Answer to Question #89509 in Physics for Bianca

Question

Does y = sin (kx − ωt) satisfies the wave equation? Justify your answer

Answer 1

The wave equation:

$v^2 \frac{\partial^2 y}{\partial x^2}=\frac{\partial^2 y}{\partial t^2}$

We have:

$\frac{\partial y}{\partial x}=k \cos{(kx − ωt)}$

$\frac{\partial^2 y}{\partial x^2}=-k^2 \sin (kx − ωt)$

$\frac{\partial y}{\partial t}=-\omega \cos{(kx − ωt)}$

$\frac{\partial^2 y}{\partial t^2}=-\omega^2 \sin (kx − ωt)$

Thus,

$v^2 (-k^2 \sin (kx − ωt) )=-\omega^2 \sin (kx − ωt)$

We have the wave equation with

$v=\frac{\omega}{k}$