Figure 17-48 shows an air-filled, acoustic interferometer, used to demonstrate the interference of sound waves. Sound source Sis an oscillating diaphragm; Dis a sound detector, such as the ear or a microphone. Path$\mathit{S}\mathit{B}\mathit{D}$can be varied in length, but path$\mathit{S}\mathit{A}\mathit{D}$is fixed. At D, the sound wave coming along pathinterferes with that coming along path$\mathit{S}\mathit{A}\mathit{D}$. In one demonstration, the sound intensity at Dhas a minimum value of 100 units at one position of the movable arm and continuously climbs to a maximum value of 900 units when that arm is shifted by $\mathbf{1}\mathbf{.65}\mathbf{\text{\hspace{0.17em}cm}}$. Find (a) the frequency of the sound emitted by the source and (b) the ratio of the amplitude at Dof the $\mathit{S}\mathit{A}\mathit{D}$wave to that of the $\mathit{S}\mathit{B}\mathit{D}$wave. (c) How can it happen that these waves have different amplitudes, considering that they originate at the same source?

1. The minimum value of sound intensity detected at D is $100\text{\hspace{0.17em}units}$.
2. The minimum value of sound intensity detected at D is $100\text{\hspace{0.17em}units}$.
3. The arm is shifted by$d=1.65\text{\hspace{0.17em}cmor}0.0165\text{\hspace{0.17em}m}$ .

Using the equation of frequency and the given conditions in the problem, we can find the frequency, and then, by using the formulae for maximum and minimum intensity in terms of amplitudes of interfering waves, we can get the ratio of amplitudes at a particular point.

Formula:

The frequency of the wave,

$\mathit{f}\mathbf{=}\frac{\mathbf{v}}{\mathbf{\lambda }}$ …(i)

The ratio of the square-root of intensities relation to maximum and minimum displacement,

$\frac{\sqrt{{\mathbf{I}}_{\mathbf{m}\mathbf{a}\mathbf{x}}}}{\sqrt{{\mathbf{I}}_{\mathbf{m}\mathbf{i}\mathbf{n}}}}\mathbf{=}\frac{{\mathbf{A}}_{\mathbf{A}}\mathbf{+}{\mathbf{A}}_{\mathbf{B}}}{{\mathbf{A}}_{\mathbf{A}}\mathbf{-}{\mathbf{A}}_{\mathbf{B}}}$ …(ii)

One condition is given, and from that condition, we can say that when the right side of the instrument is pulled by distance, then the path length for the sound waves increases by. As the interference pattern changes from minimum to next maximum, that distance will be half the wavelength of sound.

So, the wavelength of the wave is given as:

$\begin{array}{c}2d=\frac{\lambda }{2}\\ \lambda =4d\end{array}$

Using equation (i) and the value of wavelength, we get the frequency of the wave as:

$\begin{array}{c}f=\frac{v}{4d}\\ =\frac{343\text{\hspace{0.17em}m}/\text{s}}{\left(4\right)\left(0.0165\text{\hspace{0.17em}m}\right)}\\ =5.2\times {10}^3\text{\hspace{0.17em}Hz}\end{array}$

Hence, the frequency of the emitted wave is $5.2\times {10}^3\text{\hspace{0.17em}Hz}$.

Let the amplitude of SAD and SBD at D be$A_A\&A_B$respectively. Using equation (ii) can be given as:

$\begin{array}{c}\frac{A_A+A_B}{A_A-A_B}=\frac{\sqrt{900\text{\hspace{0.17em}units}}}{\sqrt{100\text{\hspace{0.17em}units}}}\\ =\frac{3}{1}\end{array}$

If we apply the componendo - dividendo rule we get the ratio of the amplitudes as:

$\begin{array}{c}\frac{A_A}{A_B}=\frac{4}{2}\\ =2\end{array}$

Hence, the value of the ratio of the amplitude is 2.

The difference in amplitudes of these waves even considering that they are of the same source is because they are covering different distances, and hence, energy loss is different.