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How to solve these kind of questions , where $|F| \propto x^2$? How to find time period and velocity type related things to the oscillatory motion?

$$m\dfrac{d^2x}{dt^2}=F=-\dfrac{dU}{dx}=-3kx|x|.$$

But after this $$m\dfrac{d^2x}{dt^2}=-3kx|x|.$$

What is general solution of this ODE? I think it would give the $x$ in terms of $t$ and I would be able to get time period from it.

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  • $\begingroup$ But still if can anyone give me solution of this DE/ $\endgroup$
    – ABC
    Commented Apr 6, 2013 at 4:25
  • $\begingroup$ Where did you get this? @Mr.007 $\endgroup$
    – Saurabh Raje
    Commented Mar 26, 2014 at 14:41

3 Answers 3

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These kinds of proportionality questions are often best answered with dimensional analysis. You want to know a form a quantity with the units of time in terms of what you have.

You have a quantity $k$ with units $\frac{\text{Energy}}{\text{Distance}^3} = \frac{\text{Mass}}{\text{Distance} \times \text{Time}^2}$. You also have the mass $m$ (units of Mass) and amplitude $a$ (units of Distance). The only way you could possibly combine these quantities to get an answer with the units of time is if the expression is some pure number times $\sqrt{\frac{m}{ak}}$. So this tells you how the period must scale with respect to the dimensionful constants.

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  • $\begingroup$ Ok.It was good for this particular problem ! +1 thanx. $\endgroup$
    – ABC
    Commented Apr 6, 2013 at 3:41
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    $\begingroup$ Just to add to zkf, the Buckingham π theorem is a formal method to compute sets of dimensionless parameters from the variables. $\endgroup$
    – Isopycnal Oscillation
    Commented Apr 6, 2013 at 3:44
  • $\begingroup$ @zkf Good answer. Welcome to the site! $\endgroup$
    – Michael
    Commented Apr 6, 2013 at 4:09
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The dimensional analysis in zkf's answer completely solves the exercise. Nevertheless, it is possible to give a closed formula for the period

$$ T~=~ 4 ~\sqrt{\frac{m}{2k}} \int_0^a\! \frac{dx}{\sqrt{a^3-x^3}} ~\stackrel{x=au}{=}~ 4 ~\sqrt{\frac{m}{2ka}} \int_0^1\! \frac{du}{\sqrt{1-u^3}}. $$

Can you see why? Unsurprisingly, this just confirms zkf's answer.

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  • $\begingroup$ Dang you, I was just writing this up! :P The one extra bit I did (well, actually Mathematica did for me): the integral works out to $\sqrt{\pi } \Gamma \left(\frac{4}{3}\right)/\Gamma \left(\frac{5}{6}\right)\approx 1.40218$. $\endgroup$
    – Michael
    Commented Apr 6, 2013 at 8:48
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zkf gives you enough to answer this question but I would like to make a few extra points:

  • The absolute value operation in the potential makes this a nonlinear problem, which are generally pretty difficult to deal with. I got impatient waiting for Mathematica to come up with a closed form solution for $x(t)$, so there probably isn't one. This is the typical situation in physics and in this case you have to resort to a numerical calculation.

  • Since this is a one dimensional problem you can still solve it using the conservation of energy (you only need one conserved quantity to solve a single particle in 1D problem). If you write the conservation of energy for this system:

$$ \frac{1}{2} m \dot{x}^2 + k |x|^3 = E = \text{constant}, $$

you can rearrange this a bit into

$$ \mathrm{d}t = \frac{\mathrm{d}x}{\sqrt{\text{stuff involving x}}}. $$

Integrating up gives you the answer QMechanic posted just now. :)

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