Relationship between Frequency and Volume in Blowing Air into a Bottle

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In summary, the problem presents a plastic bottle filled with water and a frequency f obtained by blowing into it. The goal is to prove that f=f_{0}(\frac{V}{V_{0}})^{\alpha}, where f_{0}, V_{0}, \alpha are constants. The wavelength is thought to be directly proportional to the height of the air space, but f is inversely proportional to V. However, through experimentation, it is discovered that the formula given in the problem is not correct, as when V=0, f does not equal 0. Therefore, it is concluded that the formula in the problem is not an exact answer.
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mtr
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Homework Statement


There is a plastic bottle into which we blow in such a way, that we receive a sound of freguency f. The bottle is filled with water that has volume V. Prove, that [tex]f=f_{0}(\frac{V}{V_{0}})^{\alpha}[/tex], where [tex]f_{0}, V_{0}, \alpha[/tex] are constants.

Homework Equations



The Attempt at a Solution


I always thought, that the wavelength is directly proportional to height of an empty part of the bottle, but then f is inversely proportional to V. Is there any mistake in this task?
 
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  • #2
Hi mtr,

mtr said:

Homework Statement


There is a plastic bottle into which we blow in such a way, that we receive a sound of freguency f. The bottle is filled with water that has volume V. Prove, that [tex]f=f_{0}(\frac{V}{V_{0}})^{\alpha}[/tex], where [tex]f_{0}, V_{0}, \alpha[/tex] are constants.

Homework Equations



The Attempt at a Solution


I always thought, that the wavelength is directly proportional to height of an empty part of the bottle,

As the height of the air space decreases, the wavelength will decrease also.

but then f is inversely proportional to V.

What is your reasoning for saying that?
 
  • #3
alphysicist said:
As the height of the air space decreases, the wavelength will decrease also.

That's right, but considering the wavelength, we know, that [tex]\lambda = 2L[/tex] so when L decreases, lambda decreases proportionally to the 1st power of L and inversely proportionally to the 1st power of V.


alphysicist said:
What is your reasoning for saying that?

I can see my mistake. In fact it is lambda inversly proportional to V so f is directly proportional to V.

Still, I made an experiment and it looks like there is such a formula: [tex]f=f_{0} \alpha^{\frac{V}{V_{0}}}[/tex]
Actually I'm even more confused with this. Still the formula given in the task cannot be correct, because when V=0, f does NOT equal 0.
 
  • #4
mtr said:
That's right, but considering the wavelength, we know, that [tex]\lambda = 2L[/tex] so when L decreases, lambda decreases proportionally to the 1st power of L and inversely proportionally to the 1st power of V.

Think of the bottle as a cylinder. This would be like a pipe with one end closed, so the fundamental wavelength is 4L, if L is the height of the air column.

But I don't think the wavelength is inversely proportional to V, as you've seen when you looked at the frequency. As V increases, the wavelength decreases, but not proportionally.




I can see my mistake. In fact it is lambda inversly proportional to V so f is directly proportional to V.

Still, I made an experiment and it looks like there is such a formula: [tex]f=f_{0} \alpha^{\frac{V}{V_{0}}}[/tex]
Actually I'm even more confused with this. Still the formula given in the task cannot be correct, because when V=0, f does NOT equal 0.

Was the problem that you had in the original post all that there was? There were no diagrams, or approximate ranges given? By treating the bottle as a cylinder you can get an exact result, which does not match with [itex]f\propto V^{\alpha}[/itex] from your original post.

If I'm reading the problem correctly, and you haven't overlooked any parts of the problem in your post, then I would agree that the formula in the original post does not seem to be an exact answer.
 
  • #5
alphysicist said:
But I don't think the wavelength is inversely proportional to V, as you've seen when you looked at the frequency. As V increases, the wavelength decreases, but not proportionally.

Well, I won't argue with that ;)


There were no additional information. They also do not say anything about the bottle's shape, so I suppose it can be any.
 

FAQ: Relationship between Frequency and Volume in Blowing Air into a Bottle

1. How does blowing air into a bottle create sound?

When air is blown into a bottle, it creates vibrations in the air column inside the bottle. These vibrations travel through the air and reach our ears, which interpret them as sound.

2. Why does the sound change when I blow into a bottle with different amounts of air?

The pitch of the sound produced by blowing into a bottle depends on the length of the air column inside the bottle. When more air is blown into the bottle, the air column becomes longer, resulting in a lower pitch sound. Similarly, less air produces a shorter air column and a higher pitch sound.

3. Can I make different notes by blowing into a bottle with different sizes and shapes?

Yes, the size and shape of the bottle also affect the pitch of the sound produced. A longer and narrower bottle will produce a lower pitch sound, while a shorter and wider bottle will produce a higher pitch sound.

4. Why do some bottles produce a louder sound than others when blown into?

The material and thickness of the bottle also play a role in the loudness of the sound. Thicker and denser materials will produce a louder sound compared to thinner and lighter materials.

5. Is blowing into a bottle considered a musical instrument?

Technically, blowing into a bottle can be considered a musical instrument since it produces a sound that can be manipulated to create different pitches and tones. However, it is not commonly used as a musical instrument in professional settings.

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