Q. An open pipe is in resonance in its 2$^{nd}$ harmonic with tuning fork of frequency $f_{1}.$ Now it is closed at one end. If the frequency of the tuning fork is increased slowly from $f_{1}$ then again a resonance is obtained with a frequency $f_{2}.$ If in this case the pipe vibrates $n^{th}$ harmonics then
A) $n = 3,\ f_{2} = \frac{3}{4}f_{1}$
B) $n = 3,\ f_{2} = \frac{5}{4}f_{1}$
C) $n = 5,\ f_{2} = \frac{5}{4}f_{1}$
D) $n = 5,\ f_{2} = \frac{3}{4}f_{1}$
Explanation
(c) Open pipe resonance frequency $f_{1} = \frac{2v}{2L}$
Closed pipe resonance frequency $f_{2} = \frac{nv}{4L}$
$f_{2} = \frac{n}{4}f_{1}$ (where $n$ is odd add $f_{2} > f_{1}$) $\therefore n = 5$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.