Homework 1 AM212

Solution to a nonlinear damping problem involving a mass-spring system with cubic velocity-dependent damping. The work demonstrates the non-dimensionalization process for the governing differential equation and derives the dimensionless parameter ε in terms of the system’s physical constants (mass M, spring constant Λ/L, and damping coefficient K).

Author

Affiliation

Kevin Silberberg

University of California Santa Cruz

Bush problem 4 page 23

A mass-spring system is subject to a small damping proportional to the cubed power of speed. The governing equation:

\[M\frac{d^2x}{dt^2} = -\frac{\Lambda}{L}x -K\bigg(\frac{dx}{dt}\bigg)^3 \tag{1}\]

Non-dimensionalize the Equation 1 and express \(\epsilon\) in terms of the constants of the system.

Solution

Let us non-dimensionalize Equation 1

Let \[\begin{align} \hat{x} &= \frac{x}{x_s} \Longrightarrow x = \hat{x}x_s \\ \hat{t} &= \frac{t}{t_s} \Longrightarrow t = \hat{t}t_s \end{align}\]

Plugging in for \(x\) and \(t\) into equation Equation 1 we get

\[\begin{align} \frac{Mx_s}{t_s^2}\frac{d^2\hat{x}}{d\hat{t}^2} + \frac{\Lambda x_s}{L}\hat{x} + \frac{Kx_s^3}{t_s^3}\bigg(\frac{d\hat{x}}{d\hat{t}}\bigg)^3 &= 0 \\ \frac{d^2\hat{x}}{d\hat{t}^2} + \frac{\Lambda t_s^2}{ML}\hat{x} + \frac{K}{M}\frac{x_s^2}{t_s}\bigg(\frac{d\hat{x}}{d\hat{t}}\bigg)^3 &= 0 \\ \end{align}\]

Let

\[\begin{align} \frac{\Lambda t_s^2}{ML} &= 1 \Longrightarrow t_s = \tau = \sqrt{\frac{ML}{\Lambda}} \\ \frac{x_s^2K}{M} &= 1 \Longrightarrow x_s = L_c = \sqrt{\frac{M}{K}} \end{align}\]

thus the non-dimensionalized equation becomes,

\[\frac{d^2\hat{x}}{d\hat{t}^2} + \hat{x} = -\epsilon\bigg(\frac{d\hat{x}}{d\hat{t}}\bigg)^3\]

where \(\epsilon = \frac{1}{\tau}\)