Question #163866

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

Expert's answer

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

= **→(1)**

let us find out all the derivatives

=Ak cos(wt+kx)

Again differentiate with respect to x,

= Ak^{2} sin(wt+kx)(Sinceψ(x,t)=Asin(wt+kx)

=−k^{2}ψ(x,t)→(2)

**Step 3**

Now,Let's find out time derivatives

=Aw cos(wt+kx)

Again differentiate with respect to x,

=−Aw^{2 }sin(wt+kx) (Sinceψ(x,t)=Asin(wt+kx)

= −w^{2 }ψ(x,t)→(3)

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

−k^{2} ψ(x,t)= −w^{2} ψ(x,t)

Step 4

v^{2} ψ(x,t)= ψ(x,t)

** (**Since one dimensional wave equationv= and v^{2 }= )

v^{2} ψ(x,t)=v^{2} ψ(x,t)

Hence it is proved.

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