In the SHM equation $F= -kx$, $k =mw^2$ why not use $mf^2$ where $f$ is frequency $w$ here comes out to be $1/s$ not $\text{rad}/s$?

Question

The reason I am stating this is because on I found the units of $\omega$ to be equal to $\rm{s}^{-1}$ rather than the regular $\rm{rad/s}$.

$$F=-kx\to k= -F/x$$ $$\rm{\frac Nm}=\frac{\rm{kg\cdot m}}{s^2\cdot m}=\rm{\frac{kg}{s^2}}$$

If we take the book definition of $kx=m\omega^2x$ then we get

$$k=m\omega^2\to w^2= k/m$$

And the units of $\omega$ is then

$$\left(\rm{\frac{kg}{kg\cdot s^2}}\right)^{1/2}=\rm\frac1s$$

which is the unit for frequency.

This makes more sense to me when considering a spring where applying $w$(angular velocity) seems less effective than $f$(frequency).

But I'd like to know if I made any mistakes if yes then an explanation would be very appreciated.

Answer 1

$\rm{rad/s}$ and $\rm{s^{-1}}$ are the same unit. Radians are dimensionless.

Also in this case $\omega$ is an angular frequency, not an angular velocity. So you can use either $\omega$ or $f$. It doesn't matter. They are essentially the same thing. $\omega=2\pi f$

Answer 2

From SHM we know that $$F=-kx$$ And from mechanics we know that for every non-translational motion $$F=m\omega r$$ Comparing them we get, $$k=-m \omega^2$$ Βut, we know from Newton's 2nd law of motion $$F=ma$$ So, $$-kx=ma \Longrightarrow k=-(ma)/x$$ Putting the value of k in the previous equation we get, $$a=-\omega^2 x$$

Ηοpe this helps 😊😊