A pair of in-phase stereo speakers is placed side by side, 1.26 m apart. You stand directly in front of one of the speakers, 2.79 m from the speaker. What is the lowest frequency that will produce destructive interference at your location? Group of answer choices

1264 Hz

Explanation:

We are given;

Your distance to the front of first speaker: L1 = 2.79 m

Distance between speakers; d = 1.26m

Now, your distance to second speaker can be calculated using pythagoras theorem.

Thus;

L2 = √(1.26² + 2.79²)

L2 = 3.0613 m

Let's find the wavelength from the formula;

(L1 - L2) = nλ

Where n is the order and could be; n = 0,1,2,3...

Making λ which is the wavelength the subject, we have;

(L1 - L2)/n = λ

The least frequency will occur at the wavelength when n = 1.

λ = (2.79 - 3.0613)/1

λ = -0.2713 m

We will take the absolute value and thus;

λ = 0.2713 m

Frequency is gotten from;

f = v/λ

Where v is speed of sound = 343 m/s

Thus;

f = 343/0.2713

f ≈ 1264 Hz

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