Standing Waves on a string & pipe

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Homework Help Overview

The problem involves a string fixed at both ends, vibrating in its third overtone, and causing a nearby pipe, open at both ends, to resonate in its third harmonic. The context includes calculating the length of the pipe and its fundamental frequency based on the properties of the string and the speed of sound.

Discussion Character

  • Mixed

Approaches and Questions Raised

  • Participants explore the relationship between the velocity of the vibrating string and the pipe, with some attempting to calculate frequencies based on harmonic relationships. Questions arise regarding how to connect the frequencies of the string and the pipe.

Discussion Status

Participants have made progress in calculating the frequency of the string and have identified that the frequencies of the string and pipe must be related due to resonance. Some calculations have been shared, and there is an ongoing exploration of how to derive the fundamental frequency of the pipe.

Contextual Notes

There is a focus on the harmonic relationships and the properties of the string and pipe, with some participants expressing uncertainty about the connections between their respective frequencies.

murrskeez
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Homework Statement



A string 40.0cm long of mass 8.50g is fixed at both ends and is under a tension of 425N. When the string is vibrating in its third overtone, you observe that it causes a nearby pipe, open at both ends, to resonate in its third harmonic. The speed of sound is 344m/s. a) How long is the pipe? b) What is the fundamental frequency of the pipe?

Homework Equations



Fn=(nV)/(2L)
λn=(2L)/n
V=√(F/μ) where μ=m/L

The Attempt at a Solution



Really stuck on this one. I know I can find the velocity of the string with the given information but am not sure how I can relate the velocity of the string to the velocity of the pipe. Any suggestions would be much appreciated.

m/L= 0.0085kg/0.400m = 0.0213kg/m
Vstring=√(425N/0.0213kg/m) = 141.3m/s
 
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What's the frequency of the vibrating string?
 
So the third overtone would mean n=4...

so fn=(4*141.3m/s)/(2*0.400m) = 706.5Hz

so the frequencies of the string and pipe must be related, I'm just not sure how.
 
murrskeez said:
So the third overtone would mean n=4...

so fn=(4*141.3m/s)/(2*0.400m) = 706.5Hz
Good.
so the frequencies of the string and pipe must be related, I'm just not sure how.
They are the same! (They resonate.) So what's the fundamental frequency of the pipe?
 
I think I get it :smile:

fpipe = nV/2L
706.5Hz = (3*344m/s)/(2L)
L = 0.730m

fo = v/2L
fo = (344m/s)/(2*0.730m)
fo = 236Hz

thank you so much, really appreciate it :)
 
Good! :approve:
 

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