Question

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 pushing 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.

Answer 1

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