Two loudspeakers are placed on a wall 3.00m apart. A listener stands 3.00 m from

Question

Two loudspeakers are placed on a wall 3.00m apart. A listener stands 3.00 m from thewall directly in front of one of the speakers. A single oscillatoris driving the speakers at a frequency of 700 Hz. (a) What is the phase difference between thetwo waves when they reach the observer? (Your answer should bebetween 0 and 2.)
1 rad=?

(b) What is the frequency closest to 700Hz to which the oscillator may be adjusted such that the observerhears minimal sound?
2 Hz=? (a) What is the phase difference between thetwo waves when they reach the observer? (Your answer should bebetween 0 and 2.)
1 rad=?

(b) What is the frequency closest to 700Hz to which the oscillator may be adjusted such that the observerhears minimal sound?
2 Hz=?

Explanation / Answer

Path difference, S = d2 - d1 Here, d1 = 3.0 m d2 = (9 + 3) = 3.464        S = d2 - d1=3.464 - 3.0 = 0.464 Phase difference,         =k.S = 2.S/ Given, f = 700 Hz We assume velocity of sound to be 340 m/s Wavelength,           = v/f = 340/700 = 0.4857 m             = k.S = 2.S/                 = 2x 0.464/0.4857                 = 6.002rad
For minimum sound, the pathe difference, S = '/2 Wavelength must be,                ' = 2xS =2x0.464 = 0.928 m Frequency,                f' = v/' = 340/0.928= 366.4 Hz

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