Question

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

Answer 1

Answer: (C)

In closed organ pipe. First resonance occurs at
$ \lambda /4 $ .
So, in fundamental mode of vibration of organ pipe
$ \frac{\lambda }{4}=(l+0.3d) $
where 0.3d 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 e is 0.6 r for closed organ pipe and 1.2r for an open organ pipe. where $ r $ is the radius of the pipe.