Question

Show that y=A sin (wt-kx) satisfies the equation d^2y/dc^2=1/v^2(d^2y/dt^2)

Answer 1

$$\frac{dy^2}{dx^2}=-Ak^2 \sin (wt-kx)\\ \frac{dy^2}{dt^2}=-Aw^2 \sin (wt-kx)\\v=\frac{w}{k}$$

Thus,

$$\frac{dy^2}{dx^2}=\frac{1}{v^2} \frac{dy^2}{dt^2}$$