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Homework Help: What is the value of the dampening constant?

  1. Feb 25, 2013 #1
    1. The problem statement, all variables and given/known data

    The shock absorber of a car has elastic constant k. When the car is empty, the design is such that the absorber is critically damped. At time t=0 the absorber is compressed by an amount A from its equilibrium position and released.
    a) If after 1 second the absorber compression is reduced to A/2, what is the value of the damping coefficient? (Note you will have to solve numerically an implicit equation).

    2. Relevant equations

    x(t) = (A+Bt)e-βt where β = ω0

    3. The attempt at a solution

    A(t) = Ae-βt
    1/2A = Ae-βt where t = 1s
    ln(1/2) = -β

    I know this can't be right because I'm suppose to arrive at an implicit equation. I can't seem to figure out where I'm going wrong at. Any help would be appreciated.
  2. jcsd
  3. Feb 25, 2013 #2


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    The general equation for the displacement is

    x(t) = (A+Bt)e-βt .

    At t=0, x=A. What is the value of B if the velocity is 0 at t=0 (that is, the derivative of x(t) has to be zero at t=0)?

  4. Feb 26, 2013 #3
    Thanks for you reply!

    x' = (-Aβ - Btβ + β)e-βt
    The velocity is 0 at t=0 so,

    0 = (-Aβ + β)(1)
    β=0 and A = 1

    Would I then go on to use t = 1 sec at A0 = A0/2
    where A0 = A+βt, the initial amplitude, or is A0 equal to the value I just solved for, A = 1?
  5. Feb 27, 2013 #4


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    β is the damping coefficient. It is a parameter of the problem. It does not depend on the initial condition.

    You have to fit the constant B to the initial condition V=0. There is a mistake in your derivation. Check the derivative of x(t) = (A+Bt)e-βt.

  6. Feb 27, 2013 #5
    After correcting my derivative I got,
    x' = (-Aβ - Btβ + B)e-βt
    Then using my initial conditions v = 0 at t = 0,

    0 = (-Aβ + B)(1)
    → B = Aβ

    so, x(t) = (A+Bt)e-βt where B = Aβ
    and then I can use A = A/2 when t = 1s correct?
  7. Feb 27, 2013 #6


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    Yes, it will be correct.

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