# Audio interference of sound waves from two speakers

#### DottZakapa

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
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

Related Introductory Physics Homework Help News on Phys.org

#### Doc Al

Mentor
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?

#### DottZakapa

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?

#### Doc Al

Mentor
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.

#### DottZakapa

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 ?

#### Doc Al

Mentor
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?

#### DottZakapa

that
$39.8\space = m\space \lambda$?

#### Doc Al

Mentor
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.)

#### DottZakapa

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

#### haruspex

Homework Helper
Gold Member
2018 Award
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.

#### DottZakapa

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?

#### Doc Al

Mentor
$d\space sin\alpha \space= \frac \lambda 2 \space\space$
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.

#### DottZakapa

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

#### Doc Al

Mentor
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?) #### DottZakapa 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 ?

#### Doc Al

Mentor
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.

#### DottZakapa

well thats 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

#### Doc Al

Mentor
$39.8=( m+\frac 1 2 ) \lambda$
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.)

#### DottZakapa

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
$\lambda \space = 0.398 \times \space 2$
$f=\frac {340 \frac m s} {0.389 m \times \space 2} =427.13 Hz$

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
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 :)

#### Doc Al

Mentor
Good! A little brain sweat is good for you.

"Audio interference of sound waves from two speakers"

### Physics Forums Values

We Value Quality
• Topics based on mainstream science
• Proper English grammar and spelling
We Value Civility
• Positive and compassionate attitudes
• Patience while debating
We Value Productivity
• Disciplined to remain on-topic
• Recognition of own weaknesses
• Solo and co-op problem solving