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Loudspeaker interference

by bcjochim07
Tags: interference, loudspeaker
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bcjochim07
#1
Sep7-08, 04:27 PM
P: 374
1. The problem statement, all variables and given/known data
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?


2. Relevant equations



3. 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
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alphysicist
#2
Sep7-08, 07:52 PM
HW Helper
P: 2,249
Hi bcjochim07,

Quote Quote by bcjochim07 View Post
1. The problem statement, all variables and given/known data
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?


2. Relevant equations



3. 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.
bcjochim07
#3
Sep13-08, 09:26 PM
P: 374
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?

alphysicist
#4
Sep13-08, 10:02 PM
HW Helper
P: 2,249
Loudspeaker interference

Quote Quote by bcjochim07 View Post
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.
bcjochim07
#5
Sep13-08, 11:05 PM
P: 374
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


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