Page:Elementary Text-book of Physics (Anthony, 1897).djvu/167

passing its position of equilibrium is $y=a\cos {\left({\frac {2\pi t}{T}}-{\frac {\pi }{2}}\right)},$ since when $t$ is reckoned from the position of equilibrium $\epsilon ={\frac {pi}{2}}\cdot$ Hence $y=a\sin {\frac {2\pi t}{T}}\cdot$ If $v$ represent the velocity of propagation of the wave, the particle at the distance $x$ from the origin will have a displacement equal to that of the particle at $O$ at the instant $t$, at an instant later than $t$ by the time taken for the wave to travel over the distance $x$, or ${\frac {x}{v}}$ seconds. Hence its displacement at the instant $t$ will be the same as that which existed at $O$, ${\frac {x}{v}}$ seconds earlier. But the displacement at $O$, ${\frac {x}{v}}$ seconds earlier, is

$y=a\sin {{\frac {2\pi \left(t-{\frac {x}{v}}\right)}{T}}=a\sin {2\pi \left({\frac {t}{T}}-{\frac {x}{vT}}\right)}}\cdot$

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The quantity $vT$ equals the distance through which the movement is transmitted during the time of one complete vibration of the particle at $O$. Putting this equal to $\lambda$, we have finally

$y=a\sin 2\pi \left({\frac {t}{T}}-{\frac {x}{\lambda }}\right)\cdot$

(58)

Suppose $t=0$, and give to $x$ various values. The corresponding values of $y$ will represent the displacement at that instant of the particle the distance of which from the origin is $x$. For $x=0$,