Question

Two out of phase loudspeakers are some distance apart. A person stands 5.40 m from one speaker and 2.70 m from the other. What is the fourth-lowest frequency at which constructive interference will occur? The speed of sound in air is 340 m/s.

Answer 1

Two out of phase loudspeakers are some distance apart. The third lowest frequency is 377.78 Hz.

Distance of person from one speaker = 5.4 m

Distance of person from the other = 2.7 m

The longest wavelength has the shortest frequency.

We need to calculate the difference between the speakers.

The difference between the speakers = 5.4 - 2.7 = 2.7 m

We know that, the longest wavelength is such one wavelength

λ = d

So, the third longest wavelength is

3* λ = d

Put the value of d,

λ = 2.7/3 = 0.9 m

Now, let us calculate the lowest frequency.

We know the expression for frequency as,

f = v / λ = 340/0.9 = 377.78 Hz.

Thus, the third lowest frequency is 377.78 Hz.

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