91Ó°ÊÓ

The differential equation that guides the motion of a mass-spring system is\({\bf{mx}}''{\bf{ + bx}}'{\bf{ + kx = f(t)}}\).

One has that \({\bf{m = 7\;kg,b = }}2N{\bf{ - }}\frac{{{\bf{sec}}}}{{\bf{m}}}{\bf{,k = }}\frac{{{\bf{3N}}}}{{\bf{m}}}\) and\({\bf{f(t) = 10cos10t\;N}}\), so a differential equation that guides the given mass-spring system is\(7x'' + 2x' + 3x = 10cos10t\).

One wants to find an equivalent differential equation that describes a \({\bf{R L C}}\)series circuit that\(R{\bf{ = 10\Omega }}\).

A differential equation for a \({\bf{RLC}}\) series circuit is given by\({\bf{Lq}}''{\bf{ + Rq}}'{\bf{ + }}\frac{{\bf{1}}}{{\bf{C}}}{\bf{q = E(t)}}\),

For this differential equation to be equivalent to the differential equation \({\bf{7x'' + 2x' + 3x = 10cos10t}}\) it must be;

\(\frac{L}{m}{\bf{ = }}\frac{{\bf{R}}}{{\bf{b}}}{\bf{ = }}\frac{{{\bf{1/C}}}}{{\bf{k}}}{\bf{ = }}\frac{{{\bf{E(t)}}}}{{{\bf{f(t)}}}}{\bf{ = }}A\)

One can find the constant \(A\) from \(\frac{R}{B}{\bf{ = }}A\)

\(\frac{R}{b}{\bf{ = }}\frac{{10}}{2}\; \Rightarrow \;A{\bf{ = }}5\).

One has,

\(\begin{aligned}{c}L{\bf{ = }}m \cdot A = 7 \cdot 5 \Rightarrow \;L{\bf{ = }}35{\bf{H}}\\\frac{1}{C}{\bf{ = }}k \cdot A{\bf{ = }}3 \cdot 5 \Rightarrow C{\bf{ = (15}}{{\bf{)}}^{{\bf{ - 1}}}}{\bf{\;F}}\\E(t){\bf{ = }}f(t) \cdot A{\bf{ = }}10\cos 10t \cdot 5 \Rightarrow E(t){\bf{ = }}50\cos 10t\;{\bf{V}}\end{aligned}\)

So, a\({\bf{R L C}}\)series circuit that is analog to the given mechanical system is given by\({\bf{L = 35H,10\Omega ,C = (15}}{{\bf{)}}^{{\bf{ - 1}}}}{\bf{\;F,E(t) = 50cos10t\;V}}\).