A) 100 Hz
B) 200 Hz
C) 300 Hz
D) 150 Hz
Correct Answer: C
Solution :
Let the length of closed organ pipe be 1. Then, \[{{v}_{n}}=(2n-1)\frac{v}{4l}\] Its third harmonics (put n = 2) In above question, \[{{v}_{3}}=\frac{3}{4}\frac{v}{l}\] ?(i) Now, the length of open organ pipe is \[2l,\] then \[{{v}_{n}}=n\frac{v}{2(2l)}=\frac{nv}{4l}\] Its fundamental frequency. \[{{v}_{0}}=\frac{v}{2l}=100\] ?(ii) From Eq. (i) and (ii), we get \[{{v}_{3}}=\frac{3v}{4l}=3\times 100=300\,Hz\]You need to login to perform this action.
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