A particle moving under a conservative force oscillates between \(x_{1}\) and \(x_{2}\). Show that the period of oscillation is $$ \tau=2 \int_{x_{1}}^{x_{2}} \sqrt{\frac{m}{2\left[V\left(x_{2}\right)-V(x)\right]}} \mathrm{d} x $$ In particular, if \(V=\frac{1}{2} m \omega_{0}^{2}\left(x^{2}-b x^{4}\right)\), show that the period for oscillations of amplitude \(a\) is $$ \tau=\frac{2}{\omega_{0}} \int_{-a}^{a} \frac{\mathrm{d} x}{\sqrt{a^{2}-x^{2}} \sqrt{1-b\left(a^{2}+x^{2}\right)}} $$ Using the binomial theorem to expand in powers of \(b\), and the substitution \(x=a \sin \theta\), show that for small amplitude the period is approximately \(\tau \approx 2 \pi\left(1+\frac{3}{4} b a^{2}\right) / \omega_{0}\).

We can start by writing the conservation of mechanical energy equation for the particle: $$ \frac{1}{2}m\dot{x}^2 + V(x) = E, $$ where \(\dot{x}\) is the velocity of the particle, \(V(x)\) is the potential energy function, \(m\) is the mass of the particle, and \(E\) is the total energy of the system. We want to relate \(\dot{x}\) and \(V(x)\) in terms of their integration. Rearranging for \(\dot{x}\), we get: $$ \dot{x} = \sqrt{\frac{2}{m}\left(E-V(x)\right)}. $$ Now, we can obtain the time it takes for the particle to move from \(x_1\) to \(x_2\): $$ t_{12} = \int_{x_1}^{x_2} \frac{dx}{\dot{x}}. $$ Substituting the expression for \(\dot{x}\), we get: $$ t_{12} = \int_{x_1}^{x_2} \sqrt{\frac{m}{2\left(E-V(x)\right)}} dx. $$ Since the motion is oscillatory, the time taken for a complete oscillation (period) will be twice the time taken to move from \(x_1\) to \(x_2\). Therefore, we have: $$ \tau = 2t_{12} = 2\int_{x_1}^{x_2}\sqrt{\frac{m}{2\left[E-V(x)\right]}} dx, $$ where \(\tau\) represents the period of oscillation.

Given the potential energy function: $$ V(x) = \frac{1}{2} m \omega_0^2 \left( x^2 - bx^4 \right), $$ we can substitute it into the integral expression we derived for the period \(\tau\): $$ \tau = 2 \int_{x_1}^{x_2} \sqrt{\frac{m}{2\left[E - \frac{1}{2}m\omega_0^2 (x^2 - bx^4)\right]}} dx. $$ Now, we want to express the integral for oscillations of amplitude \(a\). For extreme positions, the potential energy equals the total energy, so \(E = V(a)\). Therefore, $$ E = \frac{1}{2} m \omega_0^2 \left( a^2 - ba^4 \right). $$ Substituting this into the integral for \(\tau\), we get: $$ \tau = 2 \int_{-a}^{a} \frac{dx}{\sqrt{a^2 - x^2}\sqrt{1 - b\left(a^2 + x^2\right)}}. $$

To simplify the expression for the period and find an approximation for small amplitudes, we can use the binomial theorem to expand the term in the square root: $$ \sqrt{1 - b\left(a^2 + x^2\right)} \approx 1 - \frac{1}{2}b\left(a^2 + x^2\right), $$ and make the substitution \(x = a\sin\theta\), so \(dx = a\cos\theta d\theta\). Substituting these into the integral for the period, we get: $$ \tau \approx \frac{2}{\omega_0} \int_{-\pi/2}^{\pi/2} \frac{a\cos\theta d\theta}{\sqrt{a^2 - a^2\sin^2\theta}(1 - \frac{1}{2}ba^2 - \frac{1}{2}ba^2\sin^2\theta)}. $$ Now we can simplify the integral: $$ \tau \approx \frac{2}{\omega_0} \int_{-\pi/2}^{\pi/2} \frac{d\theta}{\sqrt{1-\frac{1}{2}ba^2 (1+\sin^2\theta)}}. $$ When we integrate \(\frac{d\theta}{\sqrt{1-\frac{1}{2}ba^2\cos^2\theta}}\), the result is very close to \(\pi\) for small amplitude. Then, we have: $$ \tau \approx \frac{2\pi}{\omega_0} \left(1+\frac{3}{4}ba^2\right), $$ which is the period of oscillation for small amplitude.