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

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To calculate the frequency of a tuning fork using water resonance, the resonances occur at two specific water levels: 56.5 cm and 17.5 cm below the rim of a graduated cylinder. The first resonance corresponds to the second harmonic (3λ/4) and the second resonance to the first harmonic (λ/4). An end correction must be considered, which adjusts the effective length of the air column. By applying the relationship between the distances and the wavelengths, the correct frequency can be determined. The discussion ultimately leads to a successful calculation of the tuning fork's frequency.
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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