Lagrangian of bead on a wire

Question

A bead of mass $m$ slides (without friction) on a wire in the shape, $y=b\cosh{\frac{x}{b}}.$

1. Write the Lagrangian for the bead.
2. Use the Lagrangian method to generate an equation of motion.
3. For small oscillations, approximate the differential equation neglecting terms higher than first order in $x$ and its derivatives.

I'm having trouble with the third part of the question. Here's my attempt at a solution:

It's pretty easy to find the Lagrangian. We have that potential energy of the bead is $$mgy=mgb\cosh{\frac{x}{b}}.$$ Kinetic energy is, $$\frac{1}{2}m(\dot{x}^2+\dot{y}^2).$$ But $$y=b\cosh{\frac{x}{b}},$$ so $$\dot{y}=\sinh{\frac{x}{b}}\dot{x}.$$ So the Lagrangian is, $$\mathcal{L}=\frac{1}{2}m(1+\sinh^2{\frac{x}{b}})\dot{x}^2-mgb\cosh{\frac{x}{b}}.$$ Simplifying gives, $$\mathcal{L}=\frac{1}{2}m\cosh^2{\frac{x}{b}}\dot{x}^2-mgb\cosh{\frac{x}{b}}.$$

To do the second part we find the necessary derivatives:

$$\frac{\partial\mathcal{L}}{\partial x}=\frac{m}{b}\cosh{\frac{x}{b}}\sinh{\frac{x}{b}}\dot{x}^2-mg\sinh{\frac{x}{b}}$$ $$\frac{\partial\mathcal{L}}{\partial\dot{x}}=m\cosh^2{\frac{x}{b}}\dot{x}$$ $$\frac{d}{dt}\frac{\partial\mathcal{L}}{\partial\dot{x}}=2\frac{m}{b}\cosh{\frac{x}{b}}\sinh{\frac{x}{b}}\dot{x}+m\cosh^2{\frac{x}{b}}\ddot{x}$$

So, the Euler-Lagrange equation says,

$$\frac{m}{b}\cosh{\frac{x}{b}}\sinh{\frac{x}{b}}\dot{x}^2-mg\sinh{\frac{x}{b}}=2\frac{m}{b}\cosh{\frac{x}{b}}\sinh{\frac{x}{b}}\dot{x}+m\cosh^2{\frac{x}{b}}\ddot{x}$$ So the equation of motion is given by, $$\frac{\frac{m}{b}\cosh{\frac{x}{b}}\sinh{\frac{x}{b}}\dot{x}^2-mg\sinh{\frac{x}{b}}-2\frac{m}{b}\cosh{\frac{x}{b}}\sinh{\frac{x}{b}}\dot{x}}{m\cosh^2{\frac{x}{b}}}$$ I'm having trouble understanding what the third part of the question wants me to do.

Answer 1

We need to study the oscillations of the bead about $x=0$, which is a stable equilibrium point. To do this, we assume $x$ is very small and close to $0$. Under these approximations, $\cosh(\frac{x}{b})=1$ and $\sinh(\frac{x}{b})=\frac{x}{b}$. We must also make an additional approximation that $\dot{x}^2$ is small compared to $\dot{x}$. Substituting these in our equations of motion, we get $$-mg\frac{x}{b}=m\ddot{x}$$ Which is that of a simple harmonic oscillator