Answer:
a) v = 126.5 m / s, b) λ₁ = 1.6 m, λ₂ = 0.8 m, λ₃ = 0.533 m, f1 = 79 Hz,
λ₂ = 0.8 m, f₃ = 237 Hz
Explanation:
This is an exercise we are going to solve in parts, let's start by looking for the speed of the wave in the string.
v = [tex]\sqrt { \frac{T}{ \mu } }[/tex]
the rope tension is T = 90 N and the density can be calculated
μ = m / l
let's calculate
μ = 4.5 10⁻³ / 0.80
μ = 5.625 10⁻³ kg/m
let's calculate the speed
v = [tex]\sqrt{ \frac{90 }{5.625 \ 10^{-3}} }[/tex]
v = 126.5 m / s
For the second part there is a process of resonance in the string. The points where it is attached are nodes so, if L is the length of the chord
L = ½ λ 1st harmonic
L = 2/2 λ 2nd harmonic
L = 3/2 λ 3rd harmonic
L = n/2 λ n harmonic
the wavelength of the first three harmonics is requested
let's calculate
λ₁ = 2L
λ₁ = 2 0.8
λ₁ = 1.6 m
λ₂ = L
λ₂ = 0.8 m
λ₃ = ⅔ L
λ₃ = ⅔ 0.8
λ₃ = 0.533 m
To calculate the frequency we use that the speed is related to the wavelength and the frequency
v = lam f
f = v / lam
we calculate
f1 = 126.5 / 1.6
f1 = 79 Hz
f2 = 126.5 / 0.8
f2 = 158 Hz
f3 = 126.5 / 0.533
f3 = 237 Hz
