Why is angular frequency $\omega = \sqrt{\frac{k}{m}}$ dimensionally correct?

Question

So I'm learning about simple harmonic motion, and I came to the part where the differential equation

$$\frac{\mathrm d^2x}{\mathrm dt^2} = -\frac{k}{m} x$$

is solved and simplified to

$$x(t) = A\cos(\omega t - \phi)$$

So here, I don't get why the angular frequency equals the following value

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

I tried to see if this has any evident reasonament to see why this is dimensionaly correct (especially with the square root). I already search for different posts here on Physics where it's explained, but the maths behind them are too complicated for me, and also they didn't answer why this is dimensionally correct.

Answer 1

Seeing that it is dimensionally correct should be easy. Just plug the units and check that they match.

$F=kx$, so $k$ must be newtons/meters. Remember that newtons are $kg\ m/s^2$. So, check that

$$ \frac{k}{m}=\frac{N/m}{kg}=\frac{kg/s^2}{kg}=\frac{1}{s^2}=s^{-2}$$

So the square root indeed has units of $s^{-1}$, which is angular frequency. Dimensionally, it fully makes sense.

If you ask about why this is like this, well, do not try to relate it to angular frequency so soon. You just check that the value $\sqrt{\frac{k}{m}}$ will appear very often, so you decide to give it a name. Let's call it $\omega$. You can call it with any other letter, but later on you will see that it is very related to an angular velocity, so it is a good name. Just that.

Answer 2

The units of $\omega$ are $s^{-1}$. One can see from hooks law that

$$ F= -kx =- \omega^2 m x \implies [\omega]^2 = \frac{N}{kg\ m} $$

Recall that $[N] = \frac{kg\ m}{s^2}$. Plugging in we get that

$$[\omega]^2 = \frac{1}{s^2} \implies [\omega] = \frac{1}{s}. $$

Answer 3

Here’s a different take that does not rely on getting the units of force, and works quite generally given a differential equation.

You are given $x(t)=A\cos(\omega t+\phi)$ and the differential equation $$ \frac{d^2x}{dt^2}=-\frac{k}{m}x \tag{1} $$ so just plug your solution into the left hand side of (1) to obtain $$ -\omega^2 x(t)=-\frac{k}{m}x(t)\, . $$ As this must hold for all values of $t$, it follows immediately that whatever the units of $\omega$, one must have $$ \omega^2=\frac{k}{m} \tag{2} $$ If (1) is dimensionally correct, then (2) must also be dimensionally correct. Now, the units of $\omega$ can be obtained directly from (1): the left hand side has units of $m s^{-2}$ since there is one power of length in the numerator, and two power of time in the denominators. The right hand side certainly has one power of length in the numerator, so it must be that, for consistency, $k/m$ has units of two powers of time in the denominator, or $s^{-2}$. Since $\omega^2$ is just $k/m$, the units of $\omega^2$ is $s^{-2}$, and thus the units of $\omega$ are $s^{-1}$