I am trying to find the period of small oscillations of the potential

$$ V(x) = \frac{1}{2}m\omega_0^2(x^2-bx^4) $$

It is given that the particle oscillates between $-a$ and $a$ for some $a < \sqrt{1/2b}$, since $x^*=\pm\sqrt{1/2b}$ are the turning points of the systems.

Since the system oscillates between $-a$ and $a$, we know that the speeds at these points are zero, i.e. $\dot{x}(|a|) = 0$. Furthermore, from conservation of energy,

$$ \frac{1}{2}m\dot{x}^2 + V(x) = E, $$

which leads to

$$ V(|a|) = E $$

From this, we can write $$ \begin{align} \dot{x} &= \sqrt{ \frac{2}{m}(V(a) - V(x))}\\ \tau &= \int_{0}^{\tau}dt = 2\int_{-a}^{a}\frac{dx}{\sqrt{ \frac{2}{m}(V(a) - V(x))}} \end{align} $$

Plugging in $V(a) = \frac{1}{2}m\omega_0^2a^2(1-ba^2)$ and $V(x)$ into the above expression gives

$$ \tau = \frac{2}{\omega_0}\int_{-a}^{a}\frac{dx}{\sqrt{a^2-x^2}\sqrt{1-b(a^2+x^2)}} $$

Here is where I am stuck. I did a Taylor expansion of the second square root term, giving

$$ \sqrt{1-b(a^2+x^2)} = \sqrt{1+\tilde{x}^2} \approx 1 + \frac{1}{2}\tilde{x}^2 = 1 - \frac{b}{2}(a^2+x^2) $$

Plugging this back into the integral and making the substitution $x = a\sin\theta$ gives

$$ \tau = \frac{2}{\omega_0}\int_{3\pi/2}^{\pi/2}\frac{d\theta}{1-\frac{ba^2}{2}(1+\sin^2\theta)} $$

However, I don't see a way to take this integral. The final answer should be the normal period for a SHO, i.e. $\tau_0 = 2\pi/\omega_0$ with a small perturbation, so that the full solution is $\tau = \frac{2\pi}{\omega_0}(1+\frac{3}{4}ba^2)$. However, I am not seeing how you go from the integral to this answer.

  • $\begingroup$ The integral is known and can be looked up in a table. If you Taylor-expand the result around a (since a needs to be much smaller than b), you get the expected result (except for a negative sign, which I assume comes from the substitution. $\endgroup$
    – march
    Sep 13, 2019 at 17:14

2 Answers 2


Have you tried approximating $$ \frac 1{1-(ba^2/2)(1+\sin^2 \theta)}= 1+ (ba^2/2)(1+\sin^2 \theta)+O[(ba^2)^2] ? $$

  • $\begingroup$ Thank you, I've done the substitution and it gives the correct results, but there is a minus sign on it. Did I make a mistake somewhere in the substitution? $\endgroup$ Sep 13, 2019 at 17:20

You are almost there. The lower limit on the integral should be $-\pi/2$, and you can use the fact that $b$ is small to approximate the integrand as $$d\theta \ (1+\frac{ba^2}{2}(1+\sin^2\theta)).$$

  • 1
    $\begingroup$ Good, except that it's a that's small and not b. (Actually, a/b is what's small here). $\endgroup$
    – march
    Sep 13, 2019 at 17:15
  • $\begingroup$ Got it, if I replace $3\pi/2$ with $-\pi/2$ I get the correct result. $\endgroup$ Sep 13, 2019 at 17:23
  • $\begingroup$ @march: Thanks! I should have stated it like this: For the $x^4$ term to be considered a small perturbation, we must have $ba^4 << a^2$, or $ba^2 << 1$, which allows you to approximate the intergral as above. $\endgroup$
    – Paul G
    Sep 13, 2019 at 17:23

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.