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## Homework Statement

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

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

## Homework Equations

$$\frac{\partial\mathcal{L}}{\partial x}=\frac{d}{dt}\frac{\partial\mathcal{L}}{\partial\dot{x}}$$

## The Attempt at a Solution

It's pretty easy to find the Lagrangian. We have that potential 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:

$$\frac{\partial\mathcal{L}}{\partial x}=\frac{m}{b}\cosh{\frac{x}{b}}\sinh{\frac{x}{b}}-mg\sinh{\frac{x}{b}}$$

$$\frac{\partial\mathcal{L}}{\partial\dot{x}}=m\cosh^2{\frac{x}{b}}\dot{x}$$

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

So, the Euler-Lagrange equation says,

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