Constructive interference with sound -- word problem

Homework Statement

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I was little bit confused about interference word problem
in an old physics exam. I managed to ace the problem in the exam, by applying a little bit common sense to it, but I feel like I didn't understand the concept of the interference completely.

(Loud)speakers A and B are 3,4m apart from each other and they are sending same-phase signal at 350Hz (ostensibly at the same pace in terms of time synchronosly or however it is said in English)
An observer closes in the distance towards speaker A along the perpendicular line, and that perpendicular line goes through the point A. The perpendicular line is perpendicular to the Line AB which is the distance between speakers A and B.

The observer notices that the sound reacher minimum and maximum values one after the other. The place which corresponds with the first maximum, is located 5m distance apart from point A.

Calculate the speed of sound based on this.

The Attempt at a Solution

I seemed to remember that constructive interference happens
when the sound is at max levels at the observer's location.

Our teacher had given us a little bit confusing formula such as

## \Delta X = ## difference between distances that interfering waves travelled (fair enough)

## \Delta X = n~~*~~ \frac{\lambda}{2} ##

In cases of constructive interference substitute n= {2, 4, 6, 8...}

In cases of destructive interference substitute n= {1, 3, 5, 7...}

well, it is little bit confusing which is the correct n= something? Which n is correct value for each different case and what is the reasoning for it?

One way I did it in the exam was to look at the constructive interference and use some common sense to gauge what is the reasonable velocity. The closest to reasonable speed of sound was calculated when I substituded n= 2

therefore the equation became

## \Delta X = 2* \frac{\lambda}{2} ##
## ~~\leftrightarrow~~ \Delta X = ~~\lambda##
## \leftrightarrow~~ \sqrt{36.56} -5 = \lambda ##

## v= (\sqrt{36.56} -5) * (350) ##
## v= 366.2703 m/s##
using that reasoning you can use the wave equation to calculate ## v= \lambda * f ##

I don't remember this stuff from physics. But I assume that the first maximum would occur when there is exactly one wavelength more in the longer path than in the shorter path - constructive interference, as you said. Based on that, I came up with the same answer you did.

kuruman
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The place which corresponds with the first maximum, is located 5m distance apart from point A.
Do you understand what this is saying to you about what kind of interference you get at this location?

vela
Staff Emeritus
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Our teacher had given us a little bit confusing formula such as

## \Delta X = ## difference between distances that interfering waves travelled (fair enough)

## \Delta X = n~~*~~ \frac{\lambda}{2} ##

In cases of constructive interference substitute n= {2, 4, 6, 8...}

In cases of destructive interference substitute n= {1, 3, 5, 7...}

I think it's better to write it as
$$\frac{\Delta x}{\lambda} = n$$ for constructive interference, and
$$\frac{\Delta x}{\lambda} = n + \frac 12$$ for destructive interference. The quantity ##\Delta x/\lambda## is the path difference measured in units of wavelengths.

Do you understand what this is saying to you about what kind of interference you get at this location?
its going to be constructive interference of some kind.

But I dont know if its one wavelength which is the delta distance.

Or is it two times the wavelength which is the delta distance.

kuruman
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The place which corresponds with the first maximum, is located 5m distance apart from point A.
First maximum is one wavelength.

1.) If you wanted to know, how would you get the distance travelled for each signal from speaker A -> observer and from speaker B->observer in the case of the second maximum.

Is there enough info?
for sure it is known what the ## \Delta x ## will be, but the individual lengths AO and BO (A to observer and B to observer)

kuruman
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Is there enough info?
for sure it is known what the Δx \Delta x will be, but the individual lengths AO and BO (A to observer and B to observer)
Yes, there is enough info.
This says that OA is 5 meters at the first maximum.
The observer notices that the sound reacher minimum and maximum values one after the other. The place which corresponds with the first maximum, is located 5m distance apart from point A.
This says that AOB is a right triangle in which AO is one of the right sides and AB = 3,4 m is the other right side.
(Loud)speakers A and B are 3,4m apart from each other and they are sending same-phase signal at 350Hz (ostensibly at the same pace in terms of time synchronosly or however it is said in English)
An observer closes in the distance towards speaker A along the perpendicular line, and that perpendicular line goes through the point A. The perpendicular line is perpendicular to the Line AB which is the distance between speakers A and B.
Use the Pythagorean theorem to find OB.

late347
Yes, there is enough info.
This says that OA is 5 meters at the first maximum.

This says that AOB is a right triangle in which AO is one of the right sides and AB = 3,4 m is the other right side.

Use the Pythagorean theorem to find OB.

I did get the lengths for direct path (5m = line from A to O = let's call it ##d_1##) in the first sound maximum case.

