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Q.
The fundamental frequency of a closed pipe is 220 Hz. If $ \frac{1}{4} $ th of the pipe is filled with water the frequency of the first overtone of the pipe now is:
EAMCETEAMCET 2000
Solution:
Fundamental frequency for a closed organ pipe is $ n=\frac{v}{4l} $ When one-fourth of the pipe is filled with water, then New length $ l=l-\frac{l}{4}=\frac{3l}{4} $ $ \therefore $ New frequency $ n=\frac{v}{4l} $ $ =\frac{v}{4\times \frac{3l}{4}}=\frac{v}{3l} $ $ \frac{n}{n}=\frac{v/3l}{v/4l} $ $ n=\frac{4}{3}n $ $ =\frac{4}{3}\times 220=\frac{830}{3}\,Hz $ For closed organ pipe, first overtone $ =3\times $ fundamental frequency = 3n $ =3\times \frac{880}{3}=880\,Hz $