Question

The physics of wind instruments is based on the concept of standing waves. When the player blows into the mouthpiece, the column of air inside the instrument vibrates, and standing waves are produced. Although the acoustics of wind instruments is complicated, a simple description in terms of open and closed tubes can help in understanding the physical phenomena related to these instruments. For example, a flute can be described as an open-open pipe because a flutist covers the mouthpiece of the flute only partially. Meanwhile, a clarinet can be described as an open-closed pipe because the mouthpiece of the clarinet is almost completely closed by the reed.

What is the frequency of the first possible harmonic after the fundamental frequency in the open-closed pipe described in Part E?
Part E was What length of open-closed pipe would you need to achieve the same fundamental frequency as the open-open pipe discussed in Part A? Half the length of the open-open pipe.
Part A was Consider a pipe of length 80.0 open at both ends. What is the lowest frequency of the sound wave produced when you blow into the pipe? frequency 214 Hz

Answer 1

a) λ = 4L, b) L_c = ½ Lₐ, c) f = 214 Hz

Step-by-step explanation:

a) in this part, the fundamental frequency (1st harmonic) is requested

In a pipe, the side that is closed has a node and the open side has a belly, so the wavelength inside the pipe is ¼ of the wavelength.

λ = 4L

to find the frequency let's use the definition of velocity

v = λ f

f = v /λ

f = v / 4L

b) in this part let's start by finding the resonance for a tube open at both ends

λ = 2 L

f = v / 2L

since they ask for the same frequency, let's match the two expressions

v / 4L_c = v / 2Lₐ

L_c = ½ Lₐ

therefore the length of the closed tube must be half the length of the open tube

c) It is indicated that the length of the open tube is L = 80 cm = 0.80 m

the frequency is

f = v / 2L

f = 343/2 0.8

f = 214 Hz