The portion of a uniform violin string that vibrates is from the "nut" to the "bridge" at the end of the finger board, and has length LA and mass m. The string tension F can only be increased or decreased by tightening or loosening the tuning pegs above the nut. Say that a string is tuned to produce a note with fundamental frequency ft. a) Then, to play a different note with fundamental frequency fp, the violinist uses her finger to push the string against the fingerboard, reducing the length of the vibrating part of the string to L2 (The string now vibrates from her finger to the bridge.)
i) When she makes this change, determine if each of the following quantities increase, decrease, or remain unchanged: (1) string tension, (2) mass density, (3) wave speed, and (4) wavelength. Explain each.
ii) Is fy greater or less than fx ? Explain.
iii) Find an equation for Lg in terms of all or some of the given parameters. Simplify.
b) Find an equation for F in terms of the given parameters. c) Calculate numerical values for L, and F if m= 2.00 g, LA = 60.0 cm, fa = 440 Hz (A note), fB = 494 Hz (B note). -Bridge

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

answer: 2×10^-2kg

explanation: density= mass per volume

vol=mass/density

converting to s.i units we have;

3÷1000kg = 0.003

1.5÷100m = 0.15

0.003/0.15 = 0.02kg