Lucy has three sources of sound that produce pure tones with wavelengths of 60cm, 100cm, and 124cm.
She also has an assortment of narrow tubes that are closed off at one end but open at the other.
After experimenting, Lucy notices that each tone will resonate with tubes of some lengths, but not with others.

For each wavelength, determine with tube lengths if any , will resonate with sound of that wavelength?

Length of tubes: 25cm, 75cm, 30cm, 93cm, 15cm, and 45cm.

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

okpalawalter8 okpalawalter8 · Answer 2 · 2021-12-07T16:16:01+01:00

Answer:

For [tex]\alpha = 60[/tex], length of the tubes = [tex]15cm, 45cm, and 75cm.[/tex]

[tex]\alpha = 100[/tex], length of the tubes = [tex]25cm and 75cm[/tex]

[tex]\alpha = 124[/tex], length of the tube = [tex]93cm[/tex]

Explanation:

we have [tex]L = (2n+1) * \frac {\alpha}{4}[/tex]

for, [tex]\alpha = 60[/tex]

[tex]L = (2n+1) * \frac{60}{4}\\\\L = (2n+1) * 15\\\\so when n = 0,1,2,3\\\\L = 15cm, 45cm, 75cm, 135cm[/tex]

for, [tex]\alpha = 100[/tex]

[tex]L = (2n+1) * \frac{100}{4}\\\\L = (2n+1) * 25\\\\so when n = 0,1,2,3\\\\L = 25cm, 75cm, 125cm, 175cm[/tex]

for, [tex]\alpha = 124[/tex]

[tex]L = (2n+1) * \frac{124}{4}\\\\L = (2n+1) * 31\\\\so when n = 0,1,2,3\\\\L = 31cm , 93cm, 155cm[/tex]

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