Question

A guitar's E-string has length \(65 \mathrm{cm}\) and is stretched to a tension of \(82 \mathrm{N}\). It vibrates at a fundamental frequency of $329.63 \mathrm{Hz}$ Determine the mass per unit length of the string.

Short answer

Answer: The mass per unit length of the guitar's E-string is approximately \(7.68\times10^{-4}\,\text{kg/m}\).

Step-by-step solution

Convert given information to SI units

We are given the string length \(L = 65\,\text{cm}\). First, we need to convert it to meters: $$ L = 65\,\text{cm} \times \frac{1\,\text{m}}{100\,\text{cm}} = 0.65\,\text{m}. $$ The tension, \(T\), is given in SI units, so \(T = 82\,\text{N}\).

Rearrange the formula for mass per unit length

To find the mass per unit length, we will rearrange the fundamental frequency formula to solve for \(μ\): $$ μ = \frac{T}{(2Lf)^2}. $$

Plug in given values and calculate

Now, we can plug in the given values for string length (\(L = 0.65\,\text{m}\)), tension (\(T = 82\,\text{N}\)), and fundamental frequency (\(f = 329.63\,\text{Hz}\)) to find the mass per unit length: $$ μ = \frac{82\,\text{N}}{(2\times0.65\,\text{m}\times329.63\,\text{Hz})^2} = \frac{82}{(1.3\times329.63)^2}\,\text{kg/m}. $$ After doing the math, we get: $$ μ \approx 7.68\times10^{-4}\,\text{kg/m}. $$ The mass per unit length of the guitar's E-string is approximately \(7.68\times10^{-4}\,\text{kg/m}\).