Formula for calculating wavelength of sound waves in interference?

In summary, the conversation discussed the discrepancy between the theoretical and experimental values in an experiment. The teacher acknowledged that the formula used was incorrect and the search for the correct formula was proving difficult. The results of the experiment were also provided, showing varying wavelengths for different frequencies. The conversation ended with a suggestion for the formula to include a third variable.
  • #1
Myphysicsteachershit
1
0
Homework Statement
We performed an experiment where 2 speakers were set up with 3 chalk lines spaced 1 metre apart from each other in front of the speakers. A 500 hz and then a 1000 hz sound was played through both speakers. We were to walk along the lines and mark the position of where the sound was the quietest (node) for both frequencies. After that we had to measure the distance between these nodes and calculate the wavelength.
Relevant Equations
For theoretical wavelength: lambda=340m/s divided by frequency
For experimental wavelength I used: lambda=2*distance between nodes
The values calculated was nowhere near the theoretical values, though I guessed they won't be as the results recorded was incredibly inaccurate. My teacher acknowledged the fact the final values won't be close to the theoretical ones but also said that my formula was wrong, that it works to find wavelength of the first line and not the other 2. I need help finding what formula to use, I have searched everyone but I don't really know what I'm looking for and our teacher said that it's not in the textbook that we use.

Here is a diagram of experiment:
_1GlvKWpV-f1ZrorClu9Cd4rIdiSOnSY2mTxxxNtrpWDz8hy35.png

Here is the results:

Average distance between nodes (cm)
Frequency (hz)Line 1Line 2Line 3
500253176105
1000333438None
My analysis:
ƛ/2=D
∴ƛ=2D where ƛ=wavelength (cm)
D=distance between nodes/antinodes (the average,cm)

500hz:
Line 1: 506cm
2: 356cm
3: 210

100hz:
Line 1: 666cm
2: 876cm
3: none

As seen above the wavelengths are not same for the different lines, that's because I used the same formula. My guess is that there should be a third value in the equation, eg 'n' where n might equal number of lines/harmonics or whatever. But I'm not sure.

Any help would be appreciated.
 
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  • #2
Hello @Myphysicsteachershit , :welcome: !

Myphysicsteachershit said:
For experimental wavelength I used: lambda=2*distance between node
This is true for two waves that travel in opposite directions. That is not the case in your setup.
There it looks (sounds) more like this: (https://en.wikipedia.org/wiki/Wave_interference)
Two_sources_interference.gif


As you see, the lines through the nodes 'move away from each other' if you go further away from the sources.
 

Related to Formula for calculating wavelength of sound waves in interference?

1. What is the formula for calculating the wavelength of sound waves in interference?

The formula for calculating the wavelength of sound waves in interference is λ = v/f, where λ represents wavelength, v represents the speed of sound, and f represents the frequency of the sound wave.

2. How do you determine the speed of sound for this formula?

The speed of sound can be determined by using the equation v = λf, where v represents the speed of sound, λ represents wavelength, and f represents frequency. This equation can also be rearranged to solve for v by dividing the wavelength by the frequency.

3. Can this formula be used for any type of sound wave interference?

Yes, this formula can be used for any type of sound wave interference, as long as the speed of sound and frequency are known. It is a universal formula that applies to all sound waves.

4. How does wavelength affect sound wave interference?

Wavelength plays a crucial role in sound wave interference. When two sound waves with the same wavelength interfere constructively, they will amplify each other and produce a louder sound. On the other hand, when two sound waves with opposite wavelengths interfere destructively, they will cancel each other out and produce a softer sound.

5. Is there a specific unit for wavelength in this formula?

Yes, the unit for wavelength in this formula is meters (m). This is because wavelength is a measure of distance, specifically the distance between two consecutive points on a sound wave that are in phase with each other.

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