Question

Let’s now apply our equations for the normal modes of open and closed pipes. On a day when the speed of sound is 345 m/s, the fundamental frequency of an open organ pipe is 690 Hz. If the n=2 mode of this pipe has the same wavelength as the n=5 mode of a stopped pipe, what is the length of each pipe?

Answer 1

L = 0.5 m , L ’= 0.625 m

Step-by-step explanation:

The equations that describe the resonant wavelength in tube are

Tube or open at both ends

λ = 2L / n n = 1, 2, 3….

Tube with one end open and one closed

λ = 4L / m m = 1, 3, 5 ...

Let's raise our case

In the first part the fundamental frequency is f = 690 Hz

Use the relationship

v = λ f

λ = v / f

The fundamental frequency occurs for n = 1 in the first equation

L = n λ / 2

L = n v / 2f

L = 2 345 / (2 690)

L = 0.5 m

In the second part they say . The n = 2 of the open pipe

λ = 2L / 2

The m = 5 of the closed pipe

λ = 4L ’/ 5

They indicate that the two wavelengths are equal, so we can match the equations

2L / 2 = 4L ’/ 5

L ’/ L = 5/4

L ’= 5/4 L

L ’= 5/4 0.5

L ’= 0.625 m