But I was wondering what would be the length of the direct path for the second sound maximum ##d_2##. That is to say the length at which the second maximum occurs along the line which goes through point A and is perpendicular to AB. I think the 90deg angle does keep the same, but the other angles must be changing I suppose. And the line AB stays the same at 3.4m. Difference between the new hypotenuse and the new side would be 2*##\lambda## but I can't get much further than that

kuruman
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But I was wondering what would be the length of the direct path for the second sound maximum ##d_2##.
Write a general expression for the path difference Δx2 between the hypotenuse BO and the right side OA in terms of the unknown OA and the fixed AB. You also have your general expression ##\Delta x = n~~*~~ \frac{\lambda}{2}## where n = {2, 4, 6, 8, ...} for constructive interference. You have already used n = 2 for the first maximum. What do you think you should use for the second maximum? How about the third maximum?

Set the two expressions for Δx equal and solve for OA, which obviously depends on your choice of n.

late347
well it looks like we have formula such as
## \Delta X = n*\frac{\lambda}{2} ##
n=4
## \leftrightarrow \Delta X = 2.093 ##

from the triangle we know that
length c = BO = hypotenuse it is going to vary
length a = AO = varying side
length AB = fixed value at 3.4m

## c =\sqrt{a^2+11.56} ##

we also know that ## \Delta X = c - a ## because c is the largest length hypotenuse
plug in c = something and we get an equation in terms of variable a only
## 2.093 = \sqrt{a^2+11.56} - a ##

## a = 1.715086001 ##

looks like the length AO=a got shorter... is it supposed to be this way?

Now I'm totally confused... how can the second maximum for sound levels be closer to the speakers?
It does appear to be the case nonetheless

kuruman
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... how can the second maximum for sound levels be closer to the speakers?
Why not? You need to understand in what way the first maximum is "first". The problem states
An observer closes in the distance towards speaker A along the perpendicular line, ...
The observer is approaching the speaker until the first maximum is heard. Any other maximum must be closer to speaker A than that, otherwise it would have to be "first".

late347
Why not? You need to understand in what way the first maximum is "first". The problem states

The observer is approaching the speaker until the first maximum is heard. Any other maximum must be closer to speaker A than that, otherwise it would have to be "first".

quite so...

I came to the same realisation after I started visualizing the process by which the difference of the length c, and length a, of the triangle changes.

When the length a = AO shrinks, then that means that the result (the difference ##\Delta X## = (c-a) is going to increase and increase. So that is going to be the correct direction to go to for the second maximum along the line AO. Towards length a "becoming smaller" or rather the observer walking closer towards the speaker A.

Another question.
Are those maximum locations one, and two (first and second) are they essentially fixed and static points along the line AO. And at those points, the sound is simply going to be at the max amplitude.

But the location of those max points is essentially going to be dictated by the current arrangement of the speakers in this particular example problem? So, of course it seems that the speakers are fixed so that the max and min locations along the line AO would seem to be some specific fixed points on that line.

kuruman
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I think you can answer your own questions if you try this. Imagine you are walking towards speaker A, locating all the maxima as you do so. The condition for the nth maximum is
$$\sqrt{y_n^2+d^2}-y_n=n \lambda$$
where yn is your distance from the speaker, d is the separation between speakers and n = 1, 2, 3, ...
Solve this equation for yn, take a good look at the solution and then answer the question, "How many maxima will you encounter until you reach speaker A?"

I think you can answer your own questions if you try this. Imagine you are walking towards speaker A, locating all the maxima as you do so. The condition for the nth maximum is
$$\sqrt{y_n^2+d^2}-y_n=n \lambda$$
where yn is your distance from the speaker, d is the separation between speakers and n = 1, 2, 3, ...
Solve this equation for yn, take a good look at the solution and then answer the question, "How many maxima will you encounter until you reach speaker A?"

After some tabulating of values and some wolfram alpha calculations

I used my formula ## \Delta X = n * \frac{\lambda}{2} ## where ## n = [2, 4, 6, 8 ...] ##

## \sqrt{a^2+11.56} -a = n * \lambda / 2 ##

##choose~~ n = 2 ; a = 5m ## first maximum
##choose~~ n= 4 ; a \approx 1.7151m ## second maximum
## choose~~ n= 6 ; a \approx 0.271351m ## third maximum

beyond that it did not seem to bring sensible results for a's value

kuruman
$$y_n=\frac{d^2-n^2 \lambda^2}{2 n \lambda}.$$
$$d = n_{max}\lambda ~~\rightarrow ~~n_{max}=\frac{d}{\lambda}=\frac{3.4}{1.05}=3.$$