Do the units in this equation make any sense? I've recently asked a question regarding the function
$$x(t)=A_0e^{-\frac{b}{2m} t} \cos{(\omega t+\varphi)}.$$
It is a solution to the damped pendulum equation. I found it here, where $A_0$ is the initial angular displacement of the pendulum, $b$ is the damping constant, $m$ is the mass of the pendulum, $\omega$ is the natural angular frequency of the damped pendulum and $\varphi$ is the phase of the function.
My question is, how do the units "add-up" here? I know $x(t)$ should return an angular displacement given time. But, I am confused as to how the unit in the exponential factor would affect the unit of the result. I've read from this post that units in the exponent don't make sense, which furthers my confusion.
Any suggestions would be great. Thanks!
 A: As the post you've linked says, any argument of functions such as $\cos$ and $\exp$ must be dimensionless. That is to say, $bt/m$ is dimensionless, as are $\omega t$ and $\varphi$. These parts of the function cannot affect the units of the result.
The dimension of $x(t)$ comes from $A_0$, which has dimensions of 'angle' (which is really dimensionless).
A: The function $\exp(x)$  is usually define as dimensionless and it is define only for dimensionless number $x$. In your case,  $−b\cdot t/2m$ is dimensionless and so is $e^{−b\cdot t/2m}$. Similarly, $ωt+φ$ and $\cos(ωt+φ)$ are dimensionless.
In conulsion, you get that the units of $x(t)$ are just the same units as the units of $A_0$.
A: A pendulum doesn't use exactly the same quantities as the link you posted. Don't use an equation without examining its context and limitations!
Examining the traditional damped HO differential equation for a simple pendulum with small angle approximation:
$$m r^2\frac{\mathrm{d}^2 x}{\mathrm{ dt}^2}+\frac{b}{2}\frac{\mathrm{d}x}{\mathrm{dt}}+mgrx=0$$
we see that $b$ must have units of torque$\times$ time / displacement.
In your situation, your displacement is radians (angular displacement), so the SI units of $b$ are newton-meter--seconds per radian.
In your exponent you have $\frac{bt}{2m}$ and the units would be newton$\cdot$meter$\cdot$seconds$^2$ per radian or meter$^2$. That's obviously incorrect.
I believe you've simply adopted some standard form equation with adapting it to a rotational (torque) situation. Consider that you need to use moment of inertia ($mr^2$) of the pendulum rather than the mass.
A: The trick is the units of $b$. The requirement that the input to the exponential is dimensionless implies that, together with the form of the exponent,
$$[b] = [MT^{-1}]$$
since $[\frac{t}{m}] = [TM^{-1}]$. So the damping constant has units of mass flow rate (e.g. kilograms or grams per second), and that cancels the units of $t$ and $m$ and leaves a dimensionless exponent.
A: An exponent cannot have a unit. It is by definition (unless you redefine it for a particular context) the number of repeated multiplications of the base number by itself. It doesn't make much sense to multiply $e$ by itself "1 metre times" or something like that.
It must be a unitless number. If it isn't, then you are missing a correction factor. It is possible when constructing  empiric relationships to end up with units in the exponent, but then a parameter ought to be introduced to "cancel out" the unit so that it is mathematically sound. If it isn't, then it's mathematically sloppy - then it has just been left our for convenience. Regardless, the unit in an exponent has no meaning and shouldn't be there.
