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Consider the 2 incoherent sound waves, $$S_1 = S_0\sin{(\omega_1 t-k_1 x)}$$ $$S_2 = S_0\sin{(\omega_2 t-k_2 x)}$$

Phase difference: $$\Delta\phi = \omega_1 t-k_1 x -\omega_2 t+k_2 x$$ Using $kv=\omega$ and $v=x/t$,

$$\Delta\phi = \omega_1x/v -\omega_1x/v - \omega_2x/v+ \omega_2x/v $$

$$\Delta\phi = 0$$

All the substitutions seem valid, yet the phase difference is $0$ according to it which is obviously incorrect


2 Answers 2


The substitution $t = x/v$ is unwarranted. Here $x$ and $t$ are independent variables and by specifying them you are asking the what the wave amplitude is at that point in space and time. Also (less importantly) notice that the terms of the different waves don’t interact in you analysis.



Your analysis fails at the point where you use $u = \frac{x}{t} \iff t = \frac{x}{u}$. As stated in another answer, $x$ and $t$ are free variables. By substituting them in your equation for the phase difference, you take away a dependent variable ($k$ that is) and end up with a solution that shows that $x$ and $t$ are the same for both $S_{1}$ and $S_{2}$ in the equations that define your waves, which is universally true.


I believe your equation should look more something like

$$ \Delta \phi \!\left( x, t \right) = \omega_{1} t - k_{1} x - \omega_{2} t + k_{2} x \tag{1} \label{1} $$

where the dependence of $\Delta \phi$ on $x$ and $t$ is clear. In that case the values $\omega_{1}$, $\omega_{2}$ and consequently $k_{1}$ and $k_{2}$ must be set as constants. Most probably these variables will be decided on physical ground or based on the modelled system.


Let’s make an example to make the suggestion clear. The value of $\omega_{1}$ and $\omega_{2}$ will be decided upon the nature of the sources used to generate the waves. For this example, I will use

$$\omega = 2 \pi f \tag{2} \label{2} $$

and arbitrarily pick $f_{1} = 100 \, \textrm{Hz}$ and $f_{2} = 150 \, \textrm{Hz}$ so that $\omega_{1} \approx 628 \, \textrm{Hz}$ and $\omega_{2} \approx 924 \, \textrm{Hz}$.

Considering a constant speed of about $c = 343 \frac{\textrm{m}}{\textrm{s}}$ (please note that I use $c$ for speed instead of $u$ but they both describe the same quantity) which corresponds to the speed of sound in the air for a temperature of about $T = 20 \, ^{o}\textrm{C}$ and humidity of about $50 \%$ and using the equation

$$ k = \frac{\omega}{c} \tag{3} \label{3} $$

the wavenumbers become $k_{1} = \frac{\omega_{1}}{c} \approx \frac{628}{343} \approx 1.83 \, \textrm{m}^{-1}$ and $k_{2} = \approx \frac{924}{343} \approx 2.69 \, \textrm{m}^{-1}$.

Now, one can choose $x$ and $t$ arbitrarily to find the phase difference at the specified distance $x$ and time $t$ between the two disturbances. In our example for position/distance (the same since we consider a $1\textrm{D}$ problem with the sources at the origin) $x = 1 \, \textrm{m}$ and time $t = 1 \, \textrm{s}$, from Equation \eqref{1} we have

$$ \begin{split} \Delta \phi \!\left( x, t \right) & = \omega_{1} t - k_{1} x - \omega_{2} t + k_{2} x \implies \\ \implies \Delta \phi \!\left( 1, 1 \right) & \approx 628 \cdot 1 - 1.83 \cdot 1 - 924 \cdot 1 + 2.69 \cdot 1 \implies \\ \implies \Delta \phi \!\left( 1, 1 \right) & \approx 628 - 1.83 - 924 + 2.69 \implies \Delta \phi \!\left( 1, 1 \right) \approx 628 - 1.83 - 924 + 2.69 \implies \\ \implies \Delta \phi \!\left( 1, 1 \right) & \approx -295.14 \end{split}$$

where the negative sign shows that $S_{1}$ lags in phase compared to $S_{2}$, which makes sense since $f_{2}$ is higher than $f_{1}$ and thus its phase changes faster.

Add a “twist”

Now to make it possibly more interesting and more generic you could set the sources $S_{1}$ and $S_{2}$ in different locations/positions and by picking a point with specific $x$ coordinate (please see at the end for a generic $3 \textrm{D}$ case) you can perform the same calculations, but in this case instead of $x$ you will have to use the distance between the chosen point and the sources given by

$$ \Delta x_{n} = \vert x - x_{n} \vert \tag{4} \label{4} $$

where $\vert \, \cdot \, \vert$ denotes the absolute value. So, Equation \eqref{1} will become

$$ \Delta \phi \!\left( x, t \right) = \omega_{1} t - k_{1} \vert x - x_{1} \vert - \omega_{2} t + k_{2} \vert x - x_{2} \vert \tag{5} \label{5} $$

where in this case $x_{1}$ and $x_{2}$ denote the position of the sources. It is worth noting that Equation \eqref{5} can readily be extended to the $3 \textrm{D}$ case simply by calculating the distance between source and point of interest as

$$ \Delta x_{n} = \lVert x - x_{n} \rVert_{2} \tag{6} \label{6} $$

where in this case $\lVert \, \cdot \, \rVert_{2}$ denotes the Euclidian norm. Plugging Equation \eqref{6} into Equation \eqref{5} we get the $3 \textrm{D}$ case for the phase difference to be

$$ \Delta \phi \!\left( x, t \right) = \omega_{1} t - k_{1} \lVert x - x_{1} \rVert_{2} - \omega_{2} t + k_{2} \lVert x - x_{2} \rVert_{2} \tag{7} \label{7} $$

Importantly, Equation \eqref{1} is a special, $1 \textrm{D}$ case of Equation \eqref{7} since the Euclidian distance can be calculated for a Euclidean space of arbitrary dimensions.


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