A cylindrical tube, open at both ends, has a fundamental frequency v . The tube is dipped vertically in water so that half of its length is inside the water. The new fundamental frequency is-

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Solution:

Initially, the tube is open at both ends, so its fundamental frequency is $v = \frac{V}{2 L}$ , where $V$ is the speed of sound and $L$ is the tube's length. When half the tube is dipped in water, it becomes a closed organ pipe of length $L /2$ . The fundamental frequency for a closed pipe is $v^{'} = \frac{V}{4 ( L /2 )} = \frac{V}{2 L}$ . Thus, $v^{'} = v$ .

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