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FINDING SOLUTIONS

241

The result, when cleared down, is:

${\begin{aligned}\int \epsilon ^{{\frac {a}{b}}t}\cdot \sin 2\pi nt\cdot dt\\&=\epsilon ^{{\frac {a}{b}}t}\left\{{\frac {ab\cdot \sin 2\pi nt-2\pi nb^{2}\cdot \cos 2\pi nt}{a^{2}+4\pi ^{2}n^{2}b^{2}}}\right\}.....{\text{[D]}}&\\\end{aligned}}$

Inserting this value in [A], we get

$y=g\left\{{\frac {a\cdot \sin 2\pi nt-2\pi nb\cdot \cos 2\pi nt}{a^{2}+4\pi ^{2}n^{2}b^{2}}}\right\}$ .

To simplify still further, let us imagine an angle $\phi$ such that $\tan \phi ={\dfrac {2\pi nb}{a}}$ . Then

$\sin \phi ={\frac {2\pi nb}{\sqrt {a^{2}+4\pi ^{2}n^{2}b^{2}}}}$ ,

and

$\cos \phi ={\frac {a}{\sqrt {a^{2}+4\pi ^{2}n^{2}b^{2}}}}$ .

Substituting these, we get:

$y=g{\frac {\cos \phi \cdot \sin 2\pi nt-\sin \phi \cdot \cos 2\pi nt}{\sqrt {a^{2}+4\pi ^{2}n^{2}b^{2}}}}$ ,

which may be written

$y=g{\frac {\sin(2\pi nt-\phi )}{\sqrt {a^{2}+4\pi ^{2}n^{2}b^{2}}}}$ ,

which is the solution desired.

This is indeed none other than the equation of an alternating electric current, where $g$ represents the amplitude of the electromotive force, $n$ the frequency, a the resistance, $b$ the coefficient of self-induction of the circuit, and $\phi$ is an angle of lag.

C.M.E.

Q

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