Solutions of damped oscillator differential equation

Question

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 $$c_1 \exp(i\omega t) + c_2 \exp(-i\omega t) ?$$

Answer 1

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.

Answer 2

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.)?

Answer 3

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.

Answer 4

The solutions are simply the complementary functions obtained on solving differential equations please refer some engineering mathematics books for this topic