Solve Problem 2.73 if the material of the helical spring is changed from music wire to aluminum with \(G=26 \mathrm{GPa}\) and \(ho=2690 \mathrm{~kg} / \mathrm{m}^{3}\).

Data From Problem 2.73:-

A helical spring, made of music wire of diameter \(d\), has a mean coil diameter \((D)\) of \(14 \mathrm{~mm}\) and \(N\) active coils (turns). It is found to have a frequency of vibration

(f) of \(193 \mathrm{~Hz}\) and a spring rate \(k\) of \(4.6 \mathrm{~N} / \mathrm{mm}\). Determine the wire diameter \(d\) and the number of coils \(N\), assuming the shear modulus \(G\) is \(80 \mathrm{GPa}\) and density \(ho\) is \(8000 \mathrm{~kg} / \mathrm{m}^{3}\). The spring rate \((k)\) and frequency ( \(f\) ) are given by

\[k=\frac{d^{4} G}{8 D^{3} N}, \quad f=\frac{1}{2} \sqrt{\frac{k g}{W}}\]

where \(W\) is the weight of the helical spring and \(g\) is the acceleration due to gravity.