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In a pipe opened at both ends n(1) and n...

In a pipe opened at both ends `n_(1)` and `n_(2)` be the frequencies corresponding to vibrating lengths `L_(1)` and `L_(2)` respectively .The end correction is

A

`(n_(1)I_(2)-n_(2)I_(2))/(2(n_(1)-n_(2))`

B

`(n_(2)I_(2)-n_(1)I_(1))/(2(n_(2)-n_(1))`

C

`(n_(2)I_(2)-n_(1)I_(1))/(2(n_(1)-n_(2))`

D

`(n_(2)I_(2)-n_(1)I_(1))/(n_(1)-n_(2))`

Text Solution

Verified by Experts

The correct Answer is:
d
Suppose pipe vibrates in fundamental mode
for first pipe
`L_(1)+=(lambda_(1))/(2)`
where d is end correction
similarly for second pipe
`L_(2)+d=(lambda_(2))/(2)`
from eqs (i) and (ii) we get
`(l_(1)+d)/(l_(3=2)+d)=(lambda_(1))/(lambda_(2))=(1)/(n_(1))xxn^(2)`
`rarr n_(1)l_(1)+n_(1)d=n_(2)l_(2)+n_(2)d`
end correction `d=(n_(2)l_(2)-n_(1)l_(1))/(n_(1)-n_(2))`
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