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Semenov [28]
2 years ago
14

The G string on a guitar is 69 cm long and has a fundamental frequency of 196 Hz. A guitarist can play different notes by pushin

g the string against various frets, which changes the string's length. The third fret from the neck gives B♭ (233.08 Hz); the fourth fret gives B (246.94 Hz).
How far apart are the third and fourth frets?

Express your answer to two significant figures and include the appropriate units.
Physics
1 answer:
Readme [11.4K]2 years ago
6 0

Answer:

 Δd = 23 cm

Explanation:

When the eta string on the guitar has nodes at its ends, so the waves produced give rise to a standing wave, which can be described by the following expressions

Fundamental      L = ½ λ

1st harmonic       L = 2 ( λ / 2)

2nd harmonic     L = 3 ( λ / 2)

Harmonic n         L = n  λ / 2

Where n is an integer

                     

The speed of the rope is given by the ratio

         v =  λ f

This speed is constant since it depends on the tension and the linear density of the rope

Let's calculate the speed with the initial data

           v = 0.69  196

           v = 135.24 m / s

Let's look for the wavelength for the two frequencies

           λ₁ = v / f₁

           λ₁ = 135.24 / 233.08

           λ₁ = 0.58022 m

           λ₂ = v / f₂

           λ₂ = 135.24 / 246.94

            λ₂ = 0.54766 m

Let's replace in the resonance equation

             Lₙ = n  λ/2

For the third fret

            m = 3

            L₃ = 3 0.58022 / 2

            L₃ = 0.87033 m

For the fourth fret

            m = 4

            L₄ = 4 0.54766 / 2

            L₄ = 1.09532 m

The distance between the two frets is

         Δd = L₄ –L₃

         Δd = 1.09532 - 0.87033

         Δd = 0.22499 m

         Δd = 22.5 cm = 23 cm

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A sphere of radius 5.00 cm carries charge 3.00 nC. Calculate the electric-field magnitude at a distance 4.00 cm from the center
OlgaM077 [116]

Answer:

a)   E = 8.63 10³ N /C,  E = 7.49 10³ N/C

b)   E= 0 N/C,  E = 7.49 10³ N/C  

Explanation:

a)  For this exercise we can use Gauss's law

         Ф = ∫ E. dA = q_{int} /ε₀

We must take a Gaussian surface in a spherical shape. In this way the line of the electric field and the radi of the sphere are parallel by which the scalar product is reduced to the algebraic product

The area of ​​a sphere is

        A = 4π r²

 

if we use the concept of density

        ρ = q_{int} / V

        q_{int} = ρ V

the volume of the sphere is

      V = 4/3 π r³

         

we substitute

         E 4π r² = ρ (4/3 π r³) /ε₀

         E = ρ r / 3ε₀

the density is

         ρ = Q / V

         V = 4/3 π a³

         E = Q 3 / (4π a³) r / 3ε₀

         k = 1 / 4π ε₀

         E = k Q r / a³

 

let's calculate

for r = 4.00cm = 0.04m

        E = 8.99 10⁹ 3.00 10⁻⁹ 0.04 / 0.05³

        E = 8.63 10³ N / c

for r = 6.00 cm

in this case the gaussine surface is outside the sphere, so all the charge is inside

         E (4π r²) = Q /ε₀

         E = k q / r²

let's calculate

         E = 8.99 10⁹ 3 10⁻⁹ / 0.06²

          E = 7.49 10³ N/C

b) We repeat in calculation for a conducting sphere.

For r = 4 cm

In this case, all the charge eta on the surface of the sphere, due to the mutual repulsion between the mobile charges, so since there is no charge inside the Gaussian surface, therefore the field is zero.

         E = 0

In the case of r = 0.06 m, in this case, all the load is inside the Gaussian surface, therefore the field is

        E = k q / r²

      E = 7.49 10³ N / C

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