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Harmonic oscillator, friction, impulse

  1. Jan 4, 2007 #1
    1. The problem statement, all variables and given/known data

    A linear harmonic oscillator is subject to friction(stokes). The Oscillator gets an Impulse at the time t = 0 at the rest position. What is the equation of motion for the time interval 0 - t0?

    2. Relevant equations

    Friction force: Fr = -a * x'(t); a = constant (stokes friction)
    Impulse: F(t) = m *v0 / t0 (for 0 <= t <= t0)

    k = spring constant
    m = mass

    x(t = 0) = 0 (rest position at the time t = 0)
    x'(t = 0) = 0 (velocity at the time t = 0 is zero)

    3. The attempt at a solution

    I have a little trouble with the impulse here. My attempt at a solution was this:
    m*x'' = -k*x - a*x' + m*v0/t0 ==> x'' + (a/m)*x' + (k/m)*x = v0 / t0

    But my book claims that the equation really goes like this:
    x'' + (a/m)*x' + (k/m)*x = v0 / t0 * (1/m)

    Does somebody know why I am wrong and the book is right? I don't understand it, because I thought I needed a force(respectively acceleration) in above equation of motion. How does v0 / t0 * (1/m) fit into the equation, doesn't it have the wrong units?
    Last edited: Jan 4, 2007
  2. jcsd
  3. Jan 4, 2007 #2


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    I don't understand your notation. You say the Impulse is at time t = 0 but then you give it as a change of momentum divided by the whole time of the motion (t0).

    I would have interpreted the "impulse at t = 0" to mean the initial velocity was v0 (not 0) and the equation for time 0 <= t <= t0 was just mx" + ax' + kx = 0.

    But I learned this stuff 30 years ago, so the standard terminology might have changed since then.

    If v0 and t0 are a velocity and a time then you are correct, your units are consistent and the book apparently is not.
  4. Jan 4, 2007 #3
    Thanks for helping!

    As far as I understand it, the impulse lasts per definition for a certain time interval. In my example here, it goes from t = 0 to t = t0. In my book it is stated as like that: F(t) = m*v0/t0 (for 0 <= t <= t0).

    The velocity at t = 0 is clearly stated as 0. Maybe it would be more correct to write the impulse like that: F(t) = m*v0/t0 (for 0 < t <= t0). But I'm not sure about that. Also, x'(t = 0) = 0 is used to determine initial conditions later in the exercise.

    Yes, v0 and t0 are velocity and time (sorry if I was not precise enough). The Problem is just, if that's an error in the book its hard to believe! Because as the exercise goes on, this has a lot of consequences, and the term 'v0 / t0 * (1/m)' is used throughout the whole exercise...
    Last edited: Jan 4, 2007
  5. Jan 4, 2007 #4


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    OK, so he's assuming the impulse comes from a constant force for the time interval 0 <= t <= t0, and if it was applied to a free mass m it would produce velocity v0 at time t0.

    In which case, sorry, I don't understand why the 1/m is there.
  6. Jan 4, 2007 #5
    Thanks anyway. Maybe it's really an error.

    I take the opportunity to ask another question about this exercise.

    At one point, I have to decide whether 'a / (2*m)' is bigger or smaller or equal to 'sqrt(k/m)'. This is relevant when deciding if the oscillator is overdamped, underdamped or critically damped. Is there a way to decide with the information given what kind of oscillator it is? The book is assuming that the oscillator is underdamped, but it does not explain why...
  7. Jan 4, 2007 #6


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    The solutions of the equation of motion mx" + ax' + kx = 0 (with no external forces) are exponentials. The exponents are either real (overdamped case) or complex (underdamped, oscillating solutions).

    A trial solution x = e^st gives the auxiliary equation ms^2 + as + k = 0 to determine s. The condition for real or imaginary roots of this quadratic equation is that a^2 - 4mk is greater or less than 0. For the critically damped case a^2 - 4mk = 0, there is only one repeated root for s, and the two independent solutions are e^st and t.e^st.

    Look for a text on solutions of second order linear differential equations if you need more on this.

    Beware - some texts on vibration write this equation in the form mx" + 2ax' + kx. Take care with the 2's.
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