Write $m\ddot{x}$ = $-k(t)x$ as a dynamical system.

Question

$m$ is the mass, $k(t)$ is the spring constant at time $t$, and $x$ is the displacement of the particle at time $t$.

This is my method:

Let

\begin{cases} x_1 = x \\ {x_2} = \dot{x} \\ {x_3} = t \end{cases}

This means we have $m\ddot{x}$ = $-k(x_3)x_1$. If we differentiate $x_1$, $x_2$ and $x_3$ we end up with:

\begin{cases} \dot{x_1} = x_2 \\ \dot{x_2} = \frac{-k(x_3)x_1}{m} \\ \dot{x_3} = 1 \end{cases}

But my textbook is saying it is

\begin{cases} \dot{x_1} = x_2 \\ \dot{x_2} = -k(x_3)x_1 \\ \dot{x_3} = 1 \end{cases}

Am I correct or is the textbook correct? Do we for some reason ignore the constant $m$ and if so why?

Answer 1

It seems you are correct and that the book has a typo.

Often we indicate $\frac k m =\omega ^2$ and in this case we have

$$\begin{cases} \dot{x_1} = x_2 \\ \dot{x_2} = -\omega^2(x_3)x_1 \\ \dot{x_3} = 1 \end{cases}$$

check whetheror not the book is assuming that notation.