Page:A Treatise on Electricity and Magnetism - Volume 2.djvu/381

Suppose that we set the magnet swinging by means of a transient current whose value is $Q_{0}$. If, for brevity, we write

${\frac {G}{H}}{\frac {\sqrt {\pi ^{2}+\lambda ^{2}}}{T_{1}}}e^{-{\frac {\lambda }{\pi }}\tan ^{-1}{\frac {\pi }{\lambda }}}=K$,

(19)

then the first elongation

$\theta _{1}=KQ_{0}=a_{1}$ (say).

(19)

The velocity instantaneously communicated to the magnet at starting is

$v_{0}={\frac {MG}{A}}Q_{0}$.

(20)

When it returns through the point of equilibrium in a negative direction its velocity will be

$v_{1}=-ve^{-\lambda }$.

(21)

The next negative elongation will be

$\theta _{2}=-\theta _{1}e^{-\lambda }=b_{1}$.

(22)

When the magnet returns to the point of equilibrium, its velocity will be

$v_{2}=v_{0}e^{-2\lambda }$.

(23)

Now let an instantaneous current, whose total quantity is $-Q$, be transmitted through the coil at the instant when the magnet is at the zero point. It will change the velocity $v_{2}$ into $v_{2}-v$, where

$v={\frac {MG}{A}}Q$.

(24)

If $Q$ is greater than $Q_{0}e^{-2\lambda }$, the new velocity will be negative and equal to

$-{\frac {MG}{A}}(Q-Q_{0}ee^{-2\lambda })$.

The motion of the magnet will thus be reversed, and the next elongation will be negative,

$\theta _{3}=-K(Q-Q_{0}e^{-2\lambda })=c_{1}=-KQ+\theta _{1}e^{-2\lambda }$.

(25)

The magnet is then allowed to come to its positive elongation

$\theta _{4}=-\theta _{3}e^{-\lambda }=d_{1}=e^{-\lambda }(KQ-a_{1}e^{-2\lambda })$,

(26)

and when it again reaches the point of equilibrium a positive current whose quantity is $Q$ is transmitted. This throws the magnet back in the positive direction to the positive elongation

$\theta _{5}=KQ-\theta _{3}e^{-2\lambda }$;

(27)

or, calling this the first elongation of a second series of four,

$a_{2}=KQ(1-e^{-2\lambda })+a_{1}e^{-4\lambda }$.

(28)

Proceeding in this way, by observing two elongations + and −, then sending a positive current and observing two elongations