Simple Harmonic Motion - What are the units for $omega_0$?

I'm trying to understand the units in:

$$mx''+kx=0$$

And the general solution is

$$x(t)=A \cos(\omega_0 t)+B \sin(\omega_0 t).$$

Let $\omega_0 =\sqrt{\frac{k}{m}}$ - the unit for the spring constant $k$ is $kgms^{-2}$ or $Nm^{-1}$, where $m$ is in $kg$, so that the units of $\omega_0$ seem to be "per second" (i.e) $1/s$.

But, later we put $\omega_0$ in to the $cos$ and $sin$ functions which will return dimensionless ratios. So, The constants $A,B$ must be in $m$, since $x$ is in $m$.

What I don't understand is why my book says $\omega_0$ has the unit $rad/s$, I get that the input for cosine is $rad$ or some other angle measure, but where did the radians come from?

My analysis of units only proved $1/s$ as the actual units..!

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I have just been informed that radians are dimensionless. So, that answers part of this question, yet I still don't know why we can't say that dimensionless one in degrees or rotations..? How do I know what kind of cosine and sine table to use with this dimensionless number?

Answer

Ah, good question. The radian is actually a "fake unit." What I mean by that is that the radian is defined as the ratio of distance around a circle (arclength) to the radius of a circle - in other words, it's a ratio of one distance to another distance. For an angle of one radian specifically, the arclength $s$ is equal to the radius $r$, so you get

$$1\text{ rad} = \frac{s}{r} = \frac{r}{r} = 1$$

The units of distance (meters or whatever) cancel out, and it turns out that "radian" is just a fancy name for 1!

Incidentally, this also implies that "degree" is just a fancy name for the number $\frac{\pi}{180}$, and "rotation" is just a fancy name for the number $2\pi$.

This actually addresses the edit to your question. Suppose that you had some object oscillating at $\omega = \pi/4\frac{\mathrm{rad}}{\mathrm{s}} = 0.785\frac{\mathrm{rad}}{\mathrm{s}}$, and you wanted to evaluate its position after 10 seconds. To get the cosine term, you would plug the numbers in, getting

$$\cos\bigl(0.785\tfrac{\mathrm{rad}}{\mathrm{s}}\times 10\mathrm{s}\bigr) = \cos(7.85\text{ rad}) = \cos(7.85)$$

and then you would go to a trig table in radians (or your calculator in radian mode) and look up 7.85.

However, suppose that you were measuring $\omega_0$ in degrees per second instead of radians per second. You would instead have

$$\cos(45^\circ/\mathrm{s}\times 10\mathrm{s}) = \cos(450^\circ)$$

If you go look this up in a trig table given in degrees, you will get the same answer as $\cos(7.85)$. Why? Well, remember that the unit "degree" is just code for $\pi/180$, so this is actually equal to

$$\cos\bigl(450\times\tfrac{\pi}{180}\bigr)$$

And $450\times\frac{\pi}{180} = 7.85$, which is just $450^\circ$ converted to radians. So now you have the same value in the cosine, $\cos(7.85)$. Trig tables listed in degrees already have this extra factor of $\frac{\pi}{180}$ built into them as a convenience for you; basically, if you look up any number $\theta$ in a table that uses degrees, what you get is actually the cosine (or sine, or whatever) of $\theta\times\frac{\pi}{180}$.