A flute which we treat as a pipe open at both ends is 60 cm long. (i) What is the fundamental frequency when…
A flute which we treat as a pipe open at both ends is 60 cm long. (i) What is the fundamental frequency when all the holes are covered? (ii) How far from the mouthpiece should a hole be uncovered for the fundamental frequency to be 330 Hz? (Take, speed of sound in air = 340 ms$^{-1}$).
Solution
Sol. (i) Fundamental frequency when the pipe is open at both ends,
$f_1 = \frac{v}{2l} = \frac{340}{2 \times 0.6} = 283.33\ \text{Hz}$
(ii) Suppose the hole is uncovered at a length $l$ from the mouthpiece, the fundamental frequency will be
$f'_1 = \frac{v}{2l}$
$\displaystyle l = \frac{v}{2f'_1} = \frac{340}{2 \times 330} = 0.515\ \text{m} = 51.5\ \text{cm}$
Note Opening holes in the side effectively shortens the length of the resonance column, thus increasing the frequency.
Answer: 283.33 Hz; 0.515 m = 51.5 cm