# Period of Small Oscillations for Perturbation on SHO

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,

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

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

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.

• 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. – march Sep 13 at 17:14

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 $$d\theta \ (1+\frac{ba^2}{2}(1+\sin^2\theta)).$$
• Good, except that it's a that's small and not b. (Actually, a/b is what's small here). – march Sep 13 at 17:15
• Got it, if I replace $3\pi/2$ with $-\pi/2$ I get the correct result. – Josh Pilipovsky Sep 13 at 17:23
• @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. – Paul G Sep 13 at 17:23
Have you tried approximating $$\frac 1{1-(ba^2/2)(1+\sin^2 \theta)}= 1+ (ba^2/2)(1+\sin^2 \theta)+O[(ba^2)^2] ?$$