Formula for calculating wavelength of sound waves in interference?

Click For Summary
The discussion centers on the inaccuracies in calculating sound wave wavelengths during an interference experiment. The initial formula used, λ = 2D, was deemed incorrect for multiple lines, as it only applies to the first harmonic. The teacher confirmed that the theoretical values would not match due to the formula's limitations. Participants suggest that a third variable, possibly related to harmonics, is necessary for accurate calculations. Clarification on the correct formula for different harmonics in sound wave interference is sought.
Myphysicsteachershit
Messages
1
Reaction score
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.
 
Physics news on Phys.org
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.
 
The book claims the answer is that all the magnitudes are the same because "the gravitational force on the penguin is the same". I'm having trouble understanding this. I thought the buoyant force was equal to the weight of the fluid displaced. Weight depends on mass which depends on density. Therefore, due to the differing densities the buoyant force will be different in each case? Is this incorrect?

Similar threads

Replies
10
Views
3K
Replies
3
Views
3K
Replies
20
Views
5K
  • · Replies 5 ·
Replies
5
Views
2K
  • · Replies 19 ·
Replies
19
Views
4K
  • · Replies 9 ·
Replies
9
Views
2K
  • · Replies 15 ·
Replies
15
Views
3K
  • · Replies 1 ·
Replies
1
Views
1K
Replies
7
Views
5K
  • · Replies 6 ·
Replies
6
Views
3K