Question

Pipe Ahas length Land one open end. Pipe Bhas length 2L and two open ends. Which harmonics of pipe Bhave a frequencythat matches a resonant frequency of pipe A?

Short answer

All odd integer harmonics of pipe B match withthecorresponding resonant frequencies of the pipe A.

Step-by-step solution

Step 1: Given

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

Determining the concept

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

Where,

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

Determining if the harmonics of pipe B match with frequency of pipe A

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,

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

$\therefore \frac{n_A}{4L_A}=\frac{n_B}{2L_B}$

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

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

$\therefore n_A=\frac{4L}{4L}{n}_B$

$\therefore n_A=n_B$

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.