question_answer

An organ pipe P1 is closed at one end and vibrating in its first overtone and another pipe P2 opened at both ends vibrating in its third overtone are in resonance with a given tuning fork. Then the ratio of P1 and P2 is :         

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 \[\therefore \]        \[\frac{3v}{4{{l}_{1}}}=\frac{4v}{2{{l}_{2}}}\] \[\Rightarrow \]          \[\frac{{{l}_{1}}}{{{l}_{2}}}=\frac{3}{8}\]