Question #254808

Determine if the function y = A sinkx cosωt is a solution to the wave equation.


1
Expert's answer
2021-10-24T18:23:23-0400

Given:

y(x,t)=Asinkxcosωty(x,t)=A\sin kx\cos \omega t


The wave equation

2yt2v22yx2=0\frac{\partial^2y}{\partial t^2}-v^2\frac{\partial^2y}{\partial x^2}=0


The speed of the wave v=ω/kv=\omega/k

We have


2yt2=ω2Asinkxcosωt\frac{\partial^2y}{\partial t^2}=-\omega^2A\sin kx\cos \omega t

2yx2=k2Asinkxcosωt\frac{\partial^2y}{\partial x^2}=-k^2A\sin kx\cos \omega t

So

ω2Asinkxcosωtv2(k2Asinkxcosωt)=-\omega^2A\sin kx\cos \omega t-v^2(-k^2A\sin kx\cos \omega t)=

k2v2Asinkxcosωt+k2v2Asinkxcosωt=0-k^2v^2A\sin kx\cos \omega t+k^2v^2A\sin kx\cos \omega t=0


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