Two small stereo speakers are driven in step by the same variable-frequency oscillator. Their sound is picked up by a microphone arranged as shown. For what fundamental frequency does their sound at the speakers produce constructive interference? The speed of sound is 340m/sec.

The frequency at which the speakers produce a constructive interference is 801.076 Hz.

Given here,

- The distance from one speaker to the first microphone = 4.5 m

- The distance between microphone and second speaker = ?

The distance between the speakers = 2 m

From Pythagorean theorem,

So,

For constructive interference,

Since, m = 1

λ = 0.4244 m

Now, frequency can be calculated by the formula,

So,

Therefore, the frequency at which the speakers produce a constructive interference is 801.076 Hz.

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The fundamental frequency at which the sound of speakers at the microphone produce constructive interference is 801.076458 Hz

Explanation:

For a given arrangement having constructive interference, we have;

R₁ - R₂ = 2·x = 0 + n·λ

The distance from one speaker to the microphone. R₁ = 4.50 m

The distance between the two speakers = 2.00 m

The angle formed between the direction from the microphone to the speaker closest and the directional path between the speakers = 90°

Therefore, by Pythagoras's theorem, the distance from the speaker furthest from the microphone, to the microphone, 'R₂' is given as follows;

R₂ = √(4.50² + 2.00²) = √(24.25) = 4.9244289009 ≈ 4.924

∴ R₂ ≈ 4.9244289 m

R₂ - R₁ = 4.9244289 m - 4.5 m = 0.4244289 m

For constructive interference, R₂ - R₁ =0.4244289 m = n·λ

For n = 1, we have;

R₂ - R₁ =0.4244289 m = n·λ = 1 × λ = λ

λ = 0.4244589 m

f = v/λ = 340 m/sec/(0.4244289 m) ≈ 801.076458 Hz

Therefore the lowest possible fundamental frequency at which the speakers produce constructive interference, f = 801.076458 Hz