Answer:
A)i) 1. constant, 2. constant, 3. constant, 4. decrease
ii) frecuency increase
iii) L = n /2f √T/μ
B) L_b = 0.534 m
Explanation:
We can approximate the violin string as a system of a fixed string at its two ends, therefore we have a node at each end and a maximum in the central part for a fundamental vibration,
λ = 2L / n
where n is an integer
The wavelength and frequency are related
v = λ f
and the speed of the wave is given by
v = √T /μ
with these expressions we can analyze the questions
A)
i) In this case the woman decreases the length of the rope L = L₂
therefore the wavelength changes
λ₂ = 2 (L₂) / n
as L₂ <L₀ the wavelength is
λ₂ < λ₀
The tension of the string is given by the force of the plug as it has not moved, the tension must not change and the density of the string is a constant that does not depend on the length of the string, therefore the speed of the string wave in the string should not change.
ii) how we analyze if the speed of the wave does not change
v = λ f
as the wavelength decreases, the frequency must increase so that the speed remains constant
fy> fx
iii) It is asked to find the length of the chord
let's use the initial equations
λ = 2L / n
v = λ f
v = 2L / n f
v = √ T /μ
we substitute
2 L / n f = √ T /μ
L = n /2f √T/μ
this is the length the string should be for each resonance
b) in this part they ask to calculate the frequency
f = n / 2L √ T /μ
the linear density is
μ = m / L
μ = 2.00 10⁻³ / 60.0 10⁻²
μ = 3.33 10⁻³ kg / m
we assume that the length is adequate to produce a fundamental frequency in each case
f_{a} = 440Hz
λ = 2La / n
λ = 2 0.60 / 1
λ = 1.20 m
v = λ f
v = 1.20 440
v = 528 m / s
v² = T /μ
T = v² μ
T = 528² 3.33 10⁻³
T = 9.28 10² N
Let's find the length of the chord for fb
f_{b} = 494 hz
L_b = 1 /(2 494) √(9.28 10² / 3.33 10⁻³)
L_b = 0.534 m
