Question

A violinist places her finger so that the vibrating section of a 1.30 g/m string has a length of 40.0 cm , then she draws her bow across it. A listener nearby in a 20∘C room hears a note with a wavelength of 60.0 cm . What is the tension in the string?

Answer 1

To solve this problem it is necessary to apply the concepts related to string vibration. A vibration in a string is a wave. Resonance causes a vibrating string to produce a sound with constant frequency, i.e. constant pitch,

The speed of propagation of the wave in a string is proportional to the square root of the tension and inversely proportional to the root of the linear density. Mathematically this can be expressed as

`v = \sqrt{(T)/(\rho)} \Rightarrow v^2 = (T)/(\rho)`

At the same time we have that the frequency is subject to the wavelength and the speed. That is to say

`f = (v)/(\lambda) \rightarrow v = 343m/s` Speed of sound at 20°C.

Therefore the frequency would be

`f = (v)/(\lambda)`

`f = (343)/(0.6)`

`f= 571.7Hz`

While that is the auditory frequency, the frequency caused in the string would be

`\lambda = 2L`

`\lambda = 2*0.4`

`\lambda = 0.8m`

So reusing the frequency formula we can find that now the velocity in the string is of

`f = (v)/(\lambda)`

`v = \lambda f`

`v = 0.8*571.7`

`v = 457.3m/s`

Finally using the vibration in the string we can find the tension, so:

`v^2 = (T)/(\rho)`

`T = v^2 \rho`

`T = 457.3 * 0.0013`

`T = 209N`