A) \[2\times \frac{4}{330}\,Hz\]
B) \[2\times \frac{330}{4}\,Hz\]
C) \[3\times \frac{330}{4}\,Hz\]
D) \[4\times \frac{330}{4}\,Hz\]
Correct Answer: C
Solution :
Given length of closed organ pipe \[=1\,m\] The frequency of closed organ pipe is \[\frac{V}{4l},\frac{3V}{4l},\frac{5V}{4l}....\] Therefore frequency of second note is \[\frac{3V}{4l}=3\times \frac{330}{4\times 1}=\frac{3\times 330}{4}\,Hz.\]You need to login to perform this action.
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