<p>Two loudspeakers are placed on a wall 3.00 m apart. A listener stands 3.00 m

Question

<p>Two loudspeakers are placed on a wall 3.00 m apart. A listener stands 3.00 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.<br /><br />(a) What is the phase difference between the two waves when they reach the observer? (Your answer should be between 0 and 2&#960;.) <br /><br /><span><em>Answer: I got 5.65 rad but i can't figure out part B!!!</em></span><br /><br />(b) What is the frequency closest to 800 to which the oscillator may be adjusted such that the observer hears minimal sound?</p>
<p>&#160;</p>
<p>Please help with part B, I keep getting it wrong, thanks!</p>

Explanation / Answer

b ) Two loudspeakers are placed on a wall r_2 = 3.00 m apart. A listener stands at a distance d = 3.00m from the wall    r_1 = r_2 ^2 + d^2           = (3.00m)^2 + (3.00m)^2           = 4.24m    wavelength is ' = 2 ( r_1 - r_2 )                                  = 2 ( 4.24 m - 3.00m)                                 = 2.48m frequency that the observer hears the minimal sound is f ' = v / '                                                                                           = 343 m/s / 2.48m                                                                                           = 138.01 Hz

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