A) n
B) n/2
C) 2 n
D) 4 n
Correct Answer: B
Solution :
Key Idea: Closed organ pipe has an antinode at its open end, and node at closed end. The distance between a node and antinode in closed pipe, \[\lambda /4=l\] \[\Rightarrow \] \[\lambda =4l\] Fundamental frequency or the first harmonic of closed pipe is \[n=\frac{v}{\lambda }\] \[\therefore \]\[n=\frac{v}{4l}\] When \[{{l}_{2}}=2{{l}_{1}},{{r}_{2}}=\frac{{{r}_{1}}}{2},{{n}_{1}}=n\] \[\Rightarrow \] \[\frac{{{n}_{1}}}{{{n}_{2}}}=\frac{{{l}_{2}}}{{{l}_{1}}}\] \[\Rightarrow \] \[{{n}_{2}}={{n}_{1}}\frac{{{l}_{1}}}{{{l}_{2}}}=n\frac{{{l}_{1}}}{2{{l}_{1}}}=\frac{n}{2}\] Note: A closed pipe produces only odd harmonics.You need to login to perform this action.
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