Physics EK 1001- #708

[Biebs]

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#708.

A 1m glass tube is stood on end and partially filled with water. The air in the tube resonates when either a 440Hz or a 1230 Hz tuning fork is vibrated at its opening. It does not resonate with a 147Hz tuning fork. If the speed is sound in air is 340m/s approximately how deep is the water in the tube?
A. 19cm
B. 42cm
C. 50cm
D. 81cm

The answer is D, 81 cm, but I am a little confused to how they derived the answer.

Any help would be appreciated.

 

[aln012]

The formula you need to know for this problem is for a tube closed at one end: Fn = nv/4L, where n = 1,3,5..., where Fn is the resonance Frequency.

you want to find the L of the tube that the wavelength of sound can travel that enables it to produce a resonance frequency of 440Hz and 1230Hz. You just need to plug one number in to find the answer.

Solving for L, you get: L= nv/4Fn. Plug in your other numbers, and starting with n=1, you get L = 340/(4*440); L = .19m. If sound can only travel through .18m of the 1m tube, then the tube must be filled with (1m - .19m) or .81m of water.

 

[mikexima]

The formula you need to know for this problem is for a tube closed at one end: Fn = nv/4L, where n = 1,3,5..., where Fn is the resonance Frequency.

you want to find the L of the tube that the wavelength of sound can travel that enables it to produce a resonance frequency of 440Hz and 1230Hz. You just need to plug one number in to find the answer.

Solving for L, you get: L= nv/4Fn. Plug in your other numbers, and starting with n=1, you get L = 340/(4*440); L = .19m. If sound can only travel through .18m of the 1m tube, then the tube must be filled with (1m - .19m) or .81m of water.

I am having difficulty with this question too. even though I know the formula. But the formula has two variables that are uknown for us (both L and n). so what you are saying is that inorder to solve this question, we have to use trial and error by plugging in values for n (1,3,5) and comparing our answer to answer choices to see if it matches? is this the only/best method to solve this question? in other words, lets say this was not a multiple choice question and rather a straight solve it question, how would you go about it?

also, you did not use the extra information included in the question about how the tube doesnt resonate @ 147hz. according to the answer in the back of the book: "if 440hz was the resonance frequency of the third harmonic, the tube would resonate at 147 hz, and each 247hz after that, since it doesnt resonate at 147, it cant be the 3rd and must be the first". this explanation completly loses me as I have no idea how they come up with the number 147 in their calculations.

 

[tn4596]

well the way i understand it...
The reason there is no resonance with the 147 Hz frequency is because 440 Hz is the fundamental frequency (first harmonic). Therefore, at that frequency, the wavelength in the air contant one node and one antinode, because it is an open-ended tube, which make 1/4 of a full wavelength.
I dont think 147 Hz is the fundamental, if it is then there would be resonnace.

 

[mikexima]

great. anyone else wanna chime in with their methods/tricks?

 

[Majik]

The first thing you need to ask yourself is, for both standing waves of different harmonic frequencies, what remains the same? The answer is the length of the tube filled with air (L). Secondly, you should realize that the because the densities between air and water are very different, the sound wave does not pass through the water, but reflects back.

We can express frequency using the equation for a standing-wave closed at 1 end: Frequency = nv/4L (where n=1,3,5,7..etc; odd numbers only).
Note that the question gives you two consecutive frequencies. Keeping in mind that these two waves are consecutive and the harmonic must be an odd number, we can find length as follows:

(440Hz) Frequency(1) = nv/4L
(1320Hz) Frequency(2) = (n+2)v/4L <-- next consecutive frequency

There's two unknowns. The length of the air in the column, and the harmonic for each standing wave. You must equate both of these expressions to Length, because length remains constant. This way we could solve for the harmonic and use that to find the length of the column by plugging it into either equation:

nv/f(1) = (n+2)v/(f2)

(1320Hz / 440Hz)n = n+2
3n = n+2
2n-n = 2
n = 1

What we just solved for was the unknown harmonic, which it turns out is the fundamental harmonic frequency. The consecutive harmonic must therefore be (n+2) or n=3. We can use either of these two values to find length. I'll use n=1 and plug it into the fundamental frequency (440Hz) to solve for length:

440Hz = nv/4L
L = (1)(340m/s) / 4(440Hz)
L = 0.19m or 19cm

The length of the entire column is 100 cm.
The length of the air in the column is 19cm.
Therefore the length of water is the difference between the two: 100cm - 19cm = 81cm (D)

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An easier and much quicker way is by equating some of the variables except the harmonic:

f = nv/4L
440 Hz = nv/4L
1320 Hz = nv/4L

Note: v/4L is a constant, always. The only thing changing here is "n." What must you multiple the frequency (440Hz) to get 1320Hz? The answer is 3. Therefore n must equal 1 and 3 respectively. Harmonics could only be a whole number.

 

[mikexima]

Note: v/4L is a constant, always. The only thing changing here is "n." What must you multiple the frequency (440Hz) to get 1320Hz? The answer is 3. Therefore n must equal 1 and 3 respectively. Harmonics could only be a whole number.

thanks for the explanation. i get it much better now. my understanding of the topic is not as in depth but is this how you were able to tell that the two frequencies are consecutive?