A) \[\frac{1}{3}\]
B) \[\frac{2}{3}\]
C) \[\frac{8}{3}\]
D) \[\frac{3}{8}\]
Correct Answer: D
Solution :
In an organ pipe closed at one end there will be an antinode at the open end and a node at closed end. The distance between an antinode and nearest node is \[\frac{\lambda }{4}.\] If \[{{l}_{1}}\]be the length of pipe, and \[\lambda \] the wavelength then, \[{{l}_{1}}=\frac{\lambda }{4},\]or \[{{\lambda }_{1}}=4{{l}_{1}}\]If n be the frequency of note emitted and v the velocity of sound in air, then \[{{n}_{1}}=\frac{v}{{{\lambda }_{1}}}=\frac{v}{4{{l}_{1}}}\]Also, closed pipe produces only odd harmonics, hence first pvertone of closed pipe is \[=3\left( \frac{v}{4{{l}_{1}}} \right)\] ?.(i) For open pipe antinodes are formed at both ends, frequency is given by \[{{n}_{2}}=\frac{v}{2{{l}_{2}}}\] Also open pipe produces both even and odd harmonics, hence we have freqency of third overtone of open pipe is \[=4\left( \frac{v}{2{{l}_{2}}} \right)\] At resonance both the frequencies are equal \[\therefore \] \[\frac{3v}{4{{l}_{1}}}=\frac{4v}{2{{l}_{2}}}\] \[\Rightarrow \] \[\frac{{{l}_{1}}}{{{l}_{2}}}=\frac{3}{8}\]You need to login to perform this action.
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