Question

A closed organ pipe of radius $r_1$ and an open pipe of radius $r_2$ and having same length 'L' resonate when excited with a given tunning fork. Closed organ pipe resonates in its fundamental mode where as open organ pipe resonates in its first overtone, then :

• A. $r_2-r_1=L$
• B. $r_2-r_1=L/2$
• C. $r_2-2r_1=2.5L$
• D. $2r_2-r_1=2.5L$

Answer 1

Answer: (C)

$r_2-2r_1=2.5L$

For closed organ pipe with end correction: Frequency of different modes,

${\nu }_n^{\prime }=\frac{(2n-1)v}{4(L+0.6r_1)}$ where $0.6r_1$ is the end correction

Now for fundamental mode i.e $n=1$, ${\nu }_1^{\prime }=\frac{v}{4(L+0.6r_1)}$

For open organ pipe with end correction: Frequency of $m^{th}$ mode,

${\nu }_m=\frac{mv}{2(L+2\times 0.6r_2)}$ where $0.6r_2$ is the end correction on one side.

Now for first overtonei.e $m=2$, ${\nu }_2=\frac{2v}{2(L+1.2r_2)}$

But ${\nu }_1^{\prime }={\nu }_2$

$\frac{v}{4(L+0.6r_1)}=\frac{2v}{2(L+1.2r_2)}$

$L+1.2r_2=4L+2.4r_1⟹r_2-2r_1=2.5L$