Q. An open pipe is in resonance in 2$^{nd}$ harmonic with frequency $v_{1}$ Now one end of the tube is closed and frequency is increased to $v_{2}$ such that the resonance again occurs in nth harmonic. Choose the correct option.
A) $n = 3,\ v_{2} = \frac{3}{4}v\_ 1$
B) $n = 3,v_{2} - \frac{5}{4}v_{1}$
C) $n = 5,v_{2} = \frac{5}{4}v_{1}$
D) $n = 5,v_{2} = \frac{3}{4}v_{1}$
Explanation
(c) $v_{1} = \frac{v}{l}$
(2$^{nd}$ harmonic of open pipe)
Here, n is odd and $v_{2} > v_{1}$
It is possible when n=5
Because with n=5
$v_{2} = \frac{5}{4}\left( \frac{v}{l} \right) > v_{1}$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.