A) 80 cm
B) 100 cm
C) 120 cm
D) 140 cm
Correct Answer: C
Solution :
Fundamental frequency of closed organ pipe \[{{f}_{c}}=\frac{v}{4{{l}_{c}}}\] Fundamental frequency of open organ pipe \[{{f}_{0}}=\frac{v}{2{{l}_{0}}}\] Second overtone frequency of open organ pipe, \[=\frac{3v}{2{{l}_{0}}}\] From question, \[\frac{v}{4{{l}_{c}}}=\frac{3v}{2{{l}_{0}}}\] \[{{l}_{0}}=6{{l}_{c}}=6\times 20=120\,cm\]You need to login to perform this action.
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