using T=2pi radical (m/k)
     
  2. jcsd
  3. Apr 8, 2009 #2
    What did you use for t when solving for b?
     
  4. Apr 8, 2009 #3
    The oscillation damping rate is BY 55% every cycle, hence we have:

    [tex]x(t+T)=x(t)-0.55\cdot x(t)=0.45\cdot x(t)[/tex]
     
  5. Apr 8, 2009 #4
    To NBAJam100: I used the T I found
    To Thaakisfox: I don't really understand what you wrote. It looks confusing. Not sure how to apply it.
     
  6. Apr 8, 2009 #5
    Well you want to find the ratio of the amplitudes between cycles.
    At time t let the position be x(t). Let the period of oscillation be T. Then the position after one cycle will be: x(t+T).

    But we know the x(t) function, so:

    [tex]\frac{x(t+T)}{x(t)}=e^{-bT/2m}[/tex]

    But it is also given that the amplitude decreases BY 55% every cycle, so:

    [tex]x(t+T)=x(t)-0.55x(t)=0.45x(t) \Longrightarrow \frac{x(t+T)}{x(t)}=0.45[/tex]

    Combining these you will get the result...
     
  7. Apr 8, 2009 #6
    Oh, I see. That makes a lot of sense. Then I just plug it in and I got 1170kg/s. My confusion with the capital and lower t's got me mixed up. Thankyou very much. =]
     
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