0

The equation of a simple harmonic motion can be $x=A \cos(\omega t)$.

$\omega$ therefore has units of $radians/sec$.

I was solving some problems when I found a statement on my notes

$x=\left(1+\omega_{0} t\right)(e)^{-\omega_{0} t}$

with the variables having the usual meanings. I believe the statement is absurd because the left hand side has dimensions of distance and the right hand side has of radians (as $\omega t $ is in radians ) not to mention that the exponential has units of radians as well.

Could anyone please point out my mistake. I'd be glad even for a hint. Thank you.

Kashmiri
  • 1,390

1 Answers1

2

"Radians" aren't really units. E.g. $\frac{\pi}{2}$ (or $90^{\circ}$) is just a dimensionless number. That's why in physics/math we prefer "radians" to degrees.

$\mathrm{radians/second}$ is thus really $\mathrm{s^{-1}}$, in $\text{SI}$ units.

Your expression:

$$x=\left(1+\omega_{0} t\right)e^{-\omega_{0} t}$$

cannot refer to a distance, unless its RHS is multiplied by some factor $A$ with units $\mathrm{m}$.

Gert
  • 35,561