# question about resonance (1 Viewer)

#### spice

##### New Member
how does resonance work in tubes and what is the formula for it, i am doing a depth study where i have to show how length affects the resonance frequency, however i have no idea how this works in tubes and how to calculate this. is the wavelength for the resonant frequency the same as the length of the pipe? thats what i can gather from most sources but they all refuse to elaborate on the topic and i am confused. if any physics students can help me out that would be much appreciated also if you could show me how to derive this equation/where it comes from that would be very helpful.

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#### MoeyNeeds90+

##### Active Member
how does resonance work in tubes and what is the formula for it, i am doing a depth study where i have to show how length affects the resonance frequency, however i have no idea how this works in tubes and how to calculate this. is the wavelength for the resonant frequency the same as the length of the pipe? thats what i can gather from most sources but they all refuse to elaborate on the topic and i am confused. if any physics students can help me out that would be much appreciated also if you could show me how to derive this equation/where it comes from that would be very helpful.
hey

Resonance in tubes, particularly in the context of pipes or cylindrical tubes, is a phenomenon where sound waves are reinforced by constructive interference, leading to an increase in the amplitude of the sound. This phenomenon occurs when the length of the tube corresponds to certain wavelengths of the sound waves.

In a closed tube (like a pipe closed at one end), the fundamental frequency (the lowest resonant frequency) occurs when the length of the tube is equal to one-fourth of the wavelength of the sound wave. In an open tube (like a pipe open at both ends), the fundamental frequency occurs when the length of the tube is equal to one-half of the wavelength of the sound wave.

The formula for the fundamental frequency of a closed tube is given by:

$f = \frac{v}{4L}$

And for an open tube:

$f = \frac{v}{2L}$

Where:
- $$f$$ is the frequency of the sound wave (the resonant frequency).
- $$v$$ is the speed of sound in the medium (e.g., air).
- $$L$$ is the length of the tube.

The derivation of these formulas involves understanding the nature of sound waves and the conditions for constructive interference within the tube.

In a closed tube, the closed end acts as a node (point of minimum vibration), while the open end acts as an antinode (point of maximum vibration). This results in the fundamental mode having a quarter-wavelength pattern within the tube.

For an open tube, both ends act as antinodes, leading to a half-wavelength pattern within the tube.

These formulas can be derived from the general equation for the fundamental frequency of a standing wave:

$f = \frac{v}{\lambda}$

Where $$\lambda$$ is the wavelength of the sound wave.

For a closed tube, $$\lambda = 4L$$ (since it's a quarter of a wavelength), and for an open tube, $$\lambda = 2L$$. Substituting these values into the general equation gives you the specific formulas mentioned earlier.

I hope this helps

The formula for calculating the fundamental frequency of a close pipe is f = v/4L

#### wizzkids

##### Active Member
Sound resonance in tubes works like this:
• Air molecules can oscillate as a longitudinal wave up and down a tube.
• An open end is always an antinode, because the air molecules are free to oscillate without any restraint.
• A closed end is always a node, because when the pressure wave arrives at the closed end, it cannot transmit the energy forward, and it is forced to reflect at the closed end (and there is a phase inversion) and a node is formed.
• Certain frequencies of oscillation create a much greater amplitude because they create standing waves in the tube, and these are the resonant frequencies or harmonics.
• The next thing you need to know is the fundamental wave equation, that links wave frequency, wavelength and wave velocity, namely v=fλ.
• The last thing you need to know is the speed of sound waves in tubes is always pretty much the same, about 317 metres per second.
Then you can study the message posted by the previous poster to figure out the rest.

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