# Spring constant of harmonic oscillator [closed]

I got a task from my lecturer to solve a differential equation for a simple harmonic oscillator: $$m{d^2\vec{r} \over dt^2}=-k^2\vec{r}.$$ So far, I have managed to find this equation only in one book. It is a bit bizarre for me as in most sources there is only $$k$$ without square. What do you think about it? Is this equation correct in some cases or does a book contain a mistake?

• It's an unusual notation but not incorrect. The author defines the spring constant as $k^2$ instead of $k$, and as long as that is clear it is OK. There's nothing wrong with that but there's nothing gained by it either. – Gert Oct 12 at 14:13
• The k is a name for a quantity to be measured. Changing the name to k^2 does not change the physics. – R.W. Bird Oct 12 at 18:26

Written as scalars, the DE looks like:

$$m\frac{\text{d}^2x}{\text{d}t^2}+k^2x=0$$

Or:

$$x''(t)+\frac{k^2}{m}x(t)=0$$

Set:

$$\omega^2=\frac{k^2}{m}$$

The DE then solves to:

$$x(t)=A\cos(\omega t+\varphi)$$

Where $$A$$ and $$\varphi$$ are determined from the initial conditions (not specified here).

Using this notation the angular velocity $$\omega$$ then becomes:

$$\omega = \frac{k}{\sqrt{m}}$$

$$\omega = \sqrt{\frac{k}{m}}$$
But as long as $$k$$ (or $$k^2$$) is well defined in each case, it doesn't alter the meaning of the expression.