The frequency of the fundamental note of a tube closed at one end is 200 Hz. The frequency of the fundamental note of a similar tube of same length but open at both ends is

400 Hz

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500 Hz

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600 Hz

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700 Hz

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Solution

The correct option is D 400 Hz
Fundamental frequency is $200 H z = \frac{v}{4 L}$ -----(1)
( $∵$ it is closed at one end it acts as closed organ pipe )
Fundamental frequency of open tube is $\frac{v}{2 L}$
So, from (1) $\Rightarrow \frac{v}{4 L} = 200 H z$
Hence, $\frac{v}{2 L} = 2 \times 200 H z = 400 H z$
So, fundamental frequency of open tube is 400Hz.

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The frequency of the fundamental note of a tube closed at one end is 200 Hz. The frequency of the fundamental note of a similar tube of same length but open at both ends is

The correct option is D 400 HzFundamental frequency is 200Hz=v4L -----(1)( ∵ it is closed at one end it acts as closed organ pipe )Fundamental frequency of open tube is v2LSo, from (1) ⇒v4L=200HzHence, v2L=2×200Hz=400HzSo, fundamental frequency of open tube is 400Hz.