Question

A particularly beautiful note reaching your ear from a rare stradivarius violin has a wavelength of 39.1 cm. the room is slightly warm, so the speed of sound is 344 m/s. part a if the string's linear density is 0.550 g/m and the tension is 160 n , how long is the vibrating section of the violin string?

Answer 1

The wavelength of the note is $\lambda = 39.1 cm = 0.391 m$. Since the speed of the wave is the speed of sound, $c=344 m/s$, the frequency of the note is
$f= \frac{c}{\lambda}=879.8 Hz$

Then, we know that the frequency of a vibrating string is related to the tension T of the string and its length L by
$f= \frac{1}{2L} \sqrt{ \frac{T}{\mu} }$
where $\mu=0.550 g/m = 0.550 \cdot 10^{-3} kg/m$ is the linear mass density of our string.
Using the value of the tension, T=160 N, and the frequency we just found, we can calculate the length of the string, L:
$L= \frac{1}{2f} \sqrt{ \frac{T}{\mu} } =0.31 m$