Walking away from two loudspeakers-destructive interference

In summary, you can hear a minimum sound intensity when you are a distance of (n+1/2) lambda away from one speaker and a distance of lambda away from the other.
  • #1
darkblue
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0

Homework Statement



You are standing 2.5 m directly in front of one of the two loudspeakers shown in the figure. They are 3.0 m apart and both are playing a 686 Hz tone in phase. As you begin to walk directly away from the speaker, at what distances from the speaker do you hear a minimum sound intensity? The room temperature is 20 degrees Celsius.

Homework Equations


v=343 m/s at 20 degrees
destructive interference: l1-l2=(n+1/2)lambda where n=0, 1, 2...
v=f*lambda

The Attempt at a Solution


First I found the wavelength which is 0.5 m. Then I drew a right angle triangle, where the distance between the speakers is 3 m and the other side is 2.5 m. Then I determined the hypotenuse, which is 3.91 m. For destructive interference, the sounds from both speakers should be out of phase and this happens when the person is a distance of (n+1/2)lambda away from one speaker and a distance of lambda away from the other (I think). So then I took the value 3.91, and added lambda/2 to get 4.16 for the hypoteneuse and solved for the other side to get 2.9 m. I did this another two times by adding 3/2 lambda and then 5/2 lambda to 3.91 (since the question is asking for the third value...this isn't explicitly stated, but I tried entering my answer with only 1 or 2 distances but I got a message saying I need to enter more values. When I entered 3 of them, a message came up saying I should re-check my calculations). I was wondering if there is something wrong with my logic.
 
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  • #2
darkblue said:
First I found the wavelength which is 0.5 m. Then I drew a right angle triangle, where the distance between the speakers is 3 m and the other side is 2.5 m. Then I determined the hypotenuse, which is 3.91 m.
OK. At that initial position, what's the path length difference between the sound coming from each speaker? That's the key. What happens to that path difference as you move?
For destructive interference, the sounds from both speakers should be out of phase and this happens when the person is a distance of (n+1/2)lambda away from one speaker and a distance of lambda away from the other (I think).
That's much too restrictive a criteria. To get destructive interference the sounds must be out of phase--all that's required for that is to have the path length difference be (n+1/2) lambda. (What's the first value of n that you reach as you walk away?)
 
  • #3
So at the initial position, one speaker is 3.91 m away while the other is 2.5 m away, right? So the difference would be 3.91-2.5 which is 1.41. As you move, the path length difference decreases.

The first value of n that you reach is 3, right?
 
  • #4
I tried something but I'm not too sure whether it makes sense.
using L1-L2=(n+1/2)lambda, I plugged in 1.41 for L1-L2 and solved for n since we know lambda is 0.5. By doing this I got n=2.32...would this confirm that the first value of n that you reach as you are walking away is 3?
 
  • #5
darkblue said:
So at the initial position, one speaker is 3.91 m away while the other is 2.5 m away, right? So the difference would be 3.91-2.5 which is 1.41. As you move, the path length difference decreases.
Sounds good.

The first value of n that you reach is 3, right?
How did you determine that?
 
  • #6
darkblue said:
I tried something but I'm not too sure whether it makes sense.
using L1-L2=(n+1/2)lambda, I plugged in 1.41 for L1-L2 and solved for n since we know lambda is 0.5. By doing this I got n=2.32...would this confirm that the first value of n that you reach as you are walking away is 3?
Remember that the path difference decreases as you walk away.
 
  • #7
I think I understand what you're saying...n should be 2, right?
 
  • #8
darkblue said:
I think I understand what you're saying...n should be 2, right?
Right. Now figure out the distance you need to be to make that happen.
 
  • #9
I see. So now I'm trying to solve for the distance...I tried doing this by first solving for the path difference by using L1-L2=(n+1/2)lambda, since we know n=2 and lambda=0.5. When I did this, I got the path difference as 1.25. I am now trying to solve for L2 (the distance from the speaker you are standing directly in front of). I tried to find a relationship between the sides using the Pythagorean theorem (since it is a right angle triangle) but nothing is working out. Is this the right approach? I can't think of any other way to find the distance without determining the path difference first.
 
  • #10
That's the right approach. Do this: Call the distance X. What's the hypotenuse in terms of X? What's the path difference? Set up and then solve the equation for the path difference.
 
  • #11
Ok, I solved it the way you said and everything worked out :) The only problem is that they're asking for three values...how would I find the other 2?
 
  • #12
I just thought of something...would I use n=1 and then n=0 to find the other 2 values?
 
  • #13
darkblue said:
I just thought of something...would I use n=1 and then n=0 to find the other 2 values?
Now you're cooking. :approve: :wink:
 
  • #14
Thanks for all the help! :)
 
  • #15
I am working on the same problem, I don't understand what to do for post #10 and beyond. Can someone please help me!
 
  • #16
LBRRIT2390 said:
I am working on the same problem, I don't understand what to do for post #10 and beyond. Can someone please help me!
Start by expressing the path difference in terms of X as described in post #10. Use the Pythagorean theorem.
 
  • #17
okay for the same question how would we solve "Suppose the speakers are reconnected so that the 686 Hz sounds they emit are exactly out of phase. At what positions are the intensity maximum and minimum now?"
 

1. What is destructive interference?

Destructive interference occurs when two sound waves with the same frequency and amplitude meet and cancel each other out, resulting in a decrease in overall sound intensity.

2. How does walking away from two loudspeakers affect destructive interference?

Walking away from two loudspeakers can increase the distance between the two sound sources, which can result in a decrease in destructive interference. This is because the sound waves have more space to spread out and may not overlap as much, reducing the canceling effect.

3. Can destructive interference be observed with only two loudspeakers?

Yes, destructive interference can occur with only two sound sources. However, the effect may be more noticeable with multiple sound sources or in a larger space.

4. How does the frequency and amplitude of the sound waves affect destructive interference?

The frequency and amplitude of the sound waves do not have a direct effect on destructive interference. However, if the two sound sources have different frequencies or amplitudes, the interference pattern may be more complex and result in areas of both constructive and destructive interference.

5. Is destructive interference always a negative effect?

Not necessarily. Destructive interference can be useful in certain situations, such as noise cancellation technology. However, in most cases, destructive interference is considered undesirable as it results in a decrease in overall sound intensity.

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