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.
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.
Best Answer
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