Interference of waves from two sources.

[nicksim117]

Homework Statement

Having a real tough time with this problem was wondering if anybody could help

Two identical loudspeakers 2 m apart are emitting 1800hz sound waves into a room where the speed of sound is 340m/s. Is the point 4m directly in front of one of the speakers, perpendicular to the plane of the speakers , a point of maximum constructive interference, perfect destructive interference, or somthing in between

Homework Equations

L=(m)xW/2 destructive
L=nxW constructive

The Attempt at a Solution

If L=4m then for constructive n would have to equal 0,1,2,3...whole #
If L is perfectly destructive then m would have to equal 1,3,5...odd #
Is that correct.

 

[kkrizka]

2. Homework Equations
L=(m)xW/2 destructive
L=nxW constructive

What is L? The equations look right, I just want to see if you know what the variables mean. From your attempt, I think you got it confused with a very similar equation for resonance.

 

[GTrax]

To get your answers for this question, you are going to have to do more than simply plug in values into equations and get answers immediately. It requires that you understand what happens

So start again - and this time, have a more fundamental equation as well.

[tex]v\,\,=\,\,f\cdot\lambda [/tex] where v is the speed of the wave, f is the frequency, and lambda is the wavelength.

Make a drawing. Show the sources 2m apart. Show the point 4m away from one where the interference is supposed to happen.

Imagine the waves leaving the speakers in phase. One travels 4m. The wave from the other source has to get to the same point over a different distance. It may or may not be in phase (constructive) by the time it arrives. Think wavelengths!

 

[nicksim117]

I did make a drawing. I have the point drawn out in front of two speakers which are 2m apartfrom each other. I can picture the waves coming at the object. but don't know how to tell if they are constructive or destructive. wave length is .188m if i plug the wl into either equation I don't get the proper values for n or m.

 

[nicksim117]

does this mean that it is somewhere in between?

 

[nicksim117]

L should be 4m is that right?

 

[nicksim117]

L should be 4m or 2m

 

[kkrizka]

does this mean that it is somewhere in between?

Yes, that is exactly what it means. Since the distance is not a multiple of the wavelength, the crests do not line up with each other and neither do the troughs. Since it is not a multiple of wavelengths plus a half of a wavelength apart, the troughs do not line-up with the crests and so do not cancel completely.

 

[kkrizka]

L should be 4m or 2m

L should be the distance between the two point sources, so L=2.

 

[nicksim117]

where would apply the 4m? should I find the distance from the second speaker to the point that is 4m directly in front of one of the speakers or am i thinking about this to hard?

 

[nicksim117]

or is the 4m not important. I am confused about the value of 4m and where to plug it into the equation.

 

[kkrizka]

where would apply the 4m? should I find the distance from the second speaker to the point that is 4m directly in front of one of the speakers or am i thinking about this to hard?

You do not need the distance in this case, because the three points (two points sources and the location you are interested in) lie in a straight line. Since it is the difference in distance that you are interested in, for a straight line this is just the distance between them.

 

[nicksim117]

Ok, I thought that they were 2m apart and so one would be 4m from point A and the other would be 4.47m from point A

 

[kkrizka]

Ok, I thought that they were 2m apart and so one would be 4m from point A and the other would be 4.47m from point A

4.47m from point A? How?

 

[nicksim117]

A2+B2=C2
If I square the distance from one point and add it to distance squared for the speakers then the square root of that would give me the distance from speaker two to point A wouldn't it.

 

[nicksim117]

It would be the hypotenuse.

 

[nicksim117]

I thought that is what the questions is asking. so you would have point A or whatever you want to call it 4m from one speaker , and 4.47m from the other. Am I reading the question wrong?

 

[kkrizka]

Umm, try to draw the picture. You don't need to use Pythagoras.

 

[alphysicist]

Hi nicksim117,

I thought that is what the questions is asking. so you would have point A or whatever you want to call it 4m from one speaker , and 4.47m from the other. Am I reading the question wrong?

That sounds right to me: the detector is 4m from one speaker and 4.47m from the other. You now want to compare the difference in these path lengths with the wavelength. If the path length difference is a integer multiple of a wavelength, then it is a constructive point. If it is an "integer plus a half" wavelength (like 0.5 [itex]\lambda[/itex],1.5 [itex]\lambda[/itex], etc.) then it is destructive interference.

 

[GTrax]

The hypotenuse is it.
That is the distance the wave from one source has to cover to get to the point where the action is.
4m is the distance the wave from the other source has to cover.

Next trick is - discover how long is a wavelength, so also you know what is a half-wavelength.

You may be disappointed to discover that the distances in the scenario do not happen to be so that you could conveniently fit an exact number of whole wavelengths, or even an odd number of half-wavelengths into either path.

But that does not matter! It is the difference in their path lengths that is important. If one wave got to travel that bit further such that its wave ended up arriving out of phase with the other, it would cancel. Maybe, in this case, it would add.

The passing of wave peaks and troughs past the observer point is a continuous thing. Do not confuse the direct pressure peaks and troughs of a sound wave passing with the maximums and minimums of the envelope of two waves mixing, either cancelling or augmenting each other depending on where the observer puts the microphone (or his ear). The sound waves themselves travel at speed, but the pattern of loud spots and dead spots do not move about. They stay put!

Although not strictly necessary, some folk like to express the difference distance in terms of wavelength degrees, because it helps visualize the values in between. You know a wavelength. Thats 360 degrees. Destructive happens most effectively at 180 degrees. I know I didn't give the full answer, but I think by now it should not be too tough to figure.