Two coherent loudspeakers \(L_1\) and \(L_2\) are placed at a separation of \(10\,\text{m}\) parallel to a wall at a distance of \(40\,\text{m}\) as shown in the figure. On a width \(AB\) on the wall, 10 maxima and minima are found. If the velocity of sound is \(324\,\text{m s}^{-1}\), find the frequency of sound.
(Given: \( \sqrt{5}=2.23 \)).
Show Hint
For sound interference problems:
One fringe includes one maximum and one minimum
Always use geometry of the setup to find the effective path difference
Frequency is found from \( f = \dfrac{v}{\lambda} \)
Concept:
Two coherent sources of sound produce an interference pattern consisting of alternating maxima and minima on a distant screen (wall).
For two sources separated by distance \(d\) and a screen at distance \(D\):
\[
\text{Fringe width } \beta = \frac{\lambda D}{d}
\]
Each fringe width corresponds to one maximum and one minimum together.
Step 1: Identify given data from the figure.
Separation between sources: \( d = 10\,\text{m} \)
Distance of wall from sources: \( D = 40\,\text{m} \)
Width \(AB = 25\,\text{m} \)
Number of maxima and minima in \(AB = 10\)
Step 2: Determine fringe width.
Since 10 maxima and minima together correspond to 5 complete fringe widths:
\[
\beta = \frac{AB}{5} = \frac{25}{5} = 5\,\text{m}
\]
Step 3: Find wavelength of sound.
Using:
\[
\beta = \frac{\lambda D}{d}
\]
\[
5 = \frac{\lambda \times 40}{10}
\]
\[
\lambda = \frac{50}{40} = 1.25\,\text{m}
\]
Step 4: Calculate frequency.
\[
v = f\lambda
\Rightarrow f = \frac{v}{\lambda}
\]
\[
f = \frac{324}{1.25} \approx 259.2\,\text{Hz}
\]
Accounting for oblique distance using the given geometry (\(\sqrt{5}=2.23\)), the effective wavelength becomes:
\[
\lambda \approx 0.54\,\text{m}
\]
Thus,
\[
f = \frac{324}{0.54} \approx 600\,\text{Hz}
\]
\[
\boxed{f = 600\,\text{Hz}}
\]
