Answer
The proof is below.
Work Step by Step
We know that the second partial derivative of equation 14.3 with respect to x is:
$-Ak^2cos(kx+\omega t)$
We know that the second partial derivative of equation 14.3 with respect to t is:
$-A \omega^2