Chapter 36: Q8P (page 1109)

Sound waves with frequency $3000 Hz$ and speed $343 m / s$ diffract through the rectangular opening of a speaker cabinet and into a large auditorium of length $d = 100 m$ . The opening, which has a horizontal width of $30.0 cm$ , faces a wall $100 m$ away ( $F i g . 36 - 36$ ). Along that wall, how far from the central axis will a listener be at the first diffraction minimum and thus have difficulty hearing the sound? (Neglect reflections.)

Expert verified

The distance between first order minima and central minima is $41.2 m$ .

Step by step solution

The frequency of the sound wave is $f = 3000 Hz$

The speed of the sound wave is $v = 343 m / s$

The length of auditorium is $D = 100 m$ .

The width of opening is $d = 30 cm$ .

The order of first minima is $m = 1$ .

The diffraction is the phenomenon in which the light bends on the edges of the slit and pattern of light is formed on a screen

The wavelength of sound wave is given as:

$v = f \cdot λ$

Substitute all the values in equation.

$343 m / s = (3000 Hz) λ λ = 0.1143 m$

The distance between first order minima and central minima is given as:

$w = \frac{D}{\sqrt{{(\frac{d}{λ})}^{2} - 1}}$

Substitute all the values in equation.

$w = \frac{100 m}{\sqrt{{(\frac{(30 cm) (\frac{1 m}{100 cm})}{(0.1143 m)})}^{2} - 1}} w = 41.2 m$

Therefore, the distance between first order minima and central minima is $41.2 m$ .

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