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Two speakers driven by the same force

  • Thread starter aaronb
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  • #1
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Homework Statement


Two loudspeakers are placed on a wall 1.00 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 800 Hz.

(a) What is the phase difference between the two waves when they reach the observer? (Your answer should be between 0 and 2π.)

(b) What is the frequency closest to 800 Hz to which the oscillator may be adjusted such that the observer hears minimal sound?

Homework Equations


delta r = (phi/2pi)lamda
frequency = velocity/lamd

The Attempt at a Solution


I found the dimensions of the triangle that is formed when the person stands in front of one speaker 4.00m away which is:
a=4,b=1,c=sqrt(17)
 

Answers and Replies

  • #2
Redbelly98
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You'll need to get the difference in the distances to each speaker.
 

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