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Damped Harmonic Motion

  1. Apr 6, 2013 #1
    So my professor was discussing the case of a mass suspended from a vertical massless spring in some viscous liquid.

    He arrives at the equation of motion which was :


    x: + [itex]\frac{b}{m}[/itex]x. + [itex]\frac{k}{m}[/itex]x = 0

    x: is the second derivative of displacement wrt time. similarly x. is the first derivative.

    He then defined b/m = gamma k/m= w^2

    He then used the trial solution x=[itex]Ae^{t\tau}[/itex] formed an auxillary equation and solved it to get :

    [itex]\frac{-\gamma}{2}[/itex]±[itex]\sqrt{\frac{\gamma^2}{4}-w^2}[/itex]

    He then examined the discriminant of the above equation to formulate the general solution for light damping.

    I understand in light damping w^2 < (gamma^2)/4

    But how does he arrive at the following general solution :

    [itex]Ae^\frac{t\gamma}{2}cos(wt + \phi)[/itex]

    Where did the e^t*gamma/2 come from ? Why is there no sine function even though we have an imaginary root case ? Why is there a phi in there ?
     
  2. jcsd
  3. Apr 6, 2013 #2
  4. Apr 6, 2013 #3
    I dont understand what you mean. Can you be more specific ?
     
  5. Apr 6, 2013 #4
    The [itex] \phi [/itex] has essentially absorbed the Sine term. It is just another way to rewrite the more general solution (Which involves both Cosine + Sine functions). The exponential comes from the solution to the differential equations. If you understand the differential equation, everything will make complete sense.

    I hope I am shedding some light on the matter and not digging you a deeper hole. . .

    This is thoroughly explained in any 2nd year classical mechanics or quantum mechanics text.
     
  6. Apr 6, 2013 #5

    WannabeNewton

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    Hi elemis. You have to be a little careful with the notation there. So for the lightly damped case, the two roots we get are given by [tex]r_1 = -\frac{\gamma}{2} + i\sqrt{\omega_{n}^{2} - \frac{\gamma^{2}}{4}}, r_2 = -\frac{\gamma}{2} - i\sqrt{\omega_{n}^{2} - \frac{\gamma^{2}}{4}}[/tex] where ##\omega_{n}## is the natural frequency. So the solution in the complex plane is going to be [tex]r(t) = c_1e^{-\frac{\gamma}{2} + i\sqrt{\omega_{n}^{2} - \frac{\gamma^{2}}{4}}t} + c_2e^{-\frac{\gamma}{2} - i\sqrt{\omega_{n}^{2} - \frac{\gamma^{2}}{4}}t} = e^{-\gamma/2}(c_1e^{i\sqrt{\omega_{n}^{2} - \frac{\gamma^{2}}{4}}t} + c_2e^{- i\sqrt{\omega_{n}^{2} - \frac{\gamma^{2}}{4}}t}) [/tex] We then define the damped frequency ##\omega_{d}:= \sqrt{\omega_{n}^{2} - \frac{\gamma^{2}}{4}}##. Then, ##r(t) = e^{(-\gamma/2)t}(c_1e^{i\omega_{d}t} + c_2e^{- i\omega_{d}t})## so if we take the real part of this we get our response ##x(t) = e^{(-\gamma/2)t}(c_1\cos\omega_{d}t + c_2\sin\omega_{d}t)##. We can rewrite this as ##x(t) = Ae^{(-\gamma/2)t}\cos(\omega_{d}t - \phi)##, after defining ##A = \sqrt{c_{1}^2 + c_{2}^2}, \phi = \arctan(c_{2}/c_{1})## and using the cosine addition formulas.
     
    Last edited: Apr 6, 2013
  7. Apr 6, 2013 #6

    AlephZero

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    Are you sure about that? You can do the math that way, but the end result is messy (as in Wannabenewton's post).

    Things work out a lot nicer if you define ##k/m = \omega^2## and then ##b/m = \gamma \omega##. ##\gamma## is then dimensionless (just a number).

    (In fact if you define ##b/m = 2 \beta \omega## the math works out even nicer still).
     
  8. Apr 6, 2013 #7

    WannabeNewton

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    This was the way my professor defined it in fact. I do agree it works out nicer and it's less work for me since I don't have to change the terms in my notes when posting here :tongue2:
     
  9. Apr 6, 2013 #8

    AlephZero

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    The only problem is that some computer systems for dynamics analysis include the "2" and others don't.

    But in real life, you often don't know the level of damping to within a factor of 2, so it might not make much difference either way :devil:
     
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