How does water affect resonance in organ pipes?

AI Thread Summary
Water affects resonance in organ pipes by reducing the effective length of the air column, which alters the resonance conditions. In a closed organ pipe of 120 cm, resonance occurs at a frequency of 340 Hz, but the introduction of water changes the dynamics, leading to confusion about the correct values of n in the resonance equation L=nλ/4. The discussion highlights that only odd harmonics are allowed in a closed pipe, which impacts the calculations for resonance length. Participants clarify that the initial length of the air column must be adjusted based on the water level, affecting the resonance frequency. Ultimately, understanding how water modifies the air column is crucial for accurately determining resonance conditions.
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Homework Statement



In a resonance tube experiment, a closed organ pipe of length 120 cm resonates ,when tuned with a tuning fork of length 120cm resonates,when tuned with a tuning fork of length 340 hz.If water is poured into the pipe ,then (speed of sound in air=340 ms^-1)

(A) minimum length of water coloumn to have resonance is 50cm

(B) the distance between two succesive nodes is 50 cm

(C)The maximum length of water coloumn to create resonance is 100cm

(D) None of the above

Homework Equations

The Attempt at a Solution



Before the water started filling

Wavelength=Velocity/frequency=1m

For a closed pipe,the resonance condition is

L=n λ/4

Substituting values

4 L/n =1

n=5

therefore - 2n+1=5
n=2

L=3λ/4


I am confused about the next part where there is water flowing into the coloumn.How will it affect the air column now[/B]
 
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palkia said:
L=n λ/4
Where does that equation come from, and is every n possible here?
palkia said:
n=5

therefore - 2n+1=5
n=2
How can you conclude n=2 from n=5?
palkia said:
L=3λ/4
That contradicts the given values.

Initially the tube won't be in resonance.
palkia said:
I am confused about the next part where there is water flowing into the coloumn.How will it affect the air column now
Where water is there is no air - you reduce the length of the air column.
 
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I made a mess there...only odd values of n will be alllowed there so I should have used a different variable for thatI n any case the value of n will be 3Initially the tube won't be in resonance. (What does this mean)
 
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palkia said:
made a mess there...only odd values of n will be alllowed there so I should have used a different variable for that
As far as I understand the problem statement both sides are closed.
palkia said:
Initially the tube won't be in resonance. (What does this mean)
I'm not sure what is unclear. A 1.2 m tube, closed on both sides, won't resonate at 340 Hz.
 
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mfb said:
As far as I understand the problem statement both sides are closed.I'm not sure what is unclear. A 1.2 m tube, closed on both sides, won't resonate at 340 Hz.
I think the organ pipe is closed at one end ,that's why n=3.The OP made a typo in my opinion
 
In that case 120 cm would be close to a resonance.
 
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harambe said:
I think the organ pipe is closed at one end ,that's why n=3.The OP made a typo in my opinion
I am so sorry...it was indeed my mistake
 
mfb said:
In that case 120 cm would be close to a resonance.

Okay so my length came 1.25 m so what does the extra length tell us??

How should I proceed to do for the water case?Also sorry for my typo...(^|^)
 
palkia said:
Okay so my length came 1.25 m so what does the extra length tell us??
Why is it 1.25 m now when it was 1.2 m before?

palkia said:
How should I proceed to do for the water case?
What are other possible lengths of the air column where you have a resonance?
 
  • #10
Since the L=5λ/4 and putting λ=1 ,I am gettting 1.25m
 
  • #11
That is the initial length, with 5 as constant in the formula. 5 is not the smallest possible value here.
 
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