Answer to Question #163866 in Other for James Woatsi

Question

Show that ¥(x,t)= Asin(wt+kx) describes a wave motion

Answer 1

Answer:

Step 1

Given:

ψ(x,t)=Asin(wt+kx)

Step 2

Calculation:

ψ(x,t)=Asin(wt+kx)

ψx,t=Asinwt+kx

We know the wave equation is

$∂^2ψ \over ∂x^2$ = $1 \over v^2$ $∂^2ψ \over ∂t^2$ →(1)

let us find out all the derivatives 

$∂ψ \over ∂x$ =Ak cos(wt+kx)

Again differentiate with respect to x,

$∂^2ψ \over ∂x^2$ = Ak²  sin(wt+kx)(Sinceψ(x,t)=Asin(wt+kx)

$∂^2ψ \over ∂x^2$ =−k²ψ(x,t)→(2)

Step 3

Now,Let's find out time derivatives 

$∂ψ \over ∂t$ =Aw cos(wt+kx)

Again differentiate with respect to x,

$∂^2ψ \over ∂t^2$ =−Aw² sin(wt+kx) (Sinceψ(x,t)=Asin(wt+kx)

$∂^2ψ \over ∂t^2$ = −w² ψ(x,t)→(3)

Put the values in(1) equation (2) & (3)

−k² ψ(x,t)= $1 \over v^2$ −w² ψ(x,t)

Step 4

v² ψ(x,t)= $w^2 \over k^2$ ψ(x,t)

(Since one dimensional wave equationv= $w \over k$ and v² = $w^2 \over k^2$ )

v² ψ(x,t)=v² ψ(x,t)

Hence it is proved.