Question

a. Two identical speakers, with the same phase constant, are arranged along a 1D track. One speaker remains at the origin. The other speaker can slide along the track to any position x. You are on the track at x=10 m. You hear interference maxima when the adjustable speaker's position is 0.6 m and 1.2 m and at no points in between. What is the frequency of the sound from the speakers in Hz?

b. The speakers are now allowed to have different phase constants. They are adjusted so that you hear interference maxima when the adjustable speaker is at x = 0.6 m and again when it is at x = 1.05 m. What is the difference in the phase constant between the two speakers in rad?

Answer 1

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a)

The maxima will occur if the phase difference is integral multiple of the wavelength.

Since it occurs at X = 0.6 m and X = 1.2 m

ΔX = mλ

1.2 - 0.6 = 1 x λ

λ = 0.6 m

Since velocity of sound in air is 343 m/s

Frequency F = v/λ

F = 343/0.6

F = 571.7 Hz

F = 572 Hz

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b)

ΔX = mλ

1.05-0.6 =λ

λ = 0.45 m

Phase difference

Δφ= 2πΔX/λ

Δφ= 2 x 3.14 x (0.45) /0.45

Δφ= 6.28 rad

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