7.6 Appplication - Mechanical System

Lecture notes

1Reading¶

Material related to this page, as well as additional exercises, can be found in ALA 10.6.

2Learning Objectives¶

By the end of this page, you should know:

• how to formulate a first-order system from Newton’s law
• the physical parameters from the equation of the system
• how to interpret the system behavior from its solution
• the different types of system behavior based on the eigenvalues and how it relates to the physical parameters

3Simple System¶

We’ll start with the simplest possible mechanical system: a single mass connected to a fixed support by a spring. Assuming no external force, Newton’s law of Force = Mass × Acceleration results in the following homogeneous second order scalar equation:

where $p(t) \in \mathbb{R}$ is the mass’ position over time, $\dot{p}$ its velocity, and $\ddot{p}$ its acceleration. Our first order of business is to convert (SM) to a first-order system. To do so, we define the vector $\vv x \in \mathbb{R}^2$ as

Then

Thus, the solutions to (SM) can be obtained by reading off the first component of $\vv x(t) = \bm p(t) \\ \dot{p}(t)\em$.

The matrix $A$ has imaginary eigenvalues $\lambda_{\pm} = \pm i \sqrt{\frac{k}{m}}$ (what physical intuition might have suggested this?), with corresponding eigenvectors

Therefore, solutions are weighted sums of

but, recalling we want real solutions, we write the general solution as

where $\vv x(0) = \begin{bmatrix} c_1 \\ c_2 \end{bmatrix} = \begin{bmatrix} p(0) \\ \dot{p}(0) \end{bmatrix}$. Thus we see that a single mass connected to a fixed support by a spring oscillates with frequency $\sqrt{\frac{k}{m}}$ in the absence of external forces like friction or damping.

4System with Friction¶

Next, we modify our problem to make it a bit more realistic to add friction, which is proportional to $\dot{p}$. Newton’s law then becomes:

Defining $\vv x = \bm p \\ \dot{p}\em$ as before, our first order reformulation becomes:

The eigenvalues of $A$ are the solutions to the characteristic equation:

There are three possible cases:

4.1Overdamped¶

If $\beta > 2\sqrt{mk}$ then this equation has two negative real roots

and thus the solution is given by a linear combination of two decaying exponentials

for $\vv v_{\pm}$ the corresponding real eigenvectors of $\lambda_{\pm}$. Such a system has so much friction that it no longer oscillates!

4.2Underdamped¶

If $0 < \beta < 2\sqrt{mk}$, then (9) has two complex-conjugate roots:

With a little bit of effort, we can compute the matrix exponential here as

so that $\vv x(t) = e^{-\mu t}\begin{bmatrix} \cos rt & \sin rt \\ -\sin rt & \cos rt \end{bmatrix} \begin{bmatrix} p(0) \\ \dot{p}(0) \end{bmatrix}$. In contrast to the overdamped setting, we see here that oscillations continue at fixed frequency $r = \sqrt{\frac{k}{m} - \frac{b^2}{4m^2}}$, while the entire trajectory also decays exponentially at rate $\mu = \frac{b}{2m}$. Thus for small friction, we eventually stop moving, but continue to oscillate as we go to the origin.

4.3Critically damped¶

The borderline case occurs when $\beta = \beta_* = 2\sqrt{mk}$, in which case our matrix $A$ has one eigenvalue $\lambda = -\frac{\beta}{2m}$, and is similar to the Jordan block

Using the techniques we saw in the previous lecture, we compute the matrix exponential

and thus our solution is