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

Steven

21 April, 18:00

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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Fatima Giles · 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