Question

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.

Answer 1

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

$d = \dfrac{n\lambda}{2}$

where, n = 1, 3,5..

we know, λ = v/f

$d = \dfrac{n v}{2f}$

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

f is the frequency

$f = \dfrac{n v}{2d}$

for n = 1

$f = \dfrac{343}{2\times 0.7}$

     f = 245 Hz

for n = 3

$f = \dfrac{3\times 343}{2\times 0.7}$

     f = 735 Hz

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