Question

studyhero A violin string has a length, from the bridge to the end of the fingerboard, of 50 cm. That section of the string has a mass of only 2 g. When the violinist plays an open string (the full length) a 440 Hz A-note is heard. Determine the length of the string needed to play a 528 Hz note without adjusting the tension in the string. This is accomplished by pressing on the fingerboard at the appropriate location.

Answer 1

Answer:

= 0.517 m

Explanation:

This is a resonance exercise where the ends of the string are fixed, therefore it has a node of them, the fundamental (longest) wavelength created has the form

                λ = 2L

wave speed is related to wavelength and frequency

               v = λ f

               v = 2L f

let's calculate

                v = 2 0.50 440

                v = 440 m / s

since they indicate that the tension of the string does not change and the linear density of the string is constant, the speed of the wave also remains constant

               f =$\frac{v}{2L}$

let's find the length for the new resonance frequency

               L = $\ \frac{v}{2f}$

let's calculate

               L = $\frac{440}{2 \ 528}$

               L = 0.5166 m