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Positions for sound intensity max/mins

  • Thread starter gmmstr827
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  • #1
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Homework Statement



Two loudspeakers are placed 3.00 m apart. They emit 494-Hz sounds, in phase. A microphone is placed 3.20 m distant from a point midway between the two speakers, where an intensity maximum is recorded. (a) How far must the microphone be moved to the right to find the intensity minimum? (b) Suppose the speakers are reconnected so that the 494-Hz sounds they emit are exactly out of phase. At what positions are the intensity maximum and minimum now?

See image for clearer description:
http://i29.photobucket.com/albums/c261/gmmstr827/speakersmic.jpg [Broken]

Homework Equations



Sound travels in air at 20° C at a rate of:
v = 343 m/s
b = beat
Γ_b = v / f_b
f_b = |f_1 - f_2 | <<< (absolute value)
f_1 = v/Γ_1
f_2 = v/Γ_2

The Attempt at a Solution



a)
f_b = 494 Hz
Γ_b = v / f_b
Γ_b = (343 m/s) / (494 Hz) = 0.69 m per wavelength beat
Moving the microphone half that distance would put it in the intensity minimum zone, so:
0.69 m / 2 = 0.35 m

The microphone must be moved about 0.35 meters to the right to reach an intensity minimum.

b)
This simply swaps the previous mins/maxes, therefore putting the default microphone position in a minimum, and about .035 m to the right would be a maximum.

^^^ Is this all correct? Thank you!
 
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Answers and Replies

  • #2
Redbelly98
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Homework Equations



Sound travels in air at 20° C at a rate of:
v = 343 m/s
b = beat
Γ_b = v / f_b
f_b = |f_1 - f_2 | <<< (absolute value)
f_1 = v/Γ_1
f_2 = v/Γ_2
But both speakers emit the same frequency (f1=f2), so there is no beating here.

What is the wavelength of the sound waves, given the frequency and speed of sound?

The Attempt at a Solution



a)
f_b = 494 Hz
Γ_b = v / f_b
Γ_b = (343 m/s) / (494 Hz) = 0.69 m per wavelength beat
This is simply the wavelength of the sound, there is no beating.
Moving the microphone half that distance would put it in the intensity minimum zone, so:
0.69 m / 2 = 0.35 m

The microphone must be moved about 0.35 meters to the right to reach an intensity minimum.
Not quite. You need to consider the distance from the microphone to each speaker. What must be true about those two distances, in order to have destructive interference? Hint: it involves the wavelength.

b)
This simply swaps the previous mins/maxes, therefore putting the default microphone position in a minimum, and about .035 m to the right would be a maximum.
Yes, once you correctly solve (a), just switch the answers to get (b).
 
  • #3
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But both speakers emit the same frequency (f1=f2), so there is no beating here.

What is the wavelength of the sound waves, given the frequency and speed of sound?


This is simply the wavelength of the sound, there is no beating.
Okay, so the wavelength will be 0.69 m according to my original answer, obtained from Γ = v / f.

Not quite. You need to consider the distance from the microphone to each speaker. What must be true about those two distances, in order to have destructive interference? Hint: it involves the wavelength.


Yes, once you correctly solve (a), just switch the answers to get (b).
Hm, I feel some geometry coming on here.
The formula to find the sides/height of an isosceles triangle is given as follows:
h = √(b^2-.25a^2) where h is the height, b is the length of each equal side, and a is the base.
I can use this to find the length of the equal sides, b.
3.20 m = √(b^2-.25(3.00 m)^2)
b = 3.53 m is the length of both side 1 and side 2.

Since those distances are equal when the microphone is in an intensity maximum, what if one distance was .69 m / 2 greater than the other, putting it halfway between wavelengths? .69 m / 2 = .35 m (rounded more accurately in context to the entire problem rather than the numbers given). However, this brings me back to my original solution of simply moving it 0.35 m to the right....
Well, since it's 3.53 m away from both speakers, and the wavelength is .69, I wonder how many wavelengths are in that distance? 3.53/.69 = 5.09 (using accurate rounding for the problem) wavelengths. For it to be a maximum, shouldn't the number of wavelengths be an integer?
What happens if I move the microphone 1.60 m to the right, so it's directly in front of the right hand speaker?
Pythagorean's Theorem: c^2 = a^2 + b^2
c^2 = 3^2 + 3.2^2
c = 19.24 m
19.24 / .69 = 27.884
Well, I don't feel like this is getting me anywhere....

Basically, it seems like I need to move it so that one speaker hits it on the wavelength, and one speaker hits it halfway through the wavelength.
 
  • #4
Redbelly98
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You need to consider the difference between the two distances. That difference must be:
  • n·λ for constructive interference, where n is an integer
  • (n+1/2)·λ for destructive interference
For locations along the "midline" the two distances are the same, hence their difference is 0·λ, and zero is an integer, so ... constructive interference.

If you move away from that midline, then the two distance are unequal. You need to find where their difference (i.e., subtract one from the other) is (½)·λ.

Hope that helps clear up what is going on.
 
  • #5
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1
You need to consider the difference between the two distances. That difference must be:
  • n·λ for constructive interference, where n is an integer
  • (n+1/2)·λ for destructive interference
For locations along the "midline" the two distances are the same, hence their difference is 0·λ, and zero is an integer, so ... constructive interference.

If you move away from that midline, then the two distance are unequal. You need to find where their difference (i.e., subtract one from the other) is (½)·λ.

Hope that helps clear up what is going on.
So, looking at the left speaker as x = 0 m, if I move the microphone from the default x = 1.5 m to x = 2.25 m, then there will be a difference of 1.5 m from the speakers to the microphone since 2.25 m - 0.75 m = 1.5 m

Therefore, I must move the microphone a total of 0.75 m to the right for an intensity minimum in part a).
 
  • #6
Redbelly98
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So, looking at the left speaker as x = 0 m, if I move the microphone from the default x = 1.5 m to x = 2.25 m, then there will be a difference of 1.5 m from the speakers to the microphone since 2.25 m - 0.75 m = 1.5 m

Therefore, I must move the microphone a total of 0.75 m to the right for an intensity minimum in part a).
That doesn't work out for two reasons:

1. Given the geometry of the figure, the distance can never be less than 3.20 m to either speaker:

speakersmic.jpg

The microphone is moving to the left or right. When the microphone is moved a distance x to the right, what are the distances to each speaker? You can use the distance formula that is taught in geometry. You can also consider the left-side speaker as located at (x=0m, y=0m), if that helps.

2. The 1.5 m you came up with is not half of the 0.69 m wavelength that we want.
 

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