Division of intensity of sound Consider a single sound source $(S)$ producing sound wave of intensity $I$ . It is connected to a detector $(D)$ through two paths with a path difference.  My doubt is that will the  intensity of sound in both paths  be $I$ or will it be divided into $I/2$ and $I/2$.
 A: Answer: The longer path has a greater intensity in it by a factor of $L_1/L_2$, where $L_1$ is the longer path and $L_2$ is the shorter path.
Solution:
I will restrict my consideration to a constant frequency source, and assume that the wavelength of the sound is sufficiently small that there are no cross-modes (only planar motion along the two paths) and that the wavelength is much shorter than either path length.  If you wish to move to higher frequencies, you will need to have more information about the paths (e.g., is this a two-dimensional problem with concentric cylinders, or is this a three-dimensional problem with tubes; what is the cross-sectional profile of the duct?), and if you wish to consider transient pulses you can use the present analysis and do a Fourier synthesis.  Furthermore, I will assume the two paths (and the inlet and exit ducts) have the same uniform cross-sectional area.  Under these assumptions, the fact that the paths are curved is somewhat irrelevant, and I will treat them as straight ducts.
In this very low-frequency limit, the system may be modeled as two simple masses (the mass of the air within the duct along the path of interest) in parallel.  For those more used to electrical circuits, there is a one-to-one correspondence between the voltage drop and the acoustical pressure, the current and the volume velocity, and an inductance and a mass per cross-sectional area squared.  Assuming the same cross-sectional area everywhere, we may then write the circuit equations as
\begin{gather}
  v_1 + v_2 = v_S = v_D, \\
  -i\omega M_1v_1 + i\omega M_2v_2 = 0,
\end{gather}
where $v$ is the volume velocity, $M$ is the air mass, $\omega$ is the angular frequency ($e^{-i\omega t}$ time convention), and $S$, $D$, 1, and 2 denote, respectively, the source, detector, the longer path, and the shorter path.  Combining these together yields
\begin{gather}
  v_1 = v_S\frac{M_1+M_2}{M_2}, \\
  v_2 = v_S\frac{M_1+M_2}{M_1}.
\end{gather}
Now, the power propagating through the paths is proportional to the volume velocity times the pressure.  Since the pressure and cross-sectional areas are the same for both paths, we may write
\begin{equation}
  \frac{P_1}{P_2} = \frac{I_1}{I_2} = \frac{v_1}{v_2} = \frac{M_1}{M_2},
\end{equation}
where $P$ is the power and $I$ is the intensity.  Finally, since $M=SL$, we conclude that
\begin{equation}
  \frac{I_1}{I_2} = \frac{L_1}{L_2} > 1.
\end{equation}
Thus, the intensity is greater in the longer path.
