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 3.50 m 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?
a. yes
b. no
Because the path difference is equal to the wavelength of the sound, the sound originating at the two speakers will interfere constructively at your location and you will perceive a louder sound.
Part B. What is the shortest distance d you need to walk forward to be at a point where you cannot hear the speakers? The forward direction is defined as being perpendicular to a line joining the two speakers and you start walking from the line that joins the two speakers.
Express your answers in meters to three significant figures.
d = m

Answers

Answer 1
Answer:

Final answer:

The sound will be louder when both loudspeakers are used compared to just one due to constructive interference.   The shortest distance you need to walk forward to not hear the sound anymore is 0.250 m, calculated using the condition for destructive interference.

Explanation:

Yes, the sound you hear will be louder than if only one speaker were in use. Given that both the loudspeakers are emitting waves in phase and the path difference is equal to the wavelength, the waves will interfere constructively at your location, which will result in a louder sound. This is because when waves meet while they're in phase, they add together to produce a greater amplitude.

The shortest distance d you would need to walk forward to a point where you can't hear the speakers can be calculated using the path difference. From the condition for destructive interference, we know that the path difference should be an odd multiple of half the wavelength (λ/2). Hence the distance would be [(1/2)*λ] which equals [(1/2)*(speed of sound/frequency)]. So, d = [(1/2)*(344/688)] = 0.250 m after joining the values.

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Answer 2
Answer:

Final answer:

Yes, the sound will be louder due to constructive interference. The shortest distance the student needs to walk forward to not hear the speakers is 0.25 m.

Explanation:

In this scenario, the student is standing equidistant from two loudspeakers that are emitting sound waves in phase at a frequency of 688 Hz. The speed of sound in air is 344 m/s. Part A of the question asks whether the sound that the student hears will be louder than if only one speaker was in use. The answer is yes. This is because the path difference between the two speakers is equal to the wavelength of the sound, resulting in constructive interference at the student's location.

Part B of the question asks for the shortest distance the student needs to walk forward to be at a point where they cannot hear the speakers. To determine this distance, we need to find the point where the path difference between the two speakers is equal to half a wavelength, resulting in destructive interference. The shortest distance the student needs to walk forward is equal to half the wavelength of the sound. Using the formula wavelength = speed of sound / frequency, we can find the wavelength and calculate the distance.

d = (344 m/s / 688 Hz) / 2 = 0.25 m

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Answers

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Explanation :

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Answers

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Answers

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Answers

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Answers

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Answers

Answer:

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Explanation:

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