A cylindrical tube open at both ends, has a fundamental frequency `f` in air. The tube is dipped vertically in air. The tube is dipped vertically in water so that half of it is in water. The fundamental frequency of the air column is now

Options

A. (a) `f`
B. (b) `f//2`
C. ( c ) `3f//4`
D. (d) `2f`

Answer

A

Explanation

(a) The fundamental frequency of open tube
`v_(0) = (upsilon)/(2l_(0))` …(i)
That of closed pipe
`v_(c) = (upsilon)/(4l_(c))` …(ii)
According to the problem `l_(c) = (l_(0)/(2)`
Thua `v_(c) = (upsilon)/(l_(0))//2)` rArr `v_(c) = (upsilon)/(2l)` ...(iii)
From equation (i) and (ii)
`v_(0) = v_(c)`
Thua, `v_(c) = f` (`because` `v_(0) = f` is given)