How Close Can You Be to Speaker B for Destructive Interference?

In summary, the problem involves two loudspeakers emitting sound waves at the same frequency, with a distance of 8 m between one speaker and the observer. The wavelength of the sound is 2 m and the condition for destructive interference is when the difference in path lengths is a half integer number of wavelengths. The closest distance to speaker B that will result in destructive interference is when it is 4 wavelengths away, or 1 m. The speaker must be out of phase by 180 degrees.
  • #1
jaded18
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[SOLVED] Interference of Sound Waves

Two loudspeakers, A and B, are driven by the same amplifier and emit sinusoidal waves in phase. The frequency of the waves emitted by each speaker is 172 Hz. You are 8.00 m from speaker A. Take the speed of sound in air to be 344 m/s.
What is the closest you can be to speaker B and be at a point of destructive interference?
__________


I am having the hardest time trying to visualize the problem. I know that destructive interference occurs when the difference in path lengths traveled by sound waves is a half integer number of wavelengths. So I need to know the wavelength of the sound which is just 2m.

I also know that in general if d_a and d_b are paths traveled by two waves of equal frequency that are originally emitted in phase, the condition for destructive interference is d_a-d_b=n(wavelength)/2 where wavelength is what I calculated it to be (2m) and n=any nonzero odd integer. I think I need to know what the value of n is that corresponds to the shortest distance d_b to solve my prob. (is d_a=8m? then what is d_b?)

I'm going around in circles and getting nowhere. Please help!
 
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  • #2
d_a=8m is given.

One has determined the wavelength of the sound, 2 m, so the distance 8 m is 4 wavelengths. The the distance to B must be out of phase by 180 degrees in order to destructively interfere.

A half wavelength is 1 m.
 
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  • #3



Hello,

I understand your confusion and I will try my best to help you solve this problem. Let's start by defining some variables:

d_a = distance from speaker A to the point of destructive interference
d_b = distance from speaker B to the point of destructive interference
f = frequency of the sound waves (172 Hz)
v = speed of sound in air (344 m/s)
λ = wavelength of the sound waves (2m)

We know that for destructive interference to occur, the path lengths traveled by the two sound waves must differ by a half integer number of wavelengths. In other words, d_a - d_b = n(λ/2), where n is any nonzero odd integer.

Since we are trying to find the closest distance to speaker B that will result in destructive interference, we can assume that d_a is fixed at 8m (since you are 8m away from speaker A). This gives us the equation:

8 - d_b = n(λ/2)

Now, we need to find the value of n that will give us the shortest distance d_b. Since n is an odd integer, let's start with n = 1:

8 - d_b = (1)(2)

d_b = 6m

This means that if you are 6m away from speaker B, you will experience destructive interference. However, this is not the closest distance possible. Let's try n = 3:

8 - d_b = (3)(2)

d_b = 2m

This is the closest distance to speaker B that will result in destructive interference. Any closer and the path lengths will not differ by a half integer number of wavelengths.

So to answer your question, the closest you can be to speaker B and experience destructive interference is 2m.

I hope this explanation helps you better understand the concept of interference of sound waves. Let me know if you have any further questions.
 

FAQ: How Close Can You Be to Speaker B for Destructive Interference?

1. What is interference of sound waves?

Interference of sound waves is a phenomenon that occurs when two or more sound waves meet and combine. This can result in either constructive interference, where the waves reinforce each other and create a louder sound, or destructive interference, where the waves cancel each other out and create a quieter sound.

2. What causes interference of sound waves?

Interference of sound waves is caused by the superposition of two or more sound waves. This means that when two sound waves meet, their amplitudes are added together, resulting in either a larger or smaller amplitude, depending on the phase relationship between the waves.

3. How does interference of sound waves affect the loudness of a sound?

The interference of sound waves can either increase or decrease the loudness of a sound. When waves interfere constructively, the resulting sound will be louder than either of the individual waves. Conversely, when waves interfere destructively, the resulting sound will be quieter than either of the individual waves.

4. Can interference of sound waves be controlled or manipulated?

Yes, interference of sound waves can be controlled and manipulated. This is commonly done in audio engineering and sound design to create effects such as reverb, chorus, and flanging. It can also be used in noise-cancelling technology to reduce unwanted sounds.

5. How does the distance between sound sources affect interference of sound waves?

The distance between sound sources can greatly affect interference of sound waves. When the sources are very close together, the interference will be more intense and the resulting sound will be louder. As the distance between sources increases, the interference becomes less intense and the resulting sound will be quieter.

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