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Two loudspeakers emit sound waves along the x-axis. A listener in front of both speakers hears...

Two loudspeakers emit sound waves along the x-axis. A listener in front of both speakers hears a maximum sound intensity when speaker 2 is at the origin and speaker 1 is at x = 0.500m . If speaker 1 is slowly moved forward, the sound intensity decreases and then increases, reaching another maximum when speaker 1 is at x =0.850m .

A)What is the frequency of the sound? Assume vsound =340m/s.

B)What is the phase difference between the speakers?

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I have solved this question earlier with different figures. Please workout using yours figures. If you need any further help just PM me. If I have helped you please rate me 5 stars first (before you rate anyone else)

Two loudspeakers emit sound waves along the x-axis. A listener in front of both speakers hears a maximum sound intensity when speaker 2 is at the origin and speaker 1 is at x=0.480m. If speaker 1 is slowly moved forward, the sound intensity decreases and then increases, reaching another maximum when speaker 1 is at x=0.950 m.

What is the frequency of the sound? Assume velocity of sound is 340m/s.
What is the phase difference between the speakers?

Speakers & listener are all located on x-axis
Let y = lambda the wavelength
m y = .48 the speakers are located m wavelengths apart for first max
(m y + y) = .95 now the speakers are (m + 1) wavelenghts apart
y = .47 subtracting first equation from second equation
f = v / y = 340 / .47 = 723 / sec frequency
Since the wavelength is .47 then speaker 2 would need to be at
.01 meters if the two were exactly in phase but speaker 2 is
1/47 wavelength ahead of speaker 1 and since 1 wavelength is
360 deg then speaker is 1/47 * 360 = 7.7 deg ahead of speaker 1

genius_generous answered 1 year ago
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