A pipe open at both ends has fundamental frequency f1. When 3/4 th of its length is in water, it produces fundamental note of f2. Then f1f2 is

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A pipe open at both ends has fundamental frequency f₁. When 3/4 th of its length is in water, it produces fundamental note of f₂. Then $\frac{f_{1}}{f_{2}}$ is

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Detailed Solution

Initially frequency of the open pipe is

$f_{1} = \frac{v}{2 l} w h e n 3 / 4 t h o f t h e l e n g t h i s i n s i d e i t b e c o m e s a c l o s e e n d t u b e o f l e n g t h l / 4 s o f r e q u e n c y b e c o m e s f_{2} = \frac{v}{4 (\frac{l}{4})} = \frac{v}{l} \frac{f_{1}}{f_{2}} = \frac{1}{2}$