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algol [13]
3 years ago
9

Imagine you are in an open field where two loudspeakers are set up and connected to the same amplifier so that they emit sound w

aves in phase at 688 Hz. Take the speed of sound in air to be 344 m/s.What is the shortest distance d you need to walk forward to be at a point where you cannot hear the speakers?

Physics
1 answer:
weeeeeb [17]3 years ago
6 0

Question:

If you are 3.00 m from speaker A directly to your right and 3.50 m from speaker B directly to your left. What is the shortest distance d you need to walk forward to be at a point where you cannot hear the speakers?

Answer:

The shortest distance d you need to walk forward to be at a point where you cannot hear the speakers is 5.625 m

Explanation:

Shortest distance is given by

For constructive interference of wave, we have

\Delta r_{construct} = 2nλ/2

For destructive interference of wave, we have

\Delta r_{destruct} = (2n+1)λ/2

That is the difference in the number of wavelength between constructive and destructive interference is λ/2

We note that our positioning is 3.0 m from the first speaker and 3.5 m from the second, Which is of the form 2n and 2n + 1

Therefore when we walk forward away from the two speakers, we cannot hear the speakers at

Δr = λ/2 = r_B - r_A

Which is the distance between the points where we have constructive and destructive interference or where there is destructive interference between the waves

Where:

r_A =  Distance from  speaker A and

r_B = Distance from  speaker B

λ/2 = (344 m/s /688 Hz)/2

= 1/4

For

λ/2 = r_B - r_A  we have

√(3.5² +d²) - √(3²+d²) = 1/4

∴ d = 5.625 m

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