A cylindrical tube closed at one end contains air. It produces the fundamental node of frequency 512 Hz. If the tube is opened at both ends, then the fundamental frequency that can be excited is:

256 Hz

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512 Hz

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1024 Hz

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128 Hz

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Solution

The correct option is C 1024 Hz
Fundamental frequency of closed pipe
$f_{0} = \frac{v}{4 L}$
$512 = \frac{v}{4 L}$ .....(1)
Fundamental frequency of open pipe
$f_{0} = \frac{v}{2 L}$ ....(2)
By (1) and (2)
$f_{0} = 1024 H z$

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Similar questions

A tube closed at one end and containing air is excited. It produces the fundamental note of frequency 512 Hz. If the same tube is open at both the ends the fundamental frequency that can be produced is

[RPET 1999]

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) :

A tube closed at one end and containing air produces, when excited, the fundamental note of frequency 512 Hz. If the tube is open at both ends the fundamental frequency that can be excited is ( in Hz ):

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. If the tube is open at both ends, the fundamental frequency will be

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A cylindrical tube closed at one end contains air. It produces the fundamental node of frequency 512 Hz. If the tube is opened at both ends, then the fundamental frequency that can be excited is:

The correct option is C 1024 HzFundamental frequency of closed pipe f0=v4L 512=v4L .....(1) Fundamental frequency of open pipe f0=v2L ....(2) By (1) and (2) f0=1024 Hz

The correct option is C 1024 Hz
Fundamental frequency of closed pipe
$f_{0} = \frac{v}{4 L}$
$512 = \frac{v}{4 L}$ .....(1)
Fundamental frequency of open pipe
$f_{0} = \frac{v}{2 L}$ ....(2)
By (1) and (2)
$f_{0} = 1024 H z$