Question

You are playing a violin, where the fundamental frequency of one of the strings is 440 Hz, as you are standing in front of the opening of a long tube that is closed at the other end. As you play, you notice that the first time you hear an echo from the tube is when the sound from the 440 Hz string is in its fourth harmonic. Assuming you are playing this string on earth, what must be the length of the tube? What is the next higher harmonic number for the same string on the violin when you hear the next higher harmonic echo from the tube?

Answer 1

a) L = 440 cm

Step-by-step explanation:

In the open tube on one side and cowbell on the other, we have a maximum in the open part and a node in the closed part, therefore the resonance frequencies are

λ₁ = 4L fundamental

λ₃ = 4L / 3 third harmonic

λ₅ = 4L / 5 five harmonic

The violin string is a fixed cure in its two extracts, so both are nodes, their length from resonance wave are

λ₁ = 2L fundamental

λ₂ = 2L / 2 second harmonic

λ₃ = 2L / 3 third harmonic

λ₄= 2L / 4 fourth harmonic

They indicate that resonance occurs in the fourth harmonic, let's look for the frequency

v =λ f

for the fundamental

v = λ₀ f₀

V = 2L f₀

for the fourth harmonica

v = λ₄ f ’

v = L / 2 f'

2L f₀ = L / 2 f ’

f ’= 4 f₀

f ’= 4 440

f ’= 1760 Hz

for this frequency it has the resonance with the tube

f ’= 4L

L = f ’/ 4

L = 1760/4

L = 440 cm

b) let's find the frequency of the next harmonic in the tube

λ₃ = 4L / 3

λ₃ = 4 400/3

λ₃ = 586.6 cm

v = λf

f = v / λlam₃

f₃3 = 340 / 586.6

f3 = 0.579

as the minimum frequency on the violin is 440 Beam there is no way to reach this value, therefore there are no higher resonances