question_answer

Two closed organ pipes when sounded simultaneously give 4 beats/s. If longer tube has a length of 1 m, then length of shorter pipe will be (v = 330 m/s ):

A)  104.8cm     

B)  95.4cm

C)  104cm       

D)  96cm

Correct Answer: B

Solution :

When I length of closed organ pipe and \[\lambda \] be the wavelength, then frequency \[n=\frac{v}{4l}\] Also, number of beats = difference of frequencies of sound-source \[{{n}_{2}}-{{n}_{1}}=\frac{v}{4{{l}_{2}}}-\frac{v}{4{{l}_{1}}}\] \[{{n}_{2}}-{{n}_{1}}=\frac{v}{4}\left( \frac{1}{{{l}_{2}}}-\frac{1}{{{l}_{1}}} \right)\] \[\Rightarrow \] \[4=\frac{330}{4}\left( \frac{1}{{{l}_{2}}}-\frac{1}{1} \right)\] \[\Rightarrow \] \[\frac{1}{{{l}_{2}}}=\frac{4\times 4}{330}+1\] \[\Rightarrow \] \[\frac{1}{{{l}_{2}}}=\frac{346}{330}\] \[\Rightarrow \] \[{{l}_{2}}=\frac{330}{346}=0.954\,\,m=95.4\,\,cm\]