Year 12 Sound Problem (Involves resonance in a closed pipe).

In summary, the conversation is about a question from a test that involves finding the velocity of sound in a tube based on its resonant frequencies. The conversation includes a discussion of finding the wavelength and using simultaneous equations to solve for v. The final solution is 232.9 m/s.
  • #1
caitie_72
4
0

Homework Statement



Hi :)
This question is from a test i had today. For some stupid reason we get to finish it off tomorrow. I figured i do what everyone else will, and sort out the what i didn't understand. Luckily this was it. Anyway, the question is typed from memory so it may seem a bit odd or inaccurate.

A pipe is in a tub of water. As it is moved down (or up, can't remember!) it resonates at 288 Hz and 512 Hz. The distance between these two points is 13.9 cm (0.139 m). What is the velocity of sound in the tube?

I'm not sure of the order of frequencies occurring is. I think it was probably the lowest frequency first. I don't really need an answer, just an understanding of how to do a problem like this. All i know is that the pipe resonates at those two frequencies, and the distance between the points where it resonates is 13.9m. If you've ever seen a question like this ... then you'd probably get what i mean, hopefully.
Do I used simultaneous questions? I'm not sure.

Homework Equations


(w=wavelength)
v=F*w

The Attempt at a Solution


What i did first, in a fit of irrational confusion, was find the wavelength. I assumed that resonance was occurring at either nodes or antinodes, half a wavelength apart (?)
So w = 2 * 13.9.

Then i got confused, because i know the fundamental wavelength of a closed pipe is w=4L.

Then i did something stupid i knew wasn't right. Found the difference between the frequencies and then multiplied it with the wavelength. Stupid stupid, i know. Makes no sense.

I'm thinking simultaneous equations?
f=v/4L.

512=3v/4L 288=v/4L

Resonance must be occurring at first and second modes, right?
I'm just confused about what the length is! Does length, in this case, mean the length of the whole pipe or the length between the points of resonance? Does resonance occur at half wavelengths or quarter wavelengths? Don't understand!

Any help or direction at ALL is appreciated. Thanks :)
 
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  • #2
Since there is no other information, you have to assume that they are resonating in the fundamental mode.
So write 512=3v/4L 288=v/4L as 512=v/4L1,and 288=v/4L2. Write these equation in terms of L1 and L2. L1 - L2 is given. Solve for v.
 
  • #3
Thank you so much! You are seriously my favourite person ever right now.
Just checking, am i doing it right?

L1=3v/2048
L2=v/1152

0.139 = 3v/2048 - v/1152
v = 232.9 m/s

I assume so. Thanks again :)
 
  • #4
L1=3v/2048
This is wrong. It should be L1=v/2048. Find L2-L1.
 

Related to Year 12 Sound Problem (Involves resonance in a closed pipe).

1. What is the cause of resonance in a closed pipe?

Resonance in a closed pipe occurs when the length of the pipe is equal to a multiple of half the wavelength of the sound waves produced. This allows for constructive interference, resulting in a louder and more sustained sound.

2. How does the length of the pipe affect resonance in a closed pipe?

The length of the pipe directly affects the resonant frequency of the sound produced. As the length of the pipe increases, the resonant frequency decreases, and vice versa.

3. Can materials of the closed pipe affect resonance?

Yes, the material of the pipe can affect resonance. Different materials have different densities and therefore can change the speed of sound, altering the resonant frequency.

4. What is the difference between resonance in a closed pipe and an open pipe?

In a closed pipe, both ends are sealed, while in an open pipe, one end is open. This means that in a closed pipe, the sound waves must reflect back and forth between the two ends, resulting in a standing wave. In an open pipe, the sound waves can escape through the open end, resulting in a different pattern of resonance.

5. How is resonance in a closed pipe used in real-world applications?

Resonance in a closed pipe is used in various musical instruments, such as flutes and pipes. It is also utilized in the design of wind instruments, where the length of the pipe can be adjusted to produce different notes. In addition, resonance in a closed pipe is used in some types of resonators, which are devices that amplify sound waves.

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