question_answer

An open pipe is in resonance in 2nd harmonic with frequency \[{{f}_{1}}\] Now one end of the tube is closed and frequency is increased to \[{{f}_{2}}\] such that the resonance again occurs in nth harmonic. Choose the correct option

A)  \[n=3,\,{{f}_{2}}=\frac{3}{4}{{f}_{1}}\]

B)  \[n=3,\,{{f}_{2}}=\frac{5}{4}{{f}_{1}}\]

C) \[n=5,\,{{f}_{2}}=\frac{3}{4}{{f}_{1}}\]

D) \[n=5,\,{{f}_{2}}=\frac{5}{4}{{f}_{1}}\]

Correct Answer: D

Solution :

[d] \[\lambda =\ell \] \[\therefore \,\,\,\,{{f}_{1}}=\frac{v}{\lambda }=\frac{v}{\ell }\]          ?. (i) \[\lambda =\frac{4\ell }{n}\] \[\therefore \,\,\,{{f}_{2}}=\frac{v}{\lambda }=\frac{nv}{4\ell }\]      ?. (ii) Here n is a odd number. From (i) and (ii) \[{{f}_{2}}=\frac{n}{4}{{f}_{1}}\] For first resonance, \[n=5,\,{{f}_{2}}=\frac{5}{4}{{f}_{1}}\]