Lagrangian for a bead on a wire

In summary, the stable point is when the bead slides along the wire without any oscillations. The bead will stay at this point as long as the wire is held at a constant temperature.
  • #1
vbrasic
73
3

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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  • #2
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.
 

FAQ: Lagrangian for a bead on a wire

1. What is the Lagrangian for a bead on a wire?

The Lagrangian for a bead on a wire is a mathematical formula that describes the motion of a bead along a wire. It takes into account the potential energy of the bead due to gravity and the kinetic energy of the bead as it moves along the wire.

2. How is the Lagrangian for a bead on a wire calculated?

The Lagrangian for a bead on a wire is calculated by taking the difference between the total kinetic energy and the total potential energy of the system. This difference is known as the Lagrangian function.

3. What are the variables in the Lagrangian for a bead on a wire?

The variables in the Lagrangian for a bead on a wire include the position of the bead along the wire, the velocity of the bead, and the parameters of the wire such as its length and tension.

4. How is the Lagrangian for a bead on a wire used in physics?

The Lagrangian for a bead on a wire is used in physics to analyze and predict the motion of the bead along the wire. It is a powerful tool in classical mechanics and is often used in the study of oscillatory motion and pendulum systems.

5. Are there any assumptions made in the calculation of the Lagrangian for a bead on a wire?

Yes, there are a few common assumptions made in the calculation of the Lagrangian for a bead on a wire. These include assuming a frictionless wire, neglecting air resistance, and assuming a uniform wire with no variations in thickness or curvature.

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