Question

the role of the mouth in sound the prodution of sound during speech or singing is a complicated process lets concentrate on the mouth a typcal depth for the huamn mouth is about 8 cm although this number can vary. (Check it against your own mouth.) We can model the mouth as an organ pipe that is open at the back of the throat. What are the wavelengths and frequencies of the first four harmonics you can produce if your mouth is (a) open, (b) closed? Use .v=354 m/s

Answer 1

When open mouth the wavelengths and the frequencies of the first four harmonics you can produce are 0.16 m and 2,212.5 Hz, 0.08 m and 4,425 Hz,  0.053 m and 6,637.5 Hz, 0.04 m and 8,850 Hz.

When closed mouth the wavelengths and the frequencies of the first four harmonics you can produce are 0.32 m and 1,106.25 Hz, 0.1067 m and 3,318.75 Hz, 0.064 m and 5,531.25 Hz, 0.0457 m and 7,743.75 Hz.

If we open our mouth it will create a tube that opens at both ends.

• The wavelength
  $\lambda = \frac{2L}{n}$
  L = the length of a tube (m) = 8 cm = 0.08 m
  λ = the wavelength (m)
  n = 1, 2, 3, 4, 5, ...
• The frequency
  f = v ÷ λ
  v = the speed of the sound (m/s)
  f = the frequency (Hz)

For the first harmonic

• $\lambda_1 = \frac{2L}{1}$
  λ₁ = 2 × 0.08 = 0.16 m
• f₁ = v ÷ λ₁ = 354 ÷ 0.16
  f₁ = 2,212.5 Hz

For the second harmonic

• $\lambda_2 = \frac{2L}{2}$
  λ₂ = 0.08 m

• f₂ = v₂ ÷ λ₂ = 354 ÷ 0.08
  f₂ = 4,425 Hz

For the third harmonic

• $\lambda_3 = \frac{2L}{3}$
  λ₃ = 2 × 0.08 m ÷ 3
  λ₃ = 0.053 m
• f₃ = v₃ ÷ λ₃ = 354 ÷ 0.053
  f₃ = 6,637.5 Hz

For the fourth harmonic

• $\lambda_4 = \frac{2L}{4}$
  λ₄ = 0.08 m ÷ 2
  λ₄ = 0.04 m
• f₄ = v₄ ÷ λ₄ = 354 ÷ 0.04
  f₄ = 8,850 Hz

If we closed our mouth it will create a tube that one end opens and the other end closed.

• The wavelength
  $\lambda = \frac{4L}{2n-1}$
  n = 1, 2, 3, 4, ...

• The frequency
  f = v ÷ λ

For the first harmonic

• $\lambda_1 = \frac{4L}{1}$
  λ₁ = 4 × 0.08 = 0.32 m
• f₁ = v ÷ λ₁ = 354 ÷ 0.32
  f₁ = 1,106.25 Hz

For the second harmonic

• $\lambda_2 = \frac{4L}{3}$
  λ₂ = 4 × 0.08 m ÷ 3
  λ₂ = 0.1067 m

• f₂ = v₂ ÷ λ₂ = 354 ÷ 0.1067
  f₂ = 3,318.75 Hz

For the third harmonic

• $\lambda_3 = \frac{4L}{5}$
  λ₃ = 4 × 0.08 m ÷ 5
  λ₃ = 0.064 m
• f₃ = v₃ ÷ λ₃ = 354 ÷ 0.064
  f₃ = 5,531.25 Hz

For the fourth harmonic

• $\lambda_4 = \frac{4L}{7}$
  λ₄ = 4 × 0.08 m ÷ 7
  λ₄ = 0.0457 m
• f₄ = v₄ ÷ λ₄ = 354 ÷ 0.0457
  f₄ = 7,743.75 Hz

Learn more about a tube that opens at both ends and a tube that one end open and the other end closed here: https://brainly.com/question/29491212

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