A) fundamental frequency
B) first overtone of pipe
C) second overtone
D) fourth overtone
Correct Answer: B
Solution :
Fundamental frequency \[{{n}_{1}}=\frac{\upsilon }{2l}\] \[=\frac{330}{2\times 33\times {{10}^{-2}}}=500Hz\] Frequency of 1st overtone \[=2\frac{\upsilon }{2l}=2\times 500=1000Hz\]You need to login to perform this action.
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