How Do You Calculate the Frequency of a Tuning Fork Using Water Resonance?

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Homework Help Overview

The problem involves calculating the frequency of a tuning fork using resonance in a graduated cylinder filled with water. The resonances occur at specific water levels, and the speed of sound is provided. The presence of an end correction is noted, which affects the effective length of the air column.

Discussion Character

  • Exploratory, Assumption checking, Problem interpretation

Approaches and Questions Raised

  • Participants discuss the relationship between the observed distances and the wavelengths associated with the resonances. There are attempts to incorporate the end correction into their calculations. Questions arise about the assignment of resonances to specific distances and the implications of negative values obtained in calculations.

Discussion Status

The discussion is ongoing, with participants exploring different interpretations of the resonance distances and their corresponding wavelengths. Some guidance has been offered regarding the end correction, and there is an acknowledgment of the need to clarify the relationship between the resonances and the distances.

Contextual Notes

Participants are working under the constraints of the problem statement, including the fixed end correction and the requirement to relate the distances to the wavelengths of the resonances. There is uncertainty regarding the correct assignment of resonances and the implications of the calculations performed.

mizzy
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Homework Statement



A tuning fork is held over the top of a graduated cylinder that is slowly filled with water. Resonances are noted when the water level is 56.5cm and 17.5cm below the rim. Note that there is an end correction , meaning that the effective length of the air column is longer than the observed length by a fixed constant amount. If the speed of sound is 343m/s, then find the frequency of the tuning fork.

Homework Equations



f = v/lambda

1st resonance = lambda/4

2nd resonance = 3lambda/4

The Attempt at a Solution



I drew the cylinders. One cylinder is 56.5 cm below the rim and the other is 17.5cm below the rim. The first cylinder is at the second resonance of 3lambda/4 and the second is lamda/4. Now i don't know what to do. Can someone guide me?

thanks
 
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Consider the resonances one at a time. For example, the first one, which is 56.5 cm below the rim, and is at 3λ/4. What is the relationship between those two distances?

What about the other resonance?
 
for the first one, 56.5cm = lambda/4

the other resonance, 17.5 = 3lambda/4

is that right?
 
mizzy said:
Note that there is an end correction , meaning that the effective length of the air column is longer than the observed length by a fixed constant amount.
I don't believe you've taken this piece of information into account.
 
oh right...

first resonance: 56.5 + D = lamda/4

second resonance: 17.5 + D = 3lamda/4
 
k. I solved these equations, but I get a negative wavelength and therefore a negative frequency.

How can that be?:confused:
 
How do you know which resonance corresponds to which distance?
 
i don't know. I'm just guessing the the first would be the second and the second distance is the first.
 
Guesses are wrong sometimes, y'know :wink:
 
  • #10
k. let me think about this one. We are dealing with a tube that is open on one end and closed in the other. Equations i know to use for this situation: lambda = 4L , f = n(v/4L) when n= 1,3,5,etc.

In the question, the tube is slowly filled with water and the resonances are noted at the distances above..the first resonance happens at 56.5cm.

am i going in the right direction?
 
  • #11
K. I tried drawing a picture so i can visualize this. The second resonance will happen at 56.5cm and the first resonance will happen at 17.5cm.

I got the answer.
 

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