Answer:
The value is [tex]w = 7.54 \ m[/tex]
Explanation:
From the question we are told that
The length of the crack is [tex]a = 0.3 \ m[/tex]
The frequency is [tex]f = 30.0 \ kHz = 30 *10^{3} \ Hz[/tex]
The distance outside the cave that is being consider is [tex]D = 100 \ m[/tex]
The speed of sound is [tex]v_s = 340 \ m/s[/tex]
Generally the wavelength of the wave is mathematically represented as
[tex]\lambda = \frac{v}f}[/tex]
=> [tex]\lambda = \frac{340 }{30*10^{3}}[/tex]
=> [tex]\lambda = 0.0113 \ m/s[/tex]
Generally for a single slit the path difference between the interference patterns of the sound wave and the center is mathematically represented as
[tex]y = \frac{ n * \lambda * D}{a}[/tex]
=> [tex]y = \frac{ 1 * 0.0113 * 100}{0.3}[/tex]
=> [tex]y = 3.77 \ m[/tex]
Generally the width of the sound beam is mathematically represented as
[tex]w = 2 * y[/tex]
=> [tex]w = 2 * 3.77[/tex]
=> [tex]w = 7.54 \ m[/tex]
