Answer:
a) We see that the tubes of lengths 15, 45 and 75 resonate with this wavelength
b) There is resonance for the lengths 25 and 75 cm
c) Resonance occurs for tubes with length 31 and 93 cm
Explanation:
To find the length of the tube that has resonance we must find the natural frequencies of the tubes, for this at the point that the tube is closed we have a node and the open point we have a belly; in this case the fundamental wave is
λ = 4L
The next resonance called first harmonic λ₃ = 4L / 3
The next fifth harmonic resonance λ₅ = 4L / 5,
WE see that the general form is λ ₙ= 4L / n n = 1, 3, 5 ...
Let's use these expressions for our problem
Let's start with the shortest wavelength.
a) Lam = 60 cm
Let's look for the tube length that this harmonica gives
L = λ n / 4
To find the shortest tube length n = 1
L = 60 1/4
L = 15 cm
For n = 3
L = 60 3/4
L = 45 cm
For n = 5
L = 60 5/4
L = 75 cm
For n = 7
L = 60 7/4
L = 105cm
We see that the tubes of lengths 15, 45 and 75 resonate with this wavelength, in different harmonics 1, 3 and 5
.b) λ = 100 cm
For n = 1
L = 100 1/4
L = 25 cm
For n = 3
L = 100 3/4
L = 75 cm
For n = 5
L = 100 5/4
L = 125 cm
There is resonance for the lengths 25 and 75 cm in the fundamental and third ammonium frequency
c) λ= 124 cm
L = 124 1/4
L = 31 cm
For the second resonance
L = 124 3/4
L = 93 cm
Resonance occurs for tubes with length 31 and 93 cm in the fundamental harmonics and third harmonics
