Question

Part 1 - Basic Equations

Answer 1

Answer:

1. λ = 2 L, 2.  v = 2L f₁ , 3.    v = √ T /μ², 4.   μ = 2,287 10⁻³ kg / m , 5.   Δv / v = 0.058 , 6.    Δμ /  μ = 0.12 , 7. Δ μ = 0.3  10⁻³ kg / m ,

8.  μ = (2.3 ±0.3)  10⁻³ kg / m

Explanation:

The speed of a wave is

            v = λ f                1

Where f is the frequency and λ the wavelength

     

The speed is given by the physical quantities of the system with the expression

            v = √ T /μ²                   2

1) The fundamental frequency of a string is when at the ends we have nodes and a maximum in the center, therefore this is

                 L = λ / 2

                 λ = 2 L

2) For this we substitute in equation 1

              v = 2L f₁

3) let's clear from equation 2

             

The speed of a wave is

            v = λ f₁

Where f is the frequency and Lam the wavelength

The speed is given by the physical quantities of the system with the expression

           v = √ T /μ²                            2

4) linear density is

           μ = T / (2 L f₁)²

           μ = 5.08 / (2 0.812 29.02)²

           μ = 2,287 10⁻³ kg / m

We maintain three significant length figures, so the result is reduced to

           μ = 2.29 10⁻³ kg / m

5) the speed of the wave is

            v = 2 L f₁

The fractional uncertainty is

         Δv / v = ΔL / L + Δf₁ / F₁

         Δv / v = 0.02 / 0.812 + 1 / 29.02

         Δv / v = 0.024 + 0.034

         Δv / v = 0.058

6) the equation for linear density is

              μ = T / (2 L f₁)²

             Δμ / μ = 2 ΔL / L + 2Δf₁ / f₁

The tension is an exact value therefore its uncertainty is zero ΔT = 0

            Δμ / μ = 2 0.02 / 0.812 + 2 1 / 29.02

             Δμ /  μ = 0.12

7) absolute uncertainty

           Δ μ = $e_{r}$   μ

           Δ μ = 0.12 2.29 10⁻³ kg / m

           Δ μ = 0.3  10⁻³ kg / m

8)

           μ = (2.3 ±0.3)  10⁻³ kg / m