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?

  L = 0.5 m ,       L ’= 0.625 m

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

does anyone know or what?