1. Not finding help here? Sign up for a free 30min tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

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

    ehild

    User Avatar
    Homework Helper
    Gold Member

    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)?


    ehild
     
  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

    ehild

    User Avatar
    Homework Helper
    Gold Member

    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.

    ehild
     
  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

    ehild

    User Avatar
    Homework Helper
    Gold Member

    Yes, it will be correct.

    ehild
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?
Draft saved Draft deleted



Similar Discussions: What is the value of the dampening constant?
Loading...