Answer:
a) λ₁ = 15.6 cm, λ₂ = 7.8 cm, λ₃ = 5.2 cm, λ₄ = 3.9 cm
b) f₁ = 2243 Hz, f₂ = 4487 Hz, f₃ = 6730 Hz, f₄ = 8974 Hz
c) λ₁ = 31.2 cm
, λ₂ = 10.4 cm, λ₃= 6.24 cm, λ₄= 4.46 cm
d) f₁ = 1121.8 Hz
, f₂ = 3365 Hz
, f₃ = 5609 Hz
, f₄ = 7848 Hz
Explanation:
a) This simulation can be approximated as a tube that is open at both ends, in this case we have wave maximums at these points, the relationship between the wavelength and the length of the wave is
λ = 2L
λ₂ = (2L) / 2
λ₃ = (2L) / 3
λₙ = 2L / n
n = 1, 2, 3 ...
Let's apply this equation to our case
λ = 2 7.80 cm
λ₁ = 15.6 cm
λ₂ = 2 7.8 / 2
λ₂ = 7.8 cm
λ₃ = 2 * 7.8 / 3
λ₃ = 5.2 cm
λ₄ = 2 7.8 / 4
λ₄ = 3.9 cm
For the frequency we use the relationship
v = λ f
f = v / λ
f1 = v / Lam
The wavelength in meters
f₁ = 350 /15.6 10-2
f₁ = 2243 Hz
f₂ = 350 / 0.078
f₂ = 4487 Hz
f₃ = 350 / 0.052
f₃ = 6730 Hz
f₄ = 350 / 0.039
f₄ = 8974 Hz
c) in this case the mouth is closed, therefore, at this point we have a node and the open part a belly (maximum)
λ = 4L
λ₂ = 4L / 3
λ₃ = 4L / 5
λₙ = 4L / n
n = 1, 3, 5, 7, ...
We calculate
λ₁ = 4 7.80 / 1
λ₁ = 31.2 cm
λ₂ = 4 7.8 / 3
λ₂ = 10.4 cm
λ₃ = 4 7.8 / 5
λ₃= 6.24 cm
λ₄ = 4 7.8 / 7
λ₄= 4.46 cm
We calculate the frequencies the wavelength in meters
f₁ = 350 / 0.312
f₁ = 1121.8 Hz
f₂ = 350 / 0.104
f₂ = 3365 Hz
f₃ = 350 /0.0624
f₃ = 5609 Hz
f₄ = 350 /.00446
f₄ = 7848 Hz