Respuesta :
The tubes with lengths of 15, 45, and 75 cm are found to resonate with this wavelength. Additionally, resonance is seen for tubes with lengths of 25 and 75 cm and 31 and 93 cm.
For this reason, at the place where the tube is closed, we have a node, and at the open point, we have a belly; in this case, the fundamental wave is = 4L. To determine the length of the tube that has resonance, we must establish the natural frequencies of the tubes.
The following resonance, known as the first harmonic, is 3 = 4L / 3.
The next fifth harmonic resonance is 5 = 4L / 5, and we can see that its general form is n= 4L / n, where n = 1, 3, and 5.
Let's apply these terms to the issue at hand.
Start with the wavelength that is the shortest.
Lam equals 60 cm.
Let's determine the length of the tube this harmonica provides.
L = λ n / 4
n = 1 L = 60 will yield the shortest tube length. 1/4\s L = 15 cm
For n = 3 L = 60 3/4\s L = 45 cm
For n = 5 L = 60 5/4\s L = 75 cm
For n = 7 L = 60 7/4\s L = 105cm
We can see that the harmonics 1, 3 and 5 of the tubes with lengths 15, 45, and 75 resonate with this wavelength.
b) λ = 100 cm
For n = 1 L = 100 1/4\s L = 25 cm
For n = 3 L = 100 3/4\s L = 75 cm
For n = 5 L = 100 5/4\s L = 125 cm
There is resonance in the fundamental and third ammonium frequencies for lengths of 25 and 75 cm. L = 124 1/4 L = 31 cm
In the case of the second resonance, L = 124 3/4 L = 93 cm.
For tubes with lengths of 31 and 93 cm, resonance occurs in the fundamental and third harmonics.
Learn more about Wavelength here:
https://brainly.com/question/14363924
#SPJ4
