Undergrad Is this allowed? - Harmonic oscillation

[APUGYael]

I divide by zero which is a no-go, but on the other hand: at resonance frequency the phase-shift is 90 degrees.

 

[BvU]

Yes, at $\omega_u = \omega_0$ you have $\cos\phi = 0$. You don't actually divide by 0, it's just that the tangent of $\phi$ is undetermined.

 

[APUGYael]

Yes, at $\omega_u = \omega_0$ you have $\cos\phi = 0$. You don't actually divide by 0, it's just that the tangent of $\phi$ is undetermined.

Is the final conclusion correct too? D=1/ω
Because it uses the same logic but with sine.

 

[BvU]

I missed that question.
The answer is: no. The denominator is zero, the numerator doesn't have to be 1, just non-zero.

$D$ is a free parameter, like $C$ and $J$.

 

[APUGYael]

I missed that question.
The answer is: no. The denominator is zero, the numerator doesn't have to be 1, just non-zero.

$D$ is a free parameter, like $C$ and $J$.

But surely
tan (x) = sin(x)/cos(x) with x = pi/2 means that
sin(pi/2)=1

 

[A.T.]

But surely
tan (x) = sin(x)/cos(x) with x = pi/2 means that
sin(pi/2)=1

Not with x = pi/2 but via the limit x-> pi/2.

 

[APUGYael]

Not with x = pi/2 but via the limit x-> pi/2.

Right. So why is D -> 1/ω as x-> pi/2 not correct?

 

[vanhees71]

Yes, at $\omega_u = \omega_0$ you have $\cos\phi = 0$. You don't actually divide by 0, it's just that the tangent of $\phi$ is undetermined.

It shows in a drastic way that it's a bad habit to use the tan function to calculate polar angles in polar coordinates (and this example of the phase shift is geometrically interpreted right this). What you really want is to calculate an angle within an interval of the length $2 \pi$ not one of the length $\pi$.

In this context an interval $\varphi \in ]-\pi,\pi]$ is most convenient. Now take the Cartesian coordinates of a point $(x,y)$ that are related to the polar coordinates by $(x,y)=r(\cos \varphi,\sin \varphi)$. Then given $(x,y)$ you first get
$$r=\sqrt{x^2+y^2}$$,
and then
$$\cos \varphi=\frac{x}{r}, \quad \sin \varphi=\frac{y}{r}.$$
You have to fulfill both (sic!) equations to get $\varphi$. The first equation alone is not sufficient, because of $\cos \varphi=\cos(-\varphi)$ you get the same angle $\varphi \in [0,\pi]$ when using the usual arccos function, i.e., the same angle $\varphi$ in this intervall for both points $(x,y)$ and $(x,-y)$. Now all you need from the second equation is the sign of $y$ since you know that for $y>0$ you must have $\varphi \in [0,\pi]$ and for $y<0$ it must be in $[-\pi,0]$. Thus you have
$$\varphi = \text{sign} y \arccos(x/r).$$
The only trouble arises if, $x<0$ and $y=0$. Then $x=-|x|=-r$, by definition we choose $\varphi=\arccos(-1)=+\pi$. So the final result is
$$\varphi=\begin{cases}
\text{sign} y \arccos(x/r) &\text{for} \quad y \neq 0, \\
0 & \text{for} x>0, \quad y=0, \\
\pi & \text{for} x<0, \quad y =0.
\end{cases} $$
Then for the special case that $x=0$ you correctly get $\varphi=\pi/2 \text{sign} y$.