Two identical loudspeakers are placed on a wall 3.00 m apart. A listener stands 4.00 m from the wall directly in front of one of the speakers. A single oscillator is driving the speakers at a frequency of 300 Hz. (a) What is the phase difference in radians between the waves from the speakers when they reach the observer? (Your answer should be between 0 and 2?.) (b) What is the frequency closest to 300 Hz to which the oscillator may be adjusted such that the observer hears minimal sound?

Question

Physics

Fernando Lamb

2 February, 16:38

Two identical loudspeakers are placed on a wall 3.00 m apart. A listener stands 4.00 m from the wall directly in front of one of the speakers. A single oscillator is driving the speakers at a frequency of 300 Hz. (a) What is the phase difference in radians between the waves from the speakers when they reach the observer? (Your answer should be between 0 and 2?.) (b) What is the frequency closest to 300 Hz to which the oscillator may be adjusted such that the observer hears minimal sound?

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Answers (1)

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Salty · Answer 1

a) ΔФ = 5.54 rad, b) f = 140 Hz

Explanation:

a) This is a sound interference exercise, which is described by

Δr / λ = ΔФ / 2π

ΔФ = Δr 2π / λ

Let's find the path difference

Δr = r₂ - r₁

r₁ = 4m

r₂ = √ (x² + y²) = √ (3² + 4²) = 5 m

Δr = 1 m

To find the wavelength we use the relation of the speed of sound

v = λ f

λ = v / f

λ = 340/300

λ = 1,133 m

We substitute

ΔФ = 2π 1 / 1.133

ΔФ = 5.54 rad

b) to have a minimum intensity the phase difference must be π radians

λ = Δr 2π / Ф

λ = 1 2π / π

λ = 2m

We look for the frequency

v = λ f

f = v / λ

f = 340/2

f = 140 Hz