91Ó°ÊÓ

The length of pipe B is $L_B=4L_A$

Use the concept of two open-ended and one open-ended pipe. With both ends of the pipe open, any harmonic can be set up in the pipe but with only one end open, only odd harmonic can be set up.

Formulae:

1. Resonant frequency for pipe with both ends open is $f=\frac{nv}{2L}$
2. Resonant frequency for pipe with both ends open is$f=\frac{nv}{4L}$

(a)

The resonant frequency for a pipe of length $L_B$ with two open ends using equation (i) is

$f=\frac{n_{B}v}{2L_B}$

For,$n=1,2,3…$

$L_B=\frac{n_{B}v}{2f}$ …… (1)

The resonant frequency for a pipe of lengthwith only one open end is

$f=\frac{n_{A}v}{4L_A}$

For,$n=1,3,5…$

$L_A=\frac{n_{A}v}{4f}$ ….. (2)

The frequency ofmatches that ofand the length of pipe B is$L_B=4L_A$. Hence, from equation (1) and equation (2), we get

$\begin{array}{l}\frac{n_{B}v}{2f}=4\left(\frac{n_{A}v}{4f}\right)\\ \frac{n_B}{2}={\left(n_A\right)}_{odd}\\ n_B=2{\left(n_A\right)}_{odd}\end{array}$ …… (a)

Using equation (a), the smallest value of$n_B$at which a harmonic frequency of B matches that of A is given as:

$\begin{array}{c}{n}_B=2\left(1\right)\\ =2\end{array}$

Hence, the smallest value of $n_B$is 2.

Using equation (a), the second smallest value of$n_B$at which a harmonic frequency of B matches that of A is given as:

$\begin{array}{c}{n}_B=2\left(3\right)\\ =6\end{array}$

Hence, the second smallest value of $n_B$is 6.

Using equation (a), the third smallest value of$n_B$at which a harmonic frequency of B matches that of A is given as:

$\begin{array}{c}{n}_B=2\left(5\right)\\ =10\end{array}$

Hence, the third smallest value of $n_B$is 10.