91Ó°ÊÓ

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

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

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

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

This frequency matches with that of pipe A. Hence

$$\begin{gathered} ∴f_A=f_B⇒\frac{3v}{2L_A}=\frac{v}{4L_B} \\ ∴L_B=\frac{2L_A}{4×3}=\frac{L_A}{6} \\ ∴L_B=\frac{L_A}{6}……………………………………………………………………………………………..\left(2\right) \end{gathered}$$

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" $$\begin{gathered} f=\frac{nv}{4L} \\ ∴f_C=\frac{3v}{4L_C} \end{gathered}$$

This frequency matches with that of pipe A. Hence

$$\begin{gathered} ∴f_A=f_C⇒\frac{3v}{2L_A}=\frac{3v}{4L_C} \\ ∴L_C=\frac{2L_A}{4}=\frac{L_A}{2} \\ ∴L_C=\frac{L_A}{2}……………………………………………………………………………………………..\left(3\right) \end{gathered}$$

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

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

This frequency matches with that of pipe A. Hence

$$\begin{gathered} ∴f_A=f_D⇒\frac{3v}{2L_A}=\frac{5v}{4L_D} \\ ∴L_D=\frac{5×2L_A}{4×3}=\frac{5L_A}{6} \\ ∴L_D=\frac{5L_A}{6}……………………………………………………………….…………………………..\left(4\right) \end{gathered}$$

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

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

This frequency matches with that of pipe A. Hence

$$\begin{gathered} ∴f_A=f_E⇒\frac{3v}{2L_A}=\frac{7v}{4L_E} \\ ∴L_E=\frac{7×2L_A}{4×3}=\frac{7L_A}{6} \\ ∴L_E=\frac{7L_A}{6}………………………………………………………………….………………………..\left(5\right) \end{gathered}$$

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