Lagrangian for a bead on a wire

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SUMMARY

The discussion focuses on deriving the Lagrangian for a bead of mass m sliding on a wire shaped as y = b cosh(x/b). The Lagrangian is established as ℒ = (1/2)m cosh²(x/b) ẋ² - mgb cosh(x/b). The Euler-Lagrange equation is applied to generate the equation of motion, which simplifies to ddot{x} = (1/b)cosh(x/b)sinh(x/b) - (g/b)sinh(x/b). The third part of the homework requires linearizing the differential equation around a stable point x = x₀ for small oscillations.

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vbrasic
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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}}.##

  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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vbrasic said:
I'm having trouble understanding what the third part of the question wants me to do.

What is the stable point with ##x = \mbox{constant}##? The problem asks you to linearise the differential equation around this point, i.e., if the stable point is ##x = x_0##, write ##x(t) = x_0 + \delta(t)## and write down the differential equation to leading (i.e, linear) order in the small quantity ##\delta(t)## and its derivatives.
 

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