Question

A pipe open at both ends resonates to a frequency '$n_1$' and a pipe closed at one end resonates to a frequency '$n_2$'. If they are joined to form a pipe closed at one end, then the fundamental frequency will be:

• A. $\frac{n_1{n}_2}{2n_2+n_1}$
• B. $\frac{2n_2{n}_1}{2n_2+n_1}$
• C. $\frac{2n_2{n}_1}{n_1+n_2}$
• D. $\frac{n_2+2n_1}{n_1{n}_2}$

Answer 1

Answer: (A)

$\frac{n_1{n}_2}{2n_2+n_1}$

Let the length of the open and closed pipe be $l_1$ and $l_2$, respectively.

Resonant frequency in open pipe $n_1=\frac{v}{2l_1}$

$⟹$ $l_1=\frac{v}{2n_1}$

Resonant frequency in closedpipe $n_2=\frac{v}{4l_2}$

$⟹$ $l_2=\frac{v}{4n_2}$

Total length of the new pipe formed closed at one end $L=l_1+l_2$

$\therefore$ $L=\frac{v}{2n_1}+\frac{v}{4n_2}=\frac{v}{4}.\frac{n_1+2n_2}{n_1{n}_2}$

Fundamental frequency of new pipe. $n=\frac{v}{4L}$

$⟹$ $n=\frac{v}{4\times \frac{v}{4}.\frac{n_1+2n_2}{n_1{n}_2}}=\frac{n_1{n}_2}{n_1+2n_2}$