Equation of circular sine waves in the water

Question

I have to write the equation of a sine wave expanding circularly from a point $P_0=(x_0,y_0)$. The wave has the form $\eta(\rho)=A\sin(\omega\rho)$ where $\rho$ is the distance from the point $P_0$. Obviously $\eta$ is the amplitude in the direction orthogonal to te plane of the circle and $A$ his amplitude. Thanks

Answer 1

What is the formula that generates a circle not centered at the origin in polar coordinates?

To begin, note from the following picture that there are two scenarios that must be handled separately: whether the polar origin is inside or outside the radius of the circle in question.

From this picture, note the specific labels $G$ and $L$ marking the Polar origin in the two cases, and $GS$ and $LT$ marking the line $\theta=0$ in each case. Further, let $A=(x_0,y_0),K=(x_1,y_1)$ be the center of each circle. Also note that $AE$ and $KR$ are the radii of the two circles, and $GH,GF$ and $LR$ are the three values that will be calculated for the value of $\rho$. There must be three values because $2\angle EGA$ is the limit for the angles where $\theta$ will be valid on the left-hand case, so we will calculate both the "far edge" of the circle away from the origin as values for when $0\le \theta\le 2\angle EGA$ and the "close edge" as negative values for when $\pi\le \theta \le \pi + 2\angle EGA$, and we will create $\phi=\angle SGE$ or $\phi=\angle TLN$ as the offset so that $\theta$ can take on values in these ranges. Then $\theta$ takes on its values as $\theta=\angle EGF$ or $\theta=\angle NLR$.

First, the value of $\phi$. We have $\phi=\angle SGE=\angle AGE-\angle AGS=\arcsin ({AE\over AG})-\arcsin({y_0\over AG})+{n\pi\over 2}$ on the left-hand side, and $\phi=\angle TLN=\arcsin({y_1\over LK})+{n\pi\over 2}$. Then, use $n\in \{0,1,2,3\}$, set for the correct offset to place $E$ and $N$ in the correct quadrant relative to $G$ and $L$.

Now we need only calculate $\rho$ based on $\theta$.

Note that $B,D,E,M,Q$ are points chosen so that the angles at each point are $90^\circ$. Then we have similar triangles $\Delta CGE,\Delta CFD,\Delta CHB$ and $\Delta LMR,\Delta NQL,\Delta PMN,\Delta PQR$.

Then we have $BH^2=AH^2-AB^2$. $AB=AE-EC-CB$, and $EC=CG\sin\theta=EG\tan\theta,{BH\over EG}={CH\over CG}, {CB\over CE}={CH\over CG}$. Applying the ratios and other substitutions, we can arrive at

$$CH=\sqrt{AE^2-(AE-EG\tan\theta)^2\cos^2\theta}+(AE-EG\tan\theta)\sin\theta$$

Then, we can say

$$\rho=EG\sec\theta+CH$$

The other two formulas are highly similar, with a $-$ on the $\sin$ term, and without a $GC$ initial offset for the right-hand case.

Applying this formula to the original question involving $\eta(\rho,\theta)$, we arrive at the following for waves further out than the polar origin:

$$\eta(\rho,\theta)=\alpha\sin\left(\omega\rho\left(\sqrt{KN^2-(KN-KL)^2\cos^2\theta}-(KN-KL)\sin\theta\right)\right)$$

I think the volume of work required to achieve this result, as well as the complexity of the equation itself, are evidence enough that it is better to translate to Cartesian coordinates prior to applying a linear translation to the center of the object being graphed.

Answer 2

When transforming to and from alternate coordinate systems, it is often optional to match the origin of a non-Cartesian system with the Cartesian origin. In fact, it is very often desirable to recenter other coordinate systems to make the equations easier in the alternate system, and thus easier to translate back to Cartesian equations. In the final translation, all that is usually required is an offset to the origin which is very easy to do in the Cartesian system, but very difficult in almost all other systems.

Considering the problem you have posed: you have essentially asked for the equation of a circle in polar coordinates (usually $r=c$ for some constant $c$ satisfies this), but with the center of the circle not at the polar origin. Then you have an exceptionally difficult equation and set of constraints to work out.

The Polar version is much more complicated than asking the question "how do I write the equation of a circle in polar coordinates?" and making a note to center the polar origin at an offset relative to the Cartesian system, then applying the offset after translating your equation to Cartesian.

With a center offset applied to a circle in Cartesian coordinates, we have the standard equation $x^2+y^2=r^2$, which is the same as saying $\sqrt{x^2+y^2}=r$. Moving the center of the circle is a simple matter of applying the offsets directly to $x$ and $y$ as $x-x_0$ and $y-y_0$.

Thus, the final form of your equation in Cartesian would be the following:

$$\eta(\rho=\{x,y\})=A\sin\left(\omega \sqrt{(x-x_0)^2+(y-y_0)^2}\right)\tag{1}$$

where $\rho$ is treated as a 2D vector.