Question: In Fig. 17-27, pipe Ais made to oscillate in its third harmonicby a small internal sound source. Sound emitted at the right endhappens to resonate four nearby pipes, each with only one openend (they are notdrawn to scale). Pipe Boscillates in its lowestharmonic, pipe Cin its second lowest harmonic, pipe Din itsthird lowest harmonic, and pipe Ein its fourth lowest harmonic.Without computation, rank all five pipes according to theirlength, greatest first. (Hint:Draw the standing waves to scale andthen draw the pipes to scale.)

1. The pipe A oscillates in third harmonic i.e n_A = 3..
2. The pipe B oscillates in the lowest harmonic.
3. The pipe A oscillates in 2^nd lowest harmonic.
4. The pipe A oscillates in 3^rd lowest harmonic.
5. The pipe A oscillates in 4^th lowest harmonic.

The sound waves of one pipe can resonate with the other pipe only when their frequencies match. Thus, calculate the resonant frequencies for both ends of open pipe A and each of the pipes B, C, D, and E which are open at one end.

The formulae are as follow:

For pipe open at both ends

$f=\frac{nv}{2L}$

For pipe open at one end

$f=\frac{nv}{4L}$

where f is frequency, L is length and vis velocity.

For Pipe A:

Let the length of the pipe A be L_A. It is open at both ends. So, its resonant frequency in third harmonic will be

$$\begin{gathered} f=\frac{nv}{2L} \\ \therefore f_A=\frac{3v}{2L_A}\dots \dots \dots \dots \dots \dots \dots \dots \dots \dots \dots \dots \dots \dots \dots \dots \dots \dots \dots \dots \dots .\dots \dots \dots \dots \dots \dots \dots \dots \dots \dots \dots \dots \dots \dots \left(1\right) \end{gathered}$$

For pipe B:

Let the length of the pipe B be L_B. It is open at one end. So, its resonant frequency in lowest harmonic will be

$$\begin{gathered} f=\frac{nv}{4L} \\ \therefore f_B=\frac{v}{4L_B} \end{gathered}$$

This frequency matches with that of pipe A. Hence

$$\begin{gathered} \therefore f_A=f_B\Rightarrow \frac{3v}{2L_A}=\frac{v}{4L_B} \\ \therefore L_B=\frac{2L_A}{4\times 3}=\frac{L_A}{6} \\ \therefore L_B=\frac{L_A}{6}\dots \dots \dots \dots \dots \dots \dots \dots \dots \dots \dots \dots \dots \dots \dots \dots \dots \dots \dots \dots \dots \dots \dots \dots \dots \dots \dots \dots \dots \dots \dots \dots \dots \dots \dots ..\left(2\right) \end{gathered}$$

For pipe C:

Let the length of pipe C be L_C. It is open at one end. So, its resonant frequency in 2^nd lowest harmonic will be

role="math" localid="1661745283061" $$\begin{gathered} f=\frac{nv}{4L} \\ \therefore f_C=\frac{3v}{4L_C} \end{gathered}$$

This frequency matches with that of pipe A. Hence

$$\begin{gathered} \therefore f_A=f_C\Rightarrow \frac{3v}{2L_A}=\frac{3v}{4L_C} \\ \therefore L_C=\frac{2L_A}{4}=\frac{L_A}{2} \\ \therefore L_C=\frac{L_A}{2}\dots \dots \dots \dots \dots \dots \dots \dots \dots \dots \dots \dots \dots \dots \dots \dots \dots \dots \dots \dots \dots \dots \dots \dots \dots \dots \dots \dots \dots \dots \dots \dots \dots \dots \dots ..\left(3\right) \end{gathered}$$

For pipe D:

Let the length of the pipe D be L_D. It is open at one end. So, its resonant frequency in 3^rd lowest harmonic will be

$$\begin{gathered} f=\frac{nv}{4L} \\ \therefore f_D=\frac{5v}{4L_D} \end{gathered}$$

This frequency matches with that of pipe A. Hence

$$\begin{gathered} \therefore f_A=f_D\Rightarrow \frac{3v}{2L_A}=\frac{5v}{4L_D} \\ \therefore L_D=\frac{5\times 2L_A}{4\times 3}=\frac{5L_A}{6} \\ \therefore L_D=\frac{5L_A}{6}\dots \dots \dots \dots \dots \dots \dots \dots \dots \dots \dots \dots \dots \dots \dots \dots \dots \dots \dots \dots \dots \dots \dots \dots .\dots \dots \dots \dots \dots \dots \dots \dots \dots \dots ..\left(4\right) \end{gathered}$$

For pipe E:

Let the length of the pipe E be L_E. It is open at one end. So, its resonant frequency in 4^th lowest harmonic will be

$$\begin{gathered} f=\frac{nv}{4L} \\ \therefore f_E=\frac{7v}{4L_E} \end{gathered}$$

This frequency matches with that of pipe A. Hence

$$\begin{gathered} \therefore f_A=f_E\Rightarrow \frac{3v}{2L_A}=\frac{7v}{4L_E} \\ \therefore L_E=\frac{7\times 2L_A}{4\times 3}=\frac{7L_A}{6} \\ \therefore L_E=\frac{7L_A}{6}\dots \dots \dots \dots \dots \dots \dots \dots \dots \dots \dots \dots \dots \dots \dots \dots \dots \dots \dots \dots \dots \dots \dots \dots \dots .\dots \dots \dots \dots \dots \dots \dots \dots \dots ..\left(5\right) \end{gathered}$$

Thus, from equations (1), (2), (3), (4) and (5), the lengths of the pipes are

$L_A,\frac{L_A}{6},\frac{L_A}{2},\frac{5L_A}{6}and\frac{7L_A}{6}$respectively for pipes A, B, C, D and E.

Thus, the length of the pipe E is the greatest while that of pipe B is the lowest.

Hence, the ranking will be E, A, D, C and B.

Therefore, the sound wave in one pipe can resonate with the other pipe when their frequencies match. The formulae can be used to determine the frequency matching the harmonics. This leads us to the information about the lengths of the pipes