Imagine you are in an open field where two loudspeakers are set up and connected

Question

Imagine you are in an open field where two loudspeakers are set up and connected to the same amplifier so that they emit sound waves in phase at 688 . Take the speed of sound in air to be 344 .What is the shortest distance you need to walk forward to be at a point where you cannot hear the speakers?
Express your answers in meters to three significant figures.
The answer is 5.62 but can you work it out completely? when I work it out I dont get the correct answer, any suggestions?
?r = ?/2 = r22 - r2

?/2 = r22 - r2
?/2 = (344 m/s / 688 Hz) / 2
?/2 = ¼
¼ = r22 - r2
d = 5.62 m

Explanation / Answer

It's important to know the distance that you are from both speakers initially, which would also tell you how far the speakers apart... I think I found the first part of your question though =P Imagine you are in an open field where two loudspeakers are set up and connected to the same amplifier so that they emit sound waves in phase at 688 Hz. Take the speed of sound in air to be 344 m/s. Part A If you are 3.00 m from speaker A directly to your right and 3.50m from speaker B directly to your left, will the sound that you hear be louder than the sound you would hear if only one speaker were in use? Yes Part B What is the shortest distance d you need to walk forward to be at a point where you cannot hear the speakers? --------------------------------------… Notice that if you walk forward, the distance from you to each speaker is the hypothenuse of a right triangle. We know that the path difference between both distances needs to create destructive interference. Since, ??=344m/s/688Hz=0.500m, the shortest distance that will create destruuctive interference is 0.5??=(1/4)m (since both speakers are in phase - remember the condition for destructive interference is (m+1/2)?? for when two sources are in phase) Remember that as the person walks forward, the distance from each speaker becomes the hypothenuse of a right triangle, So, distance from a=v(3^2+d^2 ) distance from b=v(3.5^2+d^2 ) difference in both distances must equal 1/4 v(3.5^2+d^2 )-v(3^2+d^2 )=0.25m The answer is 5.73m

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