Question

What is the frequency of the fundamental mode of vibration of a steel piano wire stretched to a tension of 440 n? the wire is 0.600 m long and has a mass of 5.60 g?

Answer 1

The fundamental frequency of vibration of a string is given by
$f= \frac{1}{2L} \sqrt{ \frac{T}{\mu} }$
where
L is the length of the string
T is the tension
$\mu= \frac{m}{L}$ is the linear mass density, where m is the mass of the string

Let's start by calculating the mass density of the piano wire in the problem. Its mass is
$m=5.60 g=5.60 \cdot 10^{-3} kg$
and its length is L=0.600 m, so its mass density is
$\mu= \frac{5.60 \cdot 10^{-3} kg}{0.600 m}=9.33 \cdot 10^{-3} kg/m$

Therefore now we have everything to calculate the fundamental frequency of the piano wire. Using T=440 N and the first equation, we find:
$f= \frac{1}{2 \cdot 0.600 m} \sqrt{ \frac{440 N}{9.33 \cdot 10^{-3} kg/m} } =180.9 Hz$