The question:

A particle of mass $m$ is executing oscillation on the $x$-axis. Its potential energy is $U(x)= K|x|^3$, where $K$ is a positive constant. If the amplitude of oscillations is $a$, then its time period $T$ is proportional to what power of its amplitude?

My approach:

I differentiated potential energy to get the force, $$F=-\frac{dU}{dx} =3K\frac{|x|^3}{x}$$

But now we have a force that depends on an even power of displacement. This means that the force is not restoring and will continue to act in the same direction for all values of $x$. Thus the motion should not have been oscillatory and there should not have been a time period. Where am I wrong?

Note: The answer given is: proportional to $a^{-1/2}$.

I have now understood why the force is restorative. I have understood the expression of force:

$$F= 3Kx|x| $$

But I know that time period of an SHM is $2\pi\sqrt\frac{x}{a}$. But this is not SHM as the force is proportional to $x^2$, so how do I proceed further?


closed as off-topic by Aaron Stevens, John Rennie, Jon Custer, ZeroTheHero, Kyle Kanos Apr 5 at 11:16

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" – Aaron Stevens, John Rennie, Jon Custer, ZeroTheHero, Kyle Kanos
If this question can be reworded to fit the rules in the help center, please edit the question.

  • 2
    $\begingroup$ Are you sure the potential energy function is not $K|x^3|$? $\endgroup$ – Farcher Apr 4 at 10:48
  • $\begingroup$ @Farcher Yes it is. My fault. Edited. I dont know how it will affect the question, though. Will the differentiation be affected? $\endgroup$ – Arnav Upadhyay Apr 4 at 11:10
  • $\begingroup$ The change I have suggested produces a minimum for the potential energy function at $x=0$ whereas the original relationship produces a point of inflection at $x=0$. With the modulus the force is always restorative. $\endgroup$ – Farcher Apr 4 at 11:22
  • $\begingroup$ @ArnavUpadhyay The direction of the force will change because of the absolute value, it will become restorative (just sketch the potential!). I think you should specify the level you are at, and what kind of tools you might be expected to use to solve this question so that we can help you with answering it. $\endgroup$ – Void Apr 4 at 11:25
  • $\begingroup$ @Farcher I studied about absolute value differentiation and plotting the U vs x graph, I have understood that the force is restorative. Thanks. But I am still not able to find the time period. $\endgroup$ – Arnav Upadhyay Apr 4 at 11:46

This is a standard result about motion under a central force.

By conservation of energy, $U(r) + \frac 1 2 m \dot r^2 = U(a)$ or $$\left(\frac{dr}{dt}\right)^2 = \frac 2 m \left(U(a) - U(r)\right).$$ Inverting this and taking the square root we get $$\frac{dt}{dr} = \sqrt{\frac m 2} \frac{1}{\sqrt{U(a) - U(r)}}.$$ The time to go from $r = 0$ to $r = a$ is therefore $$\int_0^a \frac{dt}{dr}\,dr = \sqrt{\frac m 2}\int_{0}^a \frac{dr}{\sqrt{U(a) - U(r)}}$$ and this is $1/4$ of the period of oscillation.

This is a completely general result for any central force. Plugging in the OP's potential function is left as an exercise for the OP!


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