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Two identical loudspeakers are driven in phase by the same amplifier. The speakers are positioned a distance of 3.2 m apart. A person stands 4.1 m away from one speaker and 4.8 m away from the other. Calculate the second lowest frequency that results in destructive interference at the point where the person is standing. Assume the speed of sound to be 343 m/s.

Respuesta :

Answer:

f = 735 Hz

Explanation:

given,

Person distance from speakers

r₁ = 4.1 m      r₂ = 4.8 m

Path difference

d = r₂ - r₁ = 4.8 - 4.1 = 0.7 m

For destructive interference

[tex]d = \dfrac{n\lambda}{2}[/tex]

where, n = 1, 3,5..

we know, λ = v/f

[tex]d = \dfrac{n v}{2f}[/tex]

v is the speed of the sound = 343 m/s

f is the frequency

[tex]f = \dfrac{n v}{2d}[/tex]

for n = 1

[tex]f = \dfrac{343}{2\times 0.7}[/tex]

     f = 245 Hz

for n = 3

[tex]f = \dfrac{3\times 343}{2\times 0.7}[/tex]

     f = 735 Hz

Hence,the second lowest frequency of the destructive interference is 735 Hz.

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