homework and exercises - Damped Simple Harmonic Motion Proof?

I was reading about damped simple harmonic motion but then I saw this equation:

$$-bv - kx = ma$$

$b$ is the damping constant. Then it said by substituting $dx/dt$ for $v$ and $d^2x/dt^2$ for $a$ we will have:

$$ m\frac{\mathrm d^2x}{\mathrm dt^2}+b\frac{\mathrm dx}{\mathrm dt}+kx=0 $$

Then it says the solution of the equation is: (this is my problem)

$$ x(t)=x_m \mathrm e^{-bt/2m}\cos(\omega't+\phi) $$

I don't understand the last part. How can we reach the $x(t)$? I know very little of calculus can you please explain how to solve this?

Answer

The differential equation you quote is fairly standard in university physics/engineering course but definitely requires some calculus to solve. As a first step, if you know how to differentiate products and chains, you can substitute the given solution into the differential equation and verify that it is indeed a solution. It contains two arbitrary constants (here $x_m$ and $\phi$) as you would expect of a second order differential equation (DE).

If you wanted to solve it, you still need some kind of guess as to what the function might look like; here the trial function would be:

$$ x(t) = Ae^{\lambda t} $$

and then substitute this parametrised solution into the original DE to obtain a quadratic in $\lambda$.

$$ x(t) = Ae^{\lambda t} \implies {dx\over dt} = \lambda Ae^{\lambda t} = \lambda x(t) $$ and $$ {d^2x\over dt^2} = \lambda^2 Ae^{\lambda t} = \lambda ^2x(t) $$

so that

$$ m\lambda^2 + b\lambda+k = 0. $$

as $x(t), A\neq 0$

Depending on the relative values of $m$, $k$ and $b$, you will get a quadratic with two, one or no real (i.e. one or two complex) solutions.

Given that you have two solutions $\lambda_1$ and $\lambda_2$, the intermediate result for $x(t)$ will be then

$$ x(t) = Ae^{\lambda_1 t} + Be^{\lambda_2 t} $$ For the solution you have been given, the corresponding quadratic in $\lambda$ will have no real solutions, and so the $\lambda$s will be complex, and the real part of the solution will give you the damped exponential at the front of the solution, and the imaginary parts will give you a wave-like term. $x_m$ and $\phi$ will be related to $A$ and $B$ and are determined by boundary conditions.