Q. The second overtone of an open pipe is in resonance with the first overtone of a closed pipe of length 2m. length of the open pipe is
A) 4m
B) 2m
C) 8m
D) 1m
Explanation
(a) Second overtone of open pipe of length 𝒍 is
$v_{0} = \frac{v}{2l}\ \ \ldots\ldots.(i)$
First overtone of a close pipe is
$v_{c} = \frac{v}{4l} = \frac{v}{4 \times 2}\ldots\ldots..(ii)$
Equating Eqs. (i) and (ii), we get
$\frac{v}{2l} = \frac{v}{8} \Longrightarrow l = 4m$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.