Audio interference of sound waves from two speakers

In summary: No, the path difference in this case is not an integer number of wavelengths. In fact, it would be a distance of exactly ##\lambda## wavelengths.
  • #1
DottZakapa
239
17
Homework Statement
Two identical audio speakers, connected to the same amplifier, produce monochromatic sound waves with a frequency that can be varied between 300 and 600 Hz. The speed of the sound is 340 m/s. You find that, where you are standing, you hear minimum intensity sound
a) Explain why you hear minimum-intensity sound
b) If one of the speakers is moved 39.8 cm toward you, the sound you hear has maximum intensity. What is the frequency of the sound?
c) How much closer to you from the position in part (b) must the speaker be moved to the next position where you hear maximum intensity?
Relevant Equations
interference
Homework Statement: Two identical audio speakers, connected to the same amplifier, produce monochromatic sound waves with a frequency that can be varied between 300 and 600 Hz. The speed of the sound is 340 m/s. You find that, where you are standing, you hear minimum intensity sound
a) Explain why you hear minimum-intensity sound
b) If one of the speakers is moved 39.8 cm toward you, the sound you hear has maximum intensity. What is the frequency of the sound?
c) How much closer to you from the position in part (b) must the speaker be moved to the next position where you hear maximum intensity?
Homework Equations: interference

I have no idea on how to proceed

I started with
## frequency=\frac {speed\space of\space sound} \lambda \space = \frac {340 \frac m s} \lambda ##
then
##d \space sin\alpha \space = \space \frac \lambda 2\space ##
but now I'm stuck

Any help please?
 
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  • #2
Hint: What is the condition for destructive/constructive interference? If the speakers were equidistant from you, would that produce a maximum or a minimum of intensity?
 
  • #3
i hear a minimum intensity because one of the two is not at the same distance from me, therefore I have a destructive interference.As the two waves reach me, are out of phase of half wavelength correct? if i hear max intensity there will be a phase difference of an integer wave.
correct?
 
  • #4
Yes. For maximal destructive interference, the waves must be out of phase by 180 degrees or half a wavelength. (For constructive interference, the waves must be in phase again -- the path difference must be an integer number of wavelengths.)

So far, so good. Keep going.
 
  • #5
getting the speaker closer of 39.8 cm leads to a path difference that is an integer number of wavelength , right?
but how do i get the wavelength ?
 
  • #6
DottZakapa said:
getting the speaker closer of 39.8 cm leads to a path difference that is an integer number of wavelength , right?
Right. So what does that tell you about how 39.8 cm relates to the wavelength?
 
  • #7
that
##39.8\space = m\space \lambda##?
 
  • #8
DottZakapa said:
that
##39.8\space = m\space \lambda##?
No. Realize that by moving one speaker that distance, the path difference is now an integer number of wavelengths, whereas before it must have been an odd number of half wavelengths.

Given that, how must the 39.8 cm relate to the wavelength? (There may be several possibilities, which you'll have to choose among.)
 
  • #9
Doc Al said:
Given that, how must the 39.8 cm relate to the wavelength? (There may be several possibilities, which you'll have to choose among.)

Sorry i have no clue :confused:
 
  • #10
DottZakapa said:
i hear a minimum intensity because one of the two is not at the same distance from me, therefore I have a destructive interference.As the two waves reach me, are out of phase of half wavelength correct? if i hear max intensity there will be a phase difference of an integer wave.
correct?
Right, so write each of those statements as algebraic expressions for the two distance differences.
 
  • #11
haruspex said:
Right, so write each of those statements as algebraic expressions for the two distance differences.
##d\space sin\alpha \space= \frac \lambda 2 \space\space or \space k(r_2-r_1)=\frac 1 2 \space with\space k=\frac {2\pi} \lambda##

the second would be
##(d-39.8)\space sin\alpha \space= \lambda ##

like that?
 
  • #12
DottZakapa said:
##d\space sin\alpha \space= \frac \lambda 2 \space\space##
Not sure where you're getting that one; perhaps you are mixing this up with another scenario.

Let's keep it simple. Assuming ##r_1## and ##r_2## are the distances from the speakers to you. Then the path difference is ##r_2 - r_1##. In the initial positions, the interference is deconstructive, so that path difference must equal an odd number of ##\lambda/2##.

Now you are changing one of the speaker distances so that the path difference is now an integral number of wavelengths. So, in terms of wavelengths, what must the given distance change (39.8 cm) equal? There are many possibilities; just start listing them and then you can decide which are physically possible later.
 
  • #13
Doc Al said:
Not sure where you're getting that one; perhaps you are mixing this up with another scenario.

Let's keep it simple. Assuming ##r_1## and ##r_2## are the distances from the speakers to you. Then the path difference is ##r_2 - r_1##. In the initial positions, the interference is deconstructive, so that path difference must equal an odd number of ##\lambda/2##.

