91Ó°ÊÓ

1. The pipe A has length =$L_A=L$
2. The pipe B has length =$L_B=2L$

When the resonant frequency of one pipe matches with the frequency of another pipe then a sound wave in one pipe can create a standing wave in the other pipe. The number of harmonics that matches will depend on the number of open ends of the pipe. Thus, calculate the harmonics for the two pipes by matching their resonant frequencies.

Formulae are as follow:

1. For pipe open at both ends,
   $f=\frac{nv}{2L},n=1,2,3,………$
2. For pipe open at one end,
   $f=\frac{nv}{4L},n=1,3,5,……..$

Where,

f is frequency, L is length and v is velocity.

For Pipe A:

It is open at one end. So, its resonant frequency will be,

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

For pipe B:

It is open at both ends. So, its resonant frequency will be,

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

This frequency matches with that of pipe A. Hence,

$∴f_A=f_B⇒\frac{n_{A}v}{4L_A}=\frac{n_{B}v}{2L_B}$

$∴\frac{n_A}{4L_A}=\frac{n_B}{2L_B}$

But,$L_A=L$and L_B =2L,

$∴\frac{n_A}{4L}=\frac{n_B}{2\left(2L\right)}$

$∴n_A=\frac{4L}{4L}{n}_B$

$∴n_A=n_B$

Since, pipe A is open at one end only, it has only odd integer harmonics.Hence, all odd integer harmonics of pipe B matches with corresponding resonant frequencies of pipe A.

Therefore, the sound wave in one pipe can create a standing wave in the other only when the two resonant frequencies match. Hence, the resonant frequencies of pipe A and B can be matched to determine the number of harmonics of B that match with resonant frequency of pipe A.