Confuse about phase lead and lag of a wave.

时间：03-26 整理：3721RD 点击：

A sine wave traveling in +x direction can be represented by

$3S\cos(\omega{t}-kx+\phi)$

.

As shown in the drawing attached where

$3S\phi=\frac{\pi}{2}$

:

a) Is a plot that holds t=0 and two waves along +x direction.

$3St=0\Rightarrow\;\cos(\omega{t}-kx+\phi)\;=\;\cos(-kx+\frac{\pi}{2})$

This implies maximum at

$3S-kx+\frac{\pi}{2}\;=\;0$

$3Sk=\frac{2\pi}{\lambda}\;\Rightarrow\;\frac{2\pi}{\lambda}x=\frac{\pi}{2}\;\Rightarrow\;x=\frac{\lambda}{4}$

This gives the wave form in RED that LAGs the [itex]\cos(\omega{t}-kx)[/itex].

b) Is a plot of the waveform at x=0 and the two waves vary with time t.

$3Sx=0\;\Rightarrow\;\cos(\omega{t}-kx+\phi)\;=\;\cos(\omega{t}+\frac{\pi}{2})$

This implies maximum at

$3S\omega{t}=-\frac{\pi}{2}$

$3S\omega{t}\;=\;\frac{2\pi}{T}t\;=\;-\frac{\pi}{2}\;\Rightarrow\;t=-\frac{T}{4}$

This means waveform in RED LEADs the original wave.

So the two cases give opposite result. Can anyone explain this to me?
Thanks

Alan

All depends what direction you choose in the plot. If you start with the X origin going LEFT with time for the second wave, the first being reference, then the red in your figure LAGS the black, in the second, it LEADS the black.

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