question_answer

If n, is the resonance frequency of a pipe open at both ends and \[{{n}_{2}}\]the resonance frequency of pipe open at one end only. Both are vibrating in the fundamental mode and both pipes are of same length, then :

A)  \[{{n}_{1}}={{n}_{2}}\]                                 

B)  \[2{{n}_{1}}={{n}_{2}}\]

C)  \[{{n}_{1}}=2{{n}_{2}}\]                              

D)  \[3{{n}_{1}}=4{{n}_{2}}\]

Correct Answer: C

Solution :

                Let\[l\]be the length of pipe, then fundamental frequency of open pipe is \[{{n}_{1}}=\frac{v}{2l}\]                                              ?.. (i) where v is velocity of sound. For closed pipe                 \[{{n}_{2}}=\frac{v}{4l}=\frac{1}{2}\left( \frac{v}{2l} \right)\]                       ?. (ii) From Eqs. (i) and (ii), we get \[{{n}_{2}}=\frac{1}{2}{{n}_{1}}\] \[\Rightarrow \]               \[{{n}_{1}}=2{{n}_{2}}\]