Standing sound waves created in a pipe, question about resonances

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SUMMARY

The discussion focuses on the phenomenon of sound resonance in a closed pipe, specifically analyzing the cases of resonance at frequencies of 500 Hz, 700 Hz, and 1000 Hz. The formula for resonance in a closed pipe is given as fn = n (v/4L) and wavelength = 4L/n. With a pipe length of approximately 1 meter and sound velocity at 340 m/s, the number of resonance cases corresponds to the integer multiples of λ/4 fitting within the pipe's length, resulting in 3, 4, and 6 resonance cases for the respective frequencies.

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helpmeplz!
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Hey guys i have a question for my lab report that i can't solve. We did a lab where we had a pipe and produced sound resonance in there, using a generator to vary the frequencies and we recorded our data.

the general formula is for the closed pipe, fn= n (v/4L) and wavelength= 4L/n.

The question is, explain why there are 3 cases of resonance when f= 500 hz, 4 cases of resonance when f= 700 hz and 6 cases of resonance when f= 1000 hz.

the length of the pipe was around 1 metre.
 
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Hi helpmeplease!, welcome to PF.
In a closed pipe, resonance occurs when the length of pipe is λ/4, 3λ/4, 5λ/4, and so on.
Length of the pipe is nearly one meter. If you take the velocity of the sound at temperature as 340 m/s, what is the wavelength of 500 Hz sound? Then see how many λ/4 lengths fit in one meter.
 

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