Q. Two closed organ pipes $A$ and $B$, have the same length. $A$ is wider than $B.$ They resonate in the fundamental mode at frequencies $n_{A}$ and $n_{B}$ respectively, then
A) $n_{A} = n_{B}$
B) $n_{A} > n_{B}$
C) $n_{A} < n_{B}$
D) Either (b) or (c) depending on the ratio of their diameters
Explanation
(c) In closed organ pipe. First resonance occurs at $\lambda/4.$
So, in fundamental mode of vibration of organ pipe
$\frac{\lambda}{4} = (l + 0.3\ d)$
Where $0.3\ d$ is necessary end correction
Frequency of vibration $n = \frac{v}{\lambda} = \frac{v}{4(l + 0.3d)}$
As $l$ is same, wider pipe $A$ will resonate at a lower frequency, $i.e.,\ n_{A} < n_{B}$
Note : The value of end correction $\theta$ is $0.6\ r$ for closed organ pipe and $1.2\ r$ for an open organ pipe, where $r$ is the radius of the pipe
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.