91Ó°ÊÓ

To determine the potential energy function, we can integrate the force function with respect to x. Since the force term involving \(\alpha\) is simply \(F(x) = -kx + \frac{kx^3}{\alpha^2}\), we can integrate it directly: \(U(x) = -\int F(x)\, dx = -\int (-kx + \frac{kx^3}{\alpha^2})\, dx\) Now, we integrate the individual terms: \(U(x) = k\int x\, dx - \frac{k}{\alpha^2}\int x^3\, dx\) \(U(x) = \frac{1}{2}kx^2 - \frac{1}{4\alpha^2}kx^4 + C\) where C is the integration constant.

We know that the total energy \(E\) of the particle is conserved. In this case, the total energy can be written as the sum of the kinetic energy T(x) and the potential energy U(x): \(E = T(x) + U(x)\) Since \(T(x) = \frac{1}{2}mv^2\), where \(m\) is the mass of the particle and \(v\) is its velocity, we can rewrite the conservation of energy equation as: \(\frac{1}{2}mv^2 + \frac{1}{2}kx^2 - \frac{1}{4\alpha^2}kx^4 = E\)

Let's examine the motion when the total energy is given as \(E = \frac{1}{4}k\alpha^2\). We can substitute this value into the energy conservation equation: \(\frac{1}{2}mv^2 + \frac{1}{2}kx^2 - \frac{1}{4\alpha^2}kx^4 = \frac{1}{4}k\alpha^2\) Now, we can analyze the motion in terms of the turning points. Turning points occur when the kinetic energy term is zero, i.e., when the particle momentarily comes to a stop, and changes direction. In this case, we have: \(v = 0 \Rightarrow \frac{1}{2}kx^2 - \frac{1}{4\alpha^2}kx^4 = \frac{1}{4}k\alpha^2\) We can rearrange the equation as: \(\frac{1}{2}x^2(1 - \frac{x^2}{\alpha^2}) = \frac{1}{4}\alpha^2\) From this rearranged equation, we can see that there are multiple turning points when the particle's energy level is equal to \(\frac{1}{4}k\alpha^2\). When the particle's energy level is lower than this value, the particle is in a stable equilibrium. When the energy level is higher, the particle experiences an unstable equilibrium, where it escapes towards infinity.