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Critically Damped Oscillator

  1. Mar 25, 2013 #1
    1. The problem statement, all variables and given/known data
    (A) A damped oscillator is described by the equation
    m x′′ = −b x′− kx .
    What is the condition for critical damping? Assume this condition is satisfied.
    (B) For t < 0 the mass is at rest at x = 0. The mass is set in motion by a sharp impulsive force at t = 0, so that the velocity is v0 at time t = 0. Determine the position x(t) for t > 0.
    (C) Suppose k/m = (2π rad/s)2 and v0=10 m/s. Plot, by hand, an accurate graph of x(t). Use graph paper. Use an appropriate range of t.


    2. Relevant equations
    For critically damped, β2 = w02
    where β = b/(2m) and w0 = √(k/m)

    3. The attempt at a solution
    Ok, for this problem, what I did initially was find the general form of position for a critically damped oscillator, which is:
    x(t) = (A + B*t)*e-β*t

    and the velocity function is:
    v(t) = -Aβe-βt + (Be-βt - Bβte-βt)

    Using the conditions given, I found:
    x(0) = A (obviously) which we don't know x(0)
    B = v0 + Aβ
    and x(t) can be rewritten as:
    x(t) = A(e-βt + βte-βt) + v0te-βt

    This is where I run into a wall. I can't seem to solve for A. I believe that x(0) should also be the max displacement since there is no driver for the impulse force, so A should be the max displacement, but this doesn't seem to get me anywhere. Any help on solving for A? I know how to do the rest other than that.
     
  2. jcsd
  3. Mar 25, 2013 #2

    TSny

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    Gold Member

    A "sharp impulsive force" is defined such that it instantaneously gives the mass an initial velocity without any displacement of the mass. So the mass is still at x = 0 immediately after the impulse.
     
  4. Mar 26, 2013 #3
    Ok, that makes sense. Thank you for the help.
     
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