Now you are changing one of the speaker distances so that the path difference is now an integral number of wavelengths. So, in terms of wavelengths, what must the given distance change (39.8 cm) equal? There are many possibilities; just start listing them and then you can decide which are physically possible later.
hmmm
isn't this working like a double slit ?
if all those formulas listed above aren't appropriate for the case I have no other ideas
apart saying that that distance change is equal to an integer of wave length
also subtracting the integer part and doing ##\frac {0.8} \lambda =\space integer \space number ## doesn't bring me anywhere because i also don't know the wave length.

i can't see how to use the data in other ways 😕
 
  • #14
If the original path length difference was an odd number of half wavelengths, how many wavelengths must you add (or subtract) to make the new path length difference equal a whole number of wavelengths?

(Silly analogy: You have a certain amount of money, say $12.50. How much money must you add or subtract to end up with a whole number of dollars?)
 
  • #15
Doc Al said:
If the original path length difference was an odd number of half wavelengths, how many wavelengths must you add (or subtract) to make the new path length difference equal a whole number of wavelengths?

(Silly analogy: You have a certain amount of money, say $12.50. How much money must you add or subtract to end up with a whole number of dollars?)
Getting the speaker closer to me, reducing for example ##r_2 \space by \space subtracting \space 39.8 ## the path difference between the two speakers leads to a phase difference that is an integer number of wave length,
##r_2-39.8-r_1=m\lambda##
which is like adding to the previous half wave length an other half wave length ?
 
  • #16
DottZakapa said:
which is like adding to the previous half wave length an other half wave length ?
Yes. That added distance could be ##\lambda/2## or ##3\lambda/2## or ...

For each possibility, figure out the wavelength and the corresponding frequency.
 
  • #17
well that's exactly what i can't see how to do
i know that
##f= \frac {speed \space of \space sound} \lambda##
##r_2-39.8-r_1=m\lambda##
##39.8=( m+\frac 1 2 ) \lambda##
but i don't see how to get things out
 
  • #18
DottZakapa said:
##39.8=( m+\frac 1 2 ) \lambda##
Good. (But careful with units!) m can be 0, 1, 2, and so on. So calculate the first few of those wavelengths. And their corresponding frequencies. (Note that you are given the possible range of frequencies.)
 
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  • #19
Doc Al said:
Good. (But careful with units!) m can be 0, 1, 2, and so on. So calculate the first few of those wavelengths. And their corresponding frequencies. (Note that you are given the possible range of frequencies.)
39.8 cm = 0.398 m
##39.8=( m+\frac 1 2 ) \lambda## for m=0 i get
##\lambda \space = 0.398 \times \space 2##
which leads to
##f=\frac {340 \frac m s} {0.389 m \times \space 2} =427.13 Hz##

using m = 1 leads to a frequency that is out of the range.
so i think that is correct as it is, it also matches the solution (but a solution without understanding is meaningless).

next
DottZakapa said:
c) How much closer to you from the position in part (b) must the speaker be moved to the next position where you hear maximum intensity?

being
## 0.398 = \frac {\lambda} 2 ## a quantity added to an half wave length on order to get max intensity, from this point the speaker needs to be closer 2 half wavelengths that is

##2 \cdot \space 0.398m = 0.796 m##

and that will solve the problem.

Damn it made me sweat, i appreciate your patience and time you spent , it seemed impossible for me, sometimes i get stuck in stupid reasonings.
I'm very thankful :)
 
  • #20
Good! A little brain sweat is good for you. 😄
 

FAQ: Audio interference of sound waves from two speakers

1. What is audio interference of sound waves from two speakers?

Audio interference occurs when sound waves from two speakers overlap and interfere with each other, resulting in distortion or cancellation of certain frequencies. This can happen when the speakers are too close together or when they are producing sound at similar frequencies.

2. How does audio interference affect sound quality?

Audio interference can cause a decrease in sound quality by creating unwanted distortion or making certain frequencies harder to hear. This can also make it difficult to distinguish between different sounds, leading to a muddled or unclear listening experience.

3. What can be done to reduce audio interference?

There are several steps that can be taken to reduce audio interference, including: adjusting the placement of the speakers, using speakers with directional sound output, and utilizing sound dampening materials in the room. Additionally, using speakers with a wider frequency range can help prevent overlap and interference.

4. Can audio interference cause damage to the speakers?

It is unlikely that audio interference will cause physical damage to the speakers, as most modern speakers have built-in protection to prevent this. However, prolonged exposure to high levels of interference can potentially damage the speakers over time.

5. Is audio interference a common issue with sound systems?

Audio interference can be a common issue with sound systems, especially in environments with multiple speakers or in areas with a lot of ambient noise. However, with proper equipment and placement, it can be minimized or eliminated to ensure high-quality sound.

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