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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?

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  • $\begingroup$ 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. $\endgroup$ – Gert Oct 12 at 14:13
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    $\begingroup$ The k is a name for a quantity to be measured. Changing the name to k^2 does not change the physics. $\endgroup$ – R.W. Bird Oct 12 at 18:26
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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:

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

Instead of the conventional:

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

But as long as $k$ (or $k^2$) is well defined in each case, it doesn't alter the meaning of the expression.

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