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

§ 131] 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 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 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 $$x = 0$$,