1
$\begingroup$

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?

$\endgroup$

closed as off-topic by AccidentalFourierTransform, Gert, Prahar, user36790, CuriousOne May 13 '16 at 8:41

This question appears to be off-topic. The users who voted to close gave this specific reason:

  • "Homework-like questions should ask about a specific physics concept and show some effort to work through the problem. We want our questions to be useful to the broader community, and to future users. See our meta site for more guidance on how to edit your question to make it better" – AccidentalFourierTransform, Gert, Prahar, Community, CuriousOne
If this question can be reworded to fit the rules in the help center, please edit the question.

2
$\begingroup$

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.

$\endgroup$
  • $\begingroup$ I know that $x(t) = x_m * e ^ {-\alpha t} * cos(\omega t + \phi)$. What is it's derivative with respect to t? $\endgroup$ – Soroush khoubyarian May 12 '16 at 12:54
  • $\begingroup$ of what exactly? $\endgroup$ – danimal May 12 '16 at 12:54
  • $\begingroup$ I fixed the comment. Sorry. $\endgroup$ – Soroush khoubyarian May 12 '16 at 12:58
  • $\begingroup$ Use chain and product rule to find it $\endgroup$ – Ilya Lapan May 12 '16 at 12:59
  • $\begingroup$ I think I understood most of it, although I think I should start reading a little bit calculus to fully understand it. Thank you. $\endgroup$ – Soroush khoubyarian May 12 '16 at 13:13
0
$\begingroup$

This is a linear and homogeneous differential equation. That means you can produce new solutions by adding other solutions together and multiplying solutions by a constant. That means they form a vector space. I'll try to avoid over complicating things for you, but the idea is that there is an infinite number of solutions, you can make new solutions out of the old ones, but usually you are also given boundary conditions, certain values of your function at certain points. They restrict possible solutions, and usually, out of an infinite number of possible solutions, you are left with one single solution that satisfies them.

But the point is that if you can guess the most general form of the solution, see that it works and incorporates all the possible forms of solutions in it and all the possible boundary conditions, you can then say that you have the solution to the D.E. There they already give you the form of the solution, so no need guessing it. You could go through a some mental gymnastics to arrive at that solution for yourself (and danimal outlines the idea in his post pretty well), but if they have already told you the solution, which has constants $x_m$ and $\phi$ and thus incorporates boundary conditions into it, just plug it in and see for yourself that it works.

P.S. Have you looked at first order differential equations and some simpler examples first?

$\endgroup$

Not the answer you're looking for? Browse other questions tagged or ask your own question.