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.

To know more about fundamental frequency,

link

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

answer: what are you talking about?

explanation:

a. are