Wavelength of Sound: Solve Using X & Y

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Homework Help Overview

The discussion revolves around determining the wavelength of sound based on a provided diagram and the relationship between sound intensity at specific points. The original poster attempts to use the concept of path difference and references principles from Young's double slit experiment, despite the absence of frequency and speed data.

Discussion Character

  • Exploratory, Assumption checking, Conceptual clarification

Approaches and Questions Raised

  • Participants explore the relationship between path difference and wavelength, questioning how the cancellation of sound intensity at point Y relates to wavelength values. Some suggest considering the implications of destructive interference and the mathematical reasoning behind phase differences.

Discussion Status

There is ongoing exploration of the concepts involved, with participants offering insights into the nature of sound waves and interference patterns. Some guidance has been provided regarding the relationship between path difference and wavelength, though no consensus has been reached on the exact calculations needed.

Contextual Notes

Participants note the lack of complete information, such as the frequency and speed of sound, which complicates the calculations. The original poster also mentions uncertainties about the implications of the diagram and the specific conditions at point Y.

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Homework Statement


Hello, If one can see the image I posted, the question that follows is this: Use the information provided to obtain a value for the wavelength of sound emitted. The signal is at maximum intensity at X, and minimum at Y.

http://uploads.im/7wJOq.jpg
upload_2017-12-22_10-19-42.png


Homework Equations


Now I know that the wavelength equation is v = f • λ, but in this case I don't have the frequency nor speed, so I have to use the lengths given to find the wavelength.

I know in Young's double slit experiment, the wavelength is: {Fringe Separation x Slit separation / Source to screen distance}.

The Attempt at a Solution


I believe after observing this diagram, the solution will be similar, as there is source to screen distance involved, but there are no fringes or slits as this is sound we are dealing with. I now know that the path difference is the difference in distance from the sources to the point, in this case Y, which will be 0.12m.
But I am still stuck, I don't know how to calculate the wavelength even using the phase difference formula because for that the wavelength is needed in the first place.
 

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There should be more information given. What is happening at point Y? Is this where the signal intensity is canceled out? or max intensity?
What would you know about possible values of wavelength if the signals canceled out each other at this location? Think about the differences in distance.
 
Richie Smash said:
Ah yes I know what you're saying, at Y the signal intensity is canceled out, or minimum,I believe. Well If the signals canceled each other out at this point... I'm just not making a connection, My best guess is the wavelength is the bigger distance minus the smaller distance? I'm not too sure why though, But I do know tht Y is minimum.

Edit: I've read through what you sent and I understand that the path difference for this question is 0.12m, but I see that I still cannot calculate the wavelength.
If they meet at that point 180° (½ wavelength) out of phase, then the sound intensity will be a minimum. Other possibilities are 3/2 wavelength, 5/2, etc. Use what you know about the speed of sound to see if these make sense as viable frequencies.

Do you see why 3/2 and 5/2 work as well (mathematically, maybe not practically though)?
 
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scottdave said:
If they meet at that point 180° (½ wavelength) out of phase, then the sound intensity will be a minimum. Other possibilities are 3/2 wavelength, 5/2, etc. Use what you know about the speed of sound to see if these make sense as viable frequencies.

Do you see why 3/2 and 5/2 work as well (mathematically, maybe not practically though)?

Is it that they work well because, as you add 180 degrees each time, the improper fractions such as 3/2 and 5/2 will show places of destructive interference? where as 2/2 or 4/2 would be constructive?
 
Richie Smash said:
Is it that they work well because, as you add 180 degrees each time, the improper fractions such as 3/2 and 5/2 will show places of destructive interference? where as 2/2 or 4/2 would be constructive?
Edit: I figured it out, the interference at Y is a half shift of the wavelength, which is given by the formula, R1-R equals n/2 and so on, and the wavelength of the wave is 0.24m, and from there i worked out the frequency, but the question now asks to find what distance the speakers are apart, can anyone help?
 
Richie Smash said:
Edit: I figured it out, the interference at Y is a half shift of the wavelength, which is given by the formula, R1-R equals n/2 and so on, and the wavelength of the wave is 0.24m, and from there i worked out the frequency, but the question now asks to find what distance the speakers are apart, can anyone help?
The diagram looks an awful lot like a right angled triangle - have you tried Pythagoras?
 

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