Imagine that you are a world$-$renowned maker of guitars. You have mastered the art through decades of practical experience, wisdom, and intuition won from hard work and many mistakes. But now we'll bring physics in to ruin your craft, as we always do$.$ Guitars should have the tensions of all their strings roughly equal to one another, to avoid putting too much stress on the top compared to the bottom $($for example$).$ The tension can be varied without changing the pitches by using different string gauges $($diameters$),$ through the equation v $=$ p FT $/\mu .$ 

$($a$)$ Find the relationship between velocity and frequency.

$($b$)$ The low E has a frequency of $82\mathrm{.}4$ Hz and the high E vibrates at $329\mathrm{.}6$ Hz$.$ Assuming you want to keep the tension on both strings constant, should the linear mass density of the high E string be greater or less than that of the low E string, and by a factor of how much? Also assume that any entities other than v are constant

$($c$)$ Assume $0\mathrm{.}046$ inches as the gauge $($diameter$)$ of the low E string, and a linear density of $0\mathrm{.}00002215$ lb$/$in$.$ On the other hand, the volumetric density $($our familiar density$)$ and length of our steel strings must be constant. Using this information, what should be the linear density and gauge of the high E string? For comparison, a typical electric guitar uses $10$ gauge, i$.$e$.$ the high E string has a diameter of $0\mathrm{.}010$ inches.