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I am reading about damped harmonic motion in my Physics book (Gerthsen Physik) and there are two things that irritate me:

Stokes friction

It says that Stokes friction would be $$F = -m \gamma \dot{x}.$$ To me, it does not make any sense why it depends on the mass of the object. It means that the deceleration does not depend on the mass. Or is it just since it is proportional to the volume and the volume usually depends on the mass.

linear combination of solutions

Then the differential equation is set up, I fully understand it, except the confusion about the friction.

$$ m \ddot{x} + \gamma \dot{x} + \omega_0^2 = 0 $$

With $$\delta = \frac{\gamma}{2}$$ and $$\omega = \sqrt{\omega_0^2 - \delta^2}$$ the solution to the $$x(t) = \exp(\lambda t)$$ approach are $$\lambda = -\delta \pm i \omega t$$.

One of the plus or minus variants are a valid solution, but the linear combination of both are all solutions. I understand that.

The book then says that the general solution is the combination of real and imaginary parts. Alternatively, one can add a phase angle and have one "compact" solution like this:

$$ x(t) = x_0 \exp(-\delta t) \exp(i ( \omega t + \phi ) ) $$

The $\exp(i \phi)$ certainly makes up a linear combination of real and imaginary parts, but how is it a combination like $$c_1 \exp(i\omega t) + c_2 \exp(-i\omega t) ?$$

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Can you explain a bit where you get $F = -m \gamma \dot{x}.$ from? Usually Stokes friction is defined for liquids and my issue of Gerthsen (1999) defines it as $F= 6\pi\eta r \dot{x}.$, which does not depend on the mass. –  Alexander Feb 3 '12 at 12:33
    
My edition (24th) says: “Wir betrachten eine Reibungskraft $F_R = -m \gamma$, die proportional zur Geschwindgkeit ist, also z.B. eine Stokes Reibung im zähem Öl” –  queueoverflow Feb 3 '12 at 12:41
    
As Stokes found his expression for spherical objects in fluids and there is no physical mechanism how the friction between the fluid and the sphere can depend on its mass I am pretty sure you missed something or there is a typo in the book. Also, I looked up the 24th edition and it says on page 61: $F_R = 6 \pi \eta r v$. –  Alexander Feb 3 '12 at 12:50
    
They are hiding the exact physics of the friction in $\gamma$, to simplify the equations with $\gamma \propto 1/m$. –  Alexander Feb 3 '12 at 13:03

3 Answers 3

up vote 1 down vote accepted

For the Stokes friction, the book is just choosing to normalize the coefficient with the mass. Since the coefficient of linear friction is arbitrary, you can divide it by the mass if you like, and then it has units of decay time. The parameter $\gamma$ is traditionally taken to be the decay time.

The reason the combination of the two terms into a phase-shift works is the addition law for cosines:

$$ cos(A+B) = cos(A)cos(B) - sin(A)sin(B) $$

which is simply the real part of the more obvious formula:

$$ e^{i(A+B)} = e^{iA}e^{iB} $$

and the imaginary part of the above formula reproduces the law of addition of sines. When you split cos(\omega t + \phi) using the addition rule, you find the two separate oscillations from the general solution.

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EDIT: The answer below ignores the $F = -m \omega^2_0 x $ term and thus it only applies to problems with only damping an no elasticity.

Here is the direct way of dealing with this problem without having to guess at as solution form.

$$ F = m \ddot{x} $$ $$ F = m \ddot{x} = -m \gamma \dot{x} $$ $$ \dot{v} = -\gamma v \text{ with } v = \dot{x} $$ $$ \frac{\rm d}{{\rm d} t}v=-\gamma v$$ $$ \frac{1}{v}{\rm d}v = -\gamma\, {\rm d} t$$ $$ \int{\frac{1}{v}{\rm d}v} = -\int{\gamma\, {\rm d} t}$$

If inital conditions are $v(0)=v_0$ then $$ \int_{v_0}^{v} \frac{1}{v}{\rm d}v = -\int_0^t \gamma\, {\rm d} t $$ $$ \ln\frac{v}{v_0} =-\gamma t $$ $$ v = v_0 \exp({-\gamma t}) $$ if $x(0)=x_0$ then $$ x = x_0 + \int_0^t v {\rm d} t $$ $$ x = x_0 + \frac{v_0}{\gamma}\left(1-\exp(-\gamma\, t)\right) $$

So if you can produce a system with form $\ddot{x}=-\gamma\,\dot{x}$ then the solution uses $\gamma$ of the coefficient in the exponential decay function.

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This is no good, since it is ignoring with the oscillator force. Further, it is no good because the "guessing" is not a guess at all, it is a consequence of translation invariance that a linear combination of exponentials will work to solve the problem. –  Ron Maimon Feb 16 '12 at 7:40
    
@RonMaimon - Is there a driving force here that I am missing? I does not seem from the initial posting. Also, I never implied that "guessing" is inherently wrong, just that there is an alternative for those who feel uncomfortable with it. –  ja72 Feb 16 '12 at 16:15
    
You aren't missing the driving force, you are missing the oscillator force! There is a restoring force linear in the position, not just a damping linear in the velocity. You have given the solution for a damped free motion, not a damped oscillator. Further, using exponentials to find the solution is not "guessing", it is part of a more comprehensive mathematical theory than your ad-hoc piddling around. –  Ron Maimon Feb 16 '12 at 18:51
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@RonMaimon - Chill out man. –  ja72 Feb 17 '12 at 0:25
    
I'm not freaking out, you just should edit the answer. –  Ron Maimon Feb 17 '12 at 9:08

I think that is just one solution ($e^{i\omega t}$ part, where $c=e^{i\omega t}$ or sth.)

Probably what he means is $\sin(\omega t+\phi)$ or $\cos(\omega t+\phi)$ instead of $\exp(\omega t+\phi)$ (which are real functions that relate to "real" motion.)?

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