A sound wave travels out uniformly in all directions from a point source. (a) Justify the following expression for the displacement of the transmitting medium at any distance r from the source $\mathit{S}\mathbf{(}\mathbf{r}\mathbf{,}\mathbf{t}\mathbf{)}\mathbf{=}\frac{\mathbf{b}}{\mathbf{r}}\mathit{s}\mathit{i}\mathit{n}\mathit{k}\mathbf{(}\mathbf{r}\mathbf{-}\mathbf{v}\mathbf{t}\mathbf{)}$: where b is a constant. Consider the speed, direction of propagation,periodicity, and intensity of the wave. (b) What is the dimensionof the constant ?

The sound waves travel uniformly in all directions from the point source

Use the concept of variation of intensity with distance. The intensity of the sound from an isotropic point source decreases with the square of the distance from the source. The intensity is proportional to the square of the amplitude.

Formulae:

Formula for the intensity in terms of power and area or in terms of displacement square, $I=\frac{P_s}{4\pi r^2}\text{or}I=C{S}_m^2$ (i)

The intensity at a distance from a point source that emits sound waves of power $P_s$equally in all directionsusing equation (i) is given as:

$I=\frac{P_s}{4\pi r^2}\dots \dots \dots \dots \dots \dots \dots \dots \dots \dots \dots \dots \left(1\right)$

The intensity is proportional to the square of displacement amplitude$S_m$. Hence,

$I=C{S}_m^2\dots \dots \dots \dots \dots \dots \dots \dots \dots \dots \dots \dots \left(2\right)$

Where C is the proportionality constant

Comparing both theequation (1) and equation (2), we get

$\begin{array}{c}C{S}_m^2=\frac{P_s}{4\pi r^2}\\ S_m=\sqrt{\frac{P_s}{4\pi r^{2}C}}\\ =\sqrt{\frac{P_s}{4\pi C}}\left(\frac{1}{r}\right)\mathrm{...............................}\left(a\right)\end{array}$

The displacement amplitude is inversely proportional to the distance from the source.

The sound wave travels uniformly in all directions from the source. These waves are sinusoidal of radius of the sphere in the phase. The angular frequency and angular wave number of the wave are and respectively, then the term is $\sin (kr-\omega t)$.Let

$b=\sqrt{\frac{P_s}{4\pi C}}\mathrm{..........................}\left(b\right)$

Thus, the displacement of the wave using equations (a) and (b)is given by:

$\begin{array}{c}S(r,t)=\sqrt{\frac{P_s}{4\pi C}}\left(\frac{1}{r}\right)\sin (kr-\omega t)\\ =\frac{b}{r}\sin (kr-\omega t)\end{array}$

The expression for wave number is

$k=\frac{2\pi }{\lambda }$

$\begin{array}{c}S(r,t)=\frac{b}{r}\sin k(\frac{kr}{k}-\frac{\omega }{k}t)\\ =\frac{b}{r}\sin k(r-vt)\end{array}$

Since, $\frac{\omega }{k}=v$

Hence it is proved.

The displacements of the waveand the distance from the sourceboth have dimensions of length. The trigonometric function is dimensionless. Hence, the dimensions ofis length squared. From equation (a) and equation (b), the dimension of the constant is given as:

$\begin{array}{c}b=S(r,t)r\\ \left[b\right]=\left[L^2\right]\end{array}$

Hence, the dimension of the constant b is length squared.