How Does Signal Delay Affect Interference Between Two Loudspeakers?

In summary, the two loudspeakers emitting sound waves along the x-axis with a delay of 1.47 ms have a path length difference of 2m and an initial phase difference of 5pi, resulting in complete destructive interference. The delay only needs to be accounted for once in the equation for phase difference.
  • #1
bcjochim07
374
0

Homework Statement


Two loudspeakers emit sound waves along the x axis. Speaker 2 is 2.0 m behind speaker 1. Both loudspeakers are connected to the same signal generator, which is oscillating at 340 Hz, but the wire to speaker 1 passes through a box that delays the signal by 1.47 ms.

Is the interference along the x-axis perfect constructive, perfect destructive or something in between?


Homework Equations





The Attempt at a Solution



340 m/s = (340 Hz) * [tex]\lambda[/tex]
[tex]\lambda[/tex]= 1.0 m

First, turning to the formula deltaphi= 2pi * (deltax/wavelength)


plugging in the values delta phi = 2pi*(2.0m/1.0 m) = 4pi

and so I came up with perfect constructive.

Is this correct? I am trying to think about how the delay would affect the interference of the waves, so if anyone could offer a good explanation of why the delay does or does not affect the superposition, I would greatly appreciate it.

Thanks
 
Last edited:
Physics news on Phys.org
  • #2
Hi bcjochim07,

bcjochim07 said:

Homework Statement


Two loudspeakers emit sound waves along the x axis. Speaker 2 is 2.0 m behind speaker 1. Both loudspeakers are connected to the same signal generator, which is oscillating at 340 Hz, but the wire to speaker 1 passes through a box that delays the signal by 1.47 ms.

Is the interference along the x-axis perfect constructive, perfect destructive or something in between?


Homework Equations





The Attempt at a Solution



340 m/s = (340 Hz) * [tex]\lambda[/tex]
[tex]\lambda[/tex]= 1.0 m

First, turning to the formula deltaphi= 2pi * (deltax/wavelength)


plugging in the values delta phi = 2pi*(2.0m/1.0 m) = 4pi

and so I came up with perfect constructive.

Is this correct? I am trying to think about how the delay would affect the interference of the waves, so if anyone could offer a good explanation of why the delay does or does not affect the superposition, I would greatly appreciate it.

Thanks

The delay would determine the initial phase difference of the waves as they leave the speaker, which does have to be taken into account.
 
  • #3
Ok, so the delay is .00147 s, and in .00147 s the undelayed wave from speaker two moves (.00147s)(340m/s)= .50 m.

.50m/1.0m gives an initial phase difference of half a wavelength, or pi

delta x then becomes 2m + .50 m = 2.5 m

then using the formula delta phi= 2pi*(delta x)/lambda + initial delta phi

delta phi = 2pi* (2.5)/(1.0) + pi = 6 pi. This means that complete constructive interference happens. Is this correct?
 
  • #4
bcjochim07 said:
Ok, so the delay is .00147 s, and in .00147 s the undelayed wave from speaker two moves (.00147s)(340m/s)= .50 m.

.50m/1.0m gives an initial phase difference of half a wavelength, or pi

delta x then becomes 2m + .50 m = 2.5 m

then using the formula delta phi= 2pi*(delta x)/lambda + initial delta phi

delta phi = 2pi* (2.5)/(1.0) + pi = 6 pi. This means that complete constructive interference happens. Is this correct?

I don't believe that is correct. In your equation:

delta phi= 2pi*(delta x)/lambda + initial delta phi

you have accounted for the effect of the delay twice: in the (initial delta phi) term, and also by "pretending" that the delay meant that one wave moves 2.5 meters farther than the other. Either way would actually work, but you can only include the effect of the delay once.

So I would say the term

2pi*(delta x)/lambda

accounts for the phase shift due to the actual path length difference, so (delta x) should be 2m. And the initial delay only effects the second term.

In other words, in your original post you found the phase shift due to the path length difference; now just add the phase shift due to the delay.
 
  • #5
Oh, of course, I'm not sure why I added .5 to the 2m. So, then the phase shift is 5pi, so the interference is complete destructive
 

What is loudspeaker interference?

Loudspeaker interference is when two or more loudspeakers are playing sound at the same time and their sound waves overlap, causing distortion and reduced sound quality.

What causes loudspeaker interference?

Loudspeaker interference is caused by the overlapping of sound waves from different loudspeakers. This can occur when the loudspeakers are too close together or when they are playing sound at different frequencies.

How can loudspeaker interference be prevented?

Loudspeaker interference can be prevented by properly spacing out the loudspeakers and ensuring they are not playing sound at conflicting frequencies. Using directional speakers or placing sound-absorbing materials near the speakers can also help reduce interference.

What are the effects of loudspeaker interference?

The effects of loudspeaker interference include distorted sound, reduced sound quality, and difficulty in understanding spoken words or music. It can also cause listener fatigue and discomfort.

Can loudspeaker interference be fixed?

Yes, loudspeaker interference can be fixed by adjusting the placement and settings of the loudspeakers. It may also require using specialized equipment, such as signal processors or equalizers, to adjust the sound output and eliminate interference.

Similar threads

  • Introductory Physics Homework Help
Replies
3
Views
3K
  • Introductory Physics Homework Help
Replies
19
Views
3K
Replies
3
Views
2K
  • Introductory Physics Homework Help
Replies
2
Views
3K
  • Introductory Physics Homework Help
Replies
7
Views
4K
  • Introductory Physics Homework Help
Replies
20
Views
2K
  • Introductory Physics Homework Help
Replies
3
Views
2K
  • Introductory Physics Homework Help
Replies
4
Views
2K
Replies
1
Views
1K
  • Introductory Physics Homework Help
Replies
1
Views
881
Back
Top