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

Options

A. `(n_(1)l_(1)-n_(2)l_(2))/(2(n_(1) -n_(2)))`
B. `(n_(2)l_(2)-n_(1)l_(1))/(2(n_(2) -n_(1)))`
C. `(n_(2)l_(2)-n_(1)l_(1))/(2(n_(1) -n_(2)))`
D. `(n_(2)l_(2)-n_(1)l_(1))/((n_(1) -n_(2)))`

Answer

d

Explanation

Suppose pipe vibrates in fundamental mode,
For first pipe,

`l_(1)+d=lambda_(1)/2` …(i)
where, d is end correction Similarly for second pipe,
`l_(2)+d=lambda_(2)/2` … (ii)
From (i) and (ii),
`(l_(1)+d)/(l_(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`
` rArr d[n_(1)-n_(2)]=n_(2)l_(2)-n_(1)l_(2)`
End correction, `d = (n_(2)l_(2)-n_(1)l_(1))/(n_(1)-n_(2))`