A tube of diameter d and of length l is open a both ends. Its fundamental frequnecy of resonance is found to be `f_(1)`. One end of the tube is now closed. The lowrst frequency of resonance of this closed tube is now `f_(2)`. Taking into consideration the end correction ,`(f2)/(f_(1)` is

Options

A. `((l + 0.6d))/((l + 0.3d))`
B. `(1(l + 0.3d))/(2(l + 0.6d))`
C. `(1(l + 0.6d))/(2(l + 0.3d))`
D. `(1(d + 0.3l))/(2(d + 0.6l))`

Answer

C

Explanation

`(lambda_(1))/(2) = l ++ .6d, upsilon_(1) = (V)/(lambda_(1))`

`(lambda_(2))/(4) = l + 0.3d, upsilon = (V)/(lambda_(2))`

`(upsilon_(2))/(upsilon_(1)) = (2(l + 6d))/(4(l + .3d)) = (l + .6d)/(2(l + .3d))`