Question

The minimum resonance frequency of a two-faced organ tube is 300 Hz. Dividing it into two equal parts, the minimum resonance frequency of each part will be.

Answer 1

Answer:

closed organ pipe, frequency of the fundamental note

ν

0c

=

4l

v

, where v is velocity of sound and l is the length of the pipe.

For open organ pipe, frequency of the fundamental note

ν

0o

=

2l

v

=2ν

0c

ν

0c

=

2

1

ν

0o

=

2

30

=15 Hz

Explanation:

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Answer 2

Answer:

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