Superposition of the sound waves from two speakers

In summary, Wavelength = 2L = 80cm. The wavelength of the sound is 80cm. The phase difference between the two loudspeakers is ϕ=2πx/λ. The amplitude of the sound (as a multiple of a) if the speakers are placed side by side is 2a.
  • #1
Elsi b
19
3
Homework Statement
Two loudspeakers emit sound waves of the same frequency along the x -axis. The amplitude of each wave is a. The sound intensity is minimum when speaker 2 is 10 cm behind speaker 1. The intensity increases as speaker 2 is moved forward and first reaches maximum, with amplitude 2a, when it is 30 cm in front of speaker 1. What is
a. The wavelength of the sound?
b.The phase difference between the two loudspeakers?
c.The amplitude of the sound (as a multiple of a ) if the speakers are placed side by side?
Relevant Equations
Wavelength=2L
2pix/wavelength+Phasedifference
Wavelength = 2L = 80cm
 

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  • #2
Elsi b said:
Homework Statement:: Two loudspeakers emit sound waves of the same frequency along the x -axis. The amplitude of each wave is a. The sound intensity is minimum when speaker 2 is 10 cm behind speaker 1. The intensity increases as speaker 2 is moved forward and first reaches maximum, with amplitude 2a, when it is 30 cm in front of speaker 1. What is
a. The wavelength of the sound?
b.The phase difference between the two loudspeakers?
c.The amplitude of the sound (as a multiple of a ) if the speakers are placed side by side?
Relevant Equations:: Wavelength=2L
2pix/wavelength+Phasedifference

Wavelength = 2L = 80cm
I confirm your value for wavelength.
The rest I cannot make sense of and it is truncated. Please define your variables and post as typed algebra, preferably in latex. Images are for diagrams and textbook extracts.
One symbol is unclear... could be a lowercase x or a backward α.
 
  • #3
ϕ=2πx/λ for the phase difference, how can you measure the phase difference for the speaker 2 when it moves to its maximum? I was trying to measure the speaker 2 at its minimum and then measure it to its maximum... does it make sense ...thanx
 
  • #4
Elsi b said:
ϕ=2πx/λ for the phase difference
Yes, that is a general formula for converting a distance movement to a phase shift.
With speaker 2 30cm in front, a listener hears them as in phase.
If you move speaker 2 back from its 30cm in front position to be next to speaker 1, what phase difference does that make for the listener?
 
  • #5
haruspex said:
Yes, that is a general formula for converting a distance movement to a phase shift.
With speaker 2 30cm in front, a listener hears them as in phase.
If you move speaker 2 back from its 30cm in front position to be next to speaker 1, what phase difference does that make for the listener?
It will be destructive waves when speaker 2 moves before to be in phase constructive with speaker 1
 
  • #6
Elsi b said:
It will be destructive waves when speaker 2 moves before to be in phase constructive with speaker 1
I don’t think you understood my question, and I certainly don’t understand your answer.

You have calculated the wavelength.
Suppose there is a listener somewhere in front of the speakers and speaker 2 is 30cm in front of speaker 1. The listener hears them as in phase.
Now speaker 2 is moved to be at the same place as speaker 1. How much further is that from the listener? Using the formula in post #3, what phase change does that make to what the listener hears from speaker 2?
 
  • #7
haruspex said:
I don’t think you understood my question, and I certainly don’t understand your answer.

You have calculated the wavelength.
Suppose there is a listener somewhere in front of the speakers and speaker 2 is 30cm in front of speaker 1. The listener hears them as in phase.
Now speaker 2 is moved to be at the same place as speaker 1. How much further is that from the listener? Using the formula in post #3, what phase change does that make to what the listener hears from speaker 2?
I know The two sounds waves will add up, but the formula ϕ=2πx/λ
 
  • #8
Elsi b said:
I know The two sounds waves will add up, but the formula ϕ=2πx/λ
I have no idea what you mean by "but the formula ϕ=2πx/λ". It is not a sentence.
Please try to answer my questions in post #6.
 
  • #9
haruspex said:
I don’t think you understood my question, and I certainly don’t understand your answer.

You have calculated the wavelength.
Suppose there is a listener somewhere in front of the speakers and speaker 2 is 30cm in front of speaker 1. The listener hears them as in phase.
Now speaker 2 is moved to be at the same place as speaker 1. How much further is that from the listener? Using the formula in post #3, what phase change does that make to what the listener hears from speaker 2?
Ill try to reply to ur question and this is what i understood. Let's take speaker 2 and the length is 30cm. How far is the wave (from speaker 2) can reach to the listener? This is what i understood... or maybe if u can explain to me? Thanx :-I
 
  • #10
Elsi b said:
How far is the wave (from speaker 2) can reach to the listener?
No, I'm not asking how far the wave has to travel to reach the listener. I did not specify exactly where the listener is.

Let me try a different tack.

Take a snapshot of the wave emitted from speaker 1. Draw a diagram. You have a sine curve that comes from speaker 1 out past speaker 2, 30cm in front. The snapshot is at an arbitrary instant, so it could be at any phase at speaker 1. It will be at a different phase at speaker 2

If it is at phase Φ1 at speaker 1, what phase is it at speaker 2?

Is English a second language for you? If so, please try using something like Google translate for both my questions and your answers.
 
Last edited:
  • #11
Phase 2 speaker 2
 
  • #12
Elsi b said:
Phase 2 speaker 2
The answer I am looking for to my question in post #10 is an expression involving Φ1, λ and 30cm. It uses the relevant equation you provided in post #1.
 
  • #13
Hey friend: i think u are a little bit advance in the formula. I am still new in this kind of questions. The formula that u posted , this one Φ1 its the first time is see this. What does it mean? Thanx. And if u know how to solve this question, i really appreciate it. Now I am starting to do complicated physics problems. Thanx for the help
 
  • #14
Elsi b said:
Hey friend: i think u are a little bit advance in the formula. I am still new in this kind of questions. The formula that u posted , this one Φ1 its the first time is see this. What does it mean? Thanx. And if u know how to solve this question, i really appreciate it. Now I am starting to do complicated physics problems. Thanx for the help
You posted this relevant equation:
2pix/wavelength+Phasedifference
Do you understand what it means?

If speaker 1 is emitting a sine wave ##A\sin(\omega t)## then at time t1 the phase at the speaker is ##\omega t_1##.
Its angular frequency is ##\omega## radians per second and its cycles per second frequency is ##\frac{\omega}{2\pi}##
If the wave moves at speed v then it has wavelength λ where ##v=\frac{\lambda\omega}{2\pi}##. It advances a distance one wavelength each cycle.
A part of the wave emitted at time t0 will take time x/v to travel distance x.
So at time t1 at a distance x from the speaker the phase will be ##\omega (t_1-\frac xv)=\omega (t_1-\frac{2\pi x}{\lambda\omega})=\omega t_1-\frac {2\pi x}{\lambda}##.
Meanwhile, speaker 1 is at phase ##\omega t_1##, so what is the phase difference between speaker 1's position and distance x from it?
 
  • #15
If you conclude the time for example, The phase difference at t1 is Pi rad.
 
  • #16
Elsi b said:
If you conclude the time for example, The phase difference at t1 is Pi rad.
No.
At time t1, speaker 1 is emitting at phase
haruspex said:
##\omega t_1##
Whereas at distance x in front of speaker 1, at the same instant, the wave from speaker 1 is at phase
haruspex said:
##\omega t_1-\frac {2\pi x}{\lambda}##.
So what is the difference between those two?
 
  • #17
haruspex said:
No.
At time t1, speaker 1 is emitting at phase

Whereas at distance x in front of speaker 1, at the same instant, the wave from speaker 1 is at phase

So what is the difference between those two?
ok the difference between the two formula is the difference -2πx/λ
 
  • #18
Elsi b said:
ok the difference between the two formula is the difference -2πx/λ
Right, so if x is 30cm and the wavelength is what you calculated, what is the phase difference in radians?
 
  • #19
2πx/λ
=2π0.3/0.8
=2.355
=π/1.33 rad :rolleyes:
 
  • #20
correct me if I am wrong before the speaker 2 moves to the front, can i measure the Δθ?
2πx/λ
=2π0.1/0.2


whats makes it hard is that speaker 2 moves in the front of speaker 1...
 
  • #21
ok so this is what i came up with
2πx/λ
=2π0.3/0.8
=3/4π
 
  • #22
Elsi b said:
ok so this is what i came up with
2πx/λ
=2π0.3/0.8
=3/4π
Yes.
 
  • #23
haruspex said:
Yes.
Thanx for the help
 
  • #24
Elsi b said:
Thanx for the help
You are welcome, but do you now understand it? I worry that I maybe gave too much help in post #14.
 
  • #25
yes the difference between two phases when the wave in phase moves to another position,
 

1. What is the superposition of sound waves from two speakers?

The superposition of sound waves from two speakers refers to the phenomenon where two sound waves from different sources combine to create a new resulting wave. This occurs when the two sound waves are of the same frequency and travel through the same medium.

2. How does the superposition of sound waves affect the overall sound?

The superposition of sound waves can either amplify or cancel out the overall sound depending on the phase relationship between the two waves. When the waves are in phase, they will amplify each other, resulting in a louder sound. When the waves are out of phase, they will cancel each other out, resulting in a quieter sound.

3. What is the difference between constructive and destructive interference in the superposition of sound waves?

Constructive interference occurs when two sound waves are in phase and add up to create a larger amplitude, resulting in a louder sound. Destructive interference occurs when two sound waves are out of phase and cancel each other out, resulting in a quieter sound.

4. How does the distance between the two speakers affect the superposition of sound waves?

The distance between the two speakers affects the superposition of sound waves by changing the phase relationship between the waves. As the distance between the speakers increases, the phase difference between the waves also increases, resulting in a greater chance of destructive interference and a quieter sound.

5. Can the superposition of sound waves be used to create surround sound?

Yes, the superposition of sound waves can be used to create surround sound by strategically placing multiple speakers around a room. By adjusting the distance and phase relationship between the speakers, a more immersive and realistic surround sound experience can be achieved.

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