# For derivation of equation for stationary wave, y1=Asin(kx-omega t) and y2=Asin(kx+omega t) are considered. Why y1=Asin (kx-omega t) and y2=Asin(kx+omega t+pi) cannot be used to derive?

May 2, 2018

See below

#### Explanation:

$\sin \left(k x - \omega t\right) + \sin \left(k x + \omega t + \pi\right)$

$= \sin k x \cos \omega t - \sin \omega t \cos k x + \sin k x \cos \left(\omega t + \pi\right) + \sin \left(\omega t + \pi\right) \cos k x$

Now:

$\sin \left(\alpha + \pi\right) = \sin \alpha \cos \pi + \sin \pi \cos \alpha = - \sin \alpha$

$\cos \left(\alpha + \pi\right) = \cos \alpha \cos \pi - \sin \alpha \sin \pi = - \cos \alpha$

So we now have:

$= \sin k x \cos \omega t - \sin \omega t \cos k x - \sin k x \cos \omega t - \sin \omega t \cos k x$

$= - 2 \cos k x \sin \omega t$

Whilst it's spatial and temporal phase may not suit, that certainly defines a stationary wave.