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 500 Hz. (a) What is the phase difference between thetwo waves when they reach the observer? (Your answer should bebetween 0 and 2.) rad
(b) What is the frequency closest to 500Hz to which the oscillator may be adjusted such that the observerhears minimal sound?
Hz (a) What is the phase difference between thetwo waves when they reach the observer? (Your answer should bebetween 0 and 2.) rad
(b) What is the frequency closest to 500Hz to which the oscillator may be adjusted such that the observerhears minimal sound?
Hz

Explanation / Answer

The distance between the two loudspeakers is 3.00 m The listener stands 3.00 m from the wall directly in front ofone of the speakers The oscillator driving the speakers is at a frequency of f =500 Hz (a)The phase difference between the two waves when they reachthe observer is w = 2f or w = 2 * 3.14 * 500 = 3140 rad (b)The frequency closest to 500 Hz to which the oscillator maybe adjusted such that the observer hears minimal sound is f = (v/) Here,v is the speed of sound and is the distancebetween the two loudspeakers placed on the wall. Here,v = 340 m/s and = 3.00 m or f = (340/3) = 113.3 Hz

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