# Lagrangian for a bead on a wire

## Homework Statement

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 am having trouble with the third part.

## 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.

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