Question

First overtone frequency of a closed organ pipe is equal to the first overtone frequency of an open organ pipe. Further nth harmonic of closed organ pipe is also equal to the mth harmonic of open pipe, where n and m are :

• A. 5, 4
• B. 7, 5
• C. 9, 6
• D. 7, 3

Answer 1

The correct option is C 9, 6

Let $L^{\prime }$ and $L$ be the length of the closed and open organ pipe respectively.

For closed organ pipe: Frequency of different modes, ${\nu }_n^{\prime }=\frac{(2p-1)v}{4L^{\prime }}$ where $(2p-1)$ represents nth harmonic.

Thus frequency of nth harmonic in closed pipe, ${\nu }_n^{\prime }=\frac{nv}{4L^{\prime }}$

Ist overtone i.e $p=2$, ${\nu }_3^{\prime }=\frac{3v}{4L^{\prime }}$

For open organ pipe: Frequency of mth harmonic ${\nu }_m=\frac{mv}{2L}$

Ist overtone i.e $m=2$, ${\nu }_2=\frac{2v}{2L}$

Given : ${\nu }_3^{\prime }={\nu }_2$

$\frac{3v}{4L^{\prime }}=\frac{v}{L}⟹\frac{L^{\prime }}{L}=\frac{3}{4}$ ......(1)

Also given, ${\nu }_n^{\prime }={\nu }_m$

$\frac{nv}{4L^{\prime }}=\frac{mv}{2L}$

$\frac{n}{2}=m\frac{L^{\prime }}{L}$ $⟹\frac{n}{2}=m\frac{3}{4}$

Thus $2n=3m$

Now $n=9$ and $m=6$ satisfies the relation.