Open Or Closed Organ pipe
An open organ pipe of sufficient length is dipping in water with a speed
v
vertically. If at any instant
L
is the length of tube above water. Then the rate at which fundamental frequency of pipe changes
Take
v
=
1
0
m/sec
Speed of sound =
3
6
0
m/sec
L
=
1
0
m
The answer is 9.
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2 solutions
Gauri acha sawaal tha but I solved it
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U are prakhar singh of fiitjee kanpur???
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Haan yaar galti se bangalore likh diya
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@Prakhar Singh – Yo bhaii ⌣ ¨
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@Neelesh Vij – Hey neelesh where do you live in delhi? and where do you study?
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@Prakhar Bindal – No actually i live in kanpur, Delhi is just fake.
And frequency for open organ pipe is n = 2 L v
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@Neelesh Vij – So why have they taken v/4l.
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@Prakhar Bindal – Its because if water is present then one end becomes closed so n = 4 L v
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@Neelesh Vij – Oh got it! . Although i got my answer correct i didn't though about this aspect of the problem! . Thanks for explaining :)
For An Open Organ Pipe isn't the frequency nv/2L ?
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but it is dipped in water.
Since for an open-closed pipe f 0 = 4 L c , we have d t d f 0 = − 4 L 2 c d t d L = − 4 ⋅ 1 0 2 3 6 0 ⋅ ( − 1 0 ) = 9 Hz/s .
I used a mod rather than a -ve sign :)
F= C/(4L) where F is fundamental frequency C is speed of sound L is length of pipe We have |dF/dT|=(C/(4L*L)) *(|dL/dT|)
=C V/(4 L*L)= 9