# How to calculate the overall harmonics of a system of tubes?

Let's say there is a tube closed at one and and open from the other. The open end is connected to another tube (of smaller diameter) which is open from both ends. How can I calculate the overall harmonics of this system of tubes? I have already calculated theri individual harmonic frequency but can't find a way to find the overall one. Can anyone help me regarding this?

• The overall system is more complicated than just the "addition" of those two systems. The impedance at the end of the first tube is not infinite or zero (which is the case for closed and open tube ends respectively) which has to be put into the equation. This acts like a "silencer" used in HVAC tubes and exhaust pipes of cars and bikes. I suggest you take a look at some text (such as Fundamentals of Acoustics by Kinsler et al. - Chapter 10) on this topic (acoustical filters) May 9, 2022 at 20:18

## 1 Answer

Summary: There is not a simple way to obtain the fundamental frequency of the composite tube you are describing in any sort of accurate manner. A more rigorous approach is necessary.

Longer Discussion:

As mentioned by @ZaellixA in the comments, this system is too complicated to give the simple solutions you may be used to. I will derive the solution below, but please be aware that the results are probably not generalizable.

Consider two tubes of different lengths and cross-sectional areas joined together. We will assume that the lengths are sufficiently long, the cross-sectional areas are sufficiently small, and the wavelengths are sufficiently large that we may treat the wave motion as one-dimensional (this is a standard assumption and if you think of these things as "tubes", it is almost certainly met). The solution to the one-dimensional wave equation in each tube may then be written as $$$$p_i = A_i\sin(k[x-l_i]),$$$$ where $$p_i$$ is the acoustic pressure in the $$i$$th tube, $$A_i$$ is the amplitude, $$x$$ is the position in the tube, $$k=\omega/c$$ is the wavenumber (assuming the same fluid fills both tubes), $$\omega$$ is the angular frequency, $$c$$ is the wave speed, and $$l_i$$ is a phase factor for the $$i$$th tube. Since we assume the tubes are open at the ends and we idealize the system to treat the open tubes as pressure release (zero acoustic pressure at the ends), we may then immediately identify $$l_1$$ as 0 (we will call this the left-end of the tube) and $$l_2$$ as $$L_1+L_2$$, the sum of the lengths of the two tubes. At the intersection, $$x=L_1$$, we require the pressures to be equal (this prevents infinite acceleration of the fluid at interfaces), yielding the equation $$$$p_1(L_1) = A_1\sin(kL_1) = p_2(L_1) = A_2\sin(k[L_1-L_2-L_1]) = -A_2\sin(kL_2).$$$$ We also require the volume velocity $$U$$ to be equal at the intersection. The volume velocity may be written as $$$$U_i = \frac{-S_i}{-i\omega\rho}\frac{\partial p_i}{\partial x} = \frac{-iS_i}{Z}A_i\cos(k[x-l_i]),$$$$ where $$S_i$$ is the cross-sectional area of the $$i$$th tube, $$\rho$$ is the mass density, and $$Z=\rho c$$ is the specific acoustic impedance. Matching the volume velocities then yields the equation $$$$S_1A_1\cos(kL_1) = S_2A_2\cos(kL_2).$$$$ These two interface equations may be written in matrix form as $$$$\left[\begin{matrix} \sin(kL_1) & \sin(kL_2) \\ S_1\cos(kL_1) & -S_2\cos(kL_2) \end{matrix}\right] \left[\begin{matrix} A_1 \\ A_2 \end{matrix}\right] = \left[\begin{matrix} 0 \\ 0 \end{matrix}\right].$$$$ The only way that one can have $$A_1,A_2\ne0$$ is to have the determinant of the matrix on the left be equal to zero, which yields \begin{align} S_2\sin(kL_1)\cos(kL_2) + S_1\cos(kL_1)\sin(kL_2) &= 0. \end{align} At this point, you will probably need to numerically solve for $$k$$. The fundamental frequency is the frequency associated with the smallest solution to $$k$$ that is non-zero.

There are a few special cases that yield explicit analytical solutions. The first I will consider is the trivial case of $$S_1=S_2$$ (same tube size), which yields $$$$S_1\sin(k[L_1+L_2]) = 0,$$$$ which has the solution $$$$f_n = \frac{nc}{2(L_1+L_2)};$$$$ for $$n=1$$ you have the fundamental, half-wavelength mode. Next you have the case where $$L_1=L_2$$ (same length of tube), which yields $$$$(S_1+S_2)\sin(kL_1)\cos(kL_1) = 0,$$$$ which has the solution $$$$f_n = \frac{nc}{4L_1}.$$$$ This solution is also the simple half-wavelength fundamental due to the symmetry of the problem. The final case is for small changes in the cross-sectional area. Let $$S_2=S_1+\Delta S$$, where $$\Delta S$$ is a small (compared to $$S_1$$) change in the cross-sectional area. Using a perturbation analysis (which I will forgo providing the details and just give the solution) then yields $$$$f_n \approx \frac{nc}{2(L_1+L_2)} - \frac{\Delta S}{S_1}\frac{(-1)^{n}c}{4\pi(L_1+L_2)}\sin\left(\frac{n\pi[L_1-L_2]}{L_1+L_2}\right).$$$$ If $$n(L_1-L_2)/(L_1+L_2)\ll1$$, we may further approximate this equation with $$$$f_n \approx \frac{nc}{2(L_1+L_2)} - (-1)^{n}\frac{\Delta S}{S_1}\frac{nc(L_1-L_2)}{4(L_1+L_2)^2}.$$$$ If $$L_1=5$$ m, $$L_2=4$$ m, $$S_1=5$$ cm$$^2$$, and $$S_2=4$$ cm$$^2$$, the numerical solution for $$f_1$$ is 18.83 Hz, while the first and second approximations yield 18.85 Hz and 18.84 Hz, respectively. Note that ignoring the $$\Delta S$$ correction term leads to $$f_1\approx19.06$$ Hz. Compare this with the case of $$L_2=1$$ m and $$S_2=1$$ cm$$^2$$ and tube 1 is left unchanged. In this case, the numerical solution yields $$f_1=21.95$$ Hz, while the first and second approximations yield 25.43 Hz and 20.96 Hz, respectively. Ignoring the corrections altogether yields 28.58 Hz.

Incidentally, the best you can do with only knowing the fundamental frequencies of the two tubes, say $$f_1^1$$ and $$f_1^2$$, is to take the harmonic sum: $$$$\left(\frac{1}{f_1^1} + \frac{1}{f_1^2}\right)^{-1} = \left(\frac{2L_1}{c} + \frac{2L_2}{c}\right)^{-1} = \frac{c}{2(L_1+L_2)},$$$$ which is the $$\Delta S=0$$ approximation.