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  4. A cylindrical resonance tube open at both ends, has a fundamental frequency f , in air. If half of the length is dipped vertically in water, the fundamental frequency of the air column will be:

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Q. A cylindrical resonance tube open at both ends, has a fundamental frequency $ f $ , in air. If half of the length is dipped vertically in water, the fundamental frequency of the air column will be:

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Solution:

Key Idea: When open end pipe is dipped in water vertically it becomes close end pipe. Fundamental frequency of open pipe, $ f=\frac{v}{2l} $ ???(i) When half length of tube is dipped vertically in water, then length of the air column becomes half $ \left( l=\frac{l}{2} \right) $ and the pipe becomes closed. So, new fundamental frequency of closed pipe $ f=\frac{v}{4l}=\frac{v}{4\left( \frac{l}{2} \right)}=\frac{v}{2l} $ ??(ii) From Eqs. (i) and (ii), we get, $ f=f $ Hence, there will be no change in frequency.