Question

A slide whistle is an open-closed tube with an adjustable plunger that changes the length. You are playing the slide whistle and pushing the plunger in at 8 cm/s when the tube is 15 cm long. What is the rate of change of the sound in Hz/s?

Answer 1

Answer:

df / ft = -n 12          n= 1, 3, 5, ...

Explanation:

The speed of sound is

         v = λ f

         

In a whistle that we approach by an open tube at one end and closed at the other, standing waves occur, which has a node in the closed part and a maximum in the open pate, whereby wavelength and the distance of the tube are related, the fundamental wave is

         λ₁ = 4L

   The harmonics are

        λ₃ = 4L / 3

        λ₅ = 4L / 5

The general formula

       λₙ = 4L / n              

with n = 1, 3, 5,…

We substitute and clear in the first equation

           f = v n / 4L                        n = 1, 3, 5,…

Let's use derivatives to find the frequency change

           df / dt = v n /4  dL⁻¹ / dt

          d / dt (1/L) = - 1 / L² dL / dt

Where dL / dt = 8 cm / s

We replace

         df / dt = - n v / L2 dL / dt

Let's calculate

         df / dt = - n 340/152 8

         df / ft = -n 12          n= 1, 3, 5, ...