The fundamental of a closed pipe is 220 Hz. If `(1)/(4)` of the pipe is filled with water, the frequency of the first overtone of the pipe now is

Options

A. 880 Hz
B. 1760 Hz
C. 220 Hz
D. 440 Hz

Answer

A

Explanation

In the first case, `n_(c) = 220 Hz = v/(4L) `
` :. V = 220 xx 4L` …(1)
If the tube is filled with water upto a length `L/4` , the new resonating length will be ` (3L)/4`.
`:. n'_(c) = v/(4((3L)/4)) = v/(3L) = (220 xx 4L)/(3L) = 4/3 (220)`
`:. ` First overtone of the fundamental frequency in the second case ` = 3n'_(c) = 3 (4/3 xx 220) = 880` Hz