Two closed organ pipes A and B, have the same length. A is wider than B. They resonate in the fundamental mode at frequencies nA and nB respectively, then :

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 nA and nB respectively, then : (A)  nA=nB  (B)  nA>nB  (C)  nA<nB  (D) either (b) or (c) depending on the ratio of their diameters

Tardigrade Admin · Accepted Answer

In closed organ pipe. First resonance occurs at

λ /4

.
So, in fundamental mode of vibration of organ pipe

\frac{λ}{4} = (l + 0.3 d)

where 0.3d is necessary end correction.
Frequency of vibration,

n = \frac{v}{λ} = \frac{v}{4 ( l + 0.3 d )}

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.

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 :

$n_{A} = n_{B}$

$n_{A} > n_{B}$

$n_{A} < n_{B}$

either (b) or (c) depending on the ratio of their diameters

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 nA​ and nB​ respectively, then :

nA​=nB​

nA​>nB​

nA​<nB​

either (b) or (c) depending on the ratio of their diameters

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 :

$n_{A} = n_{B}$

$n_{A} > n_{B}$

$n_{A} < n_{B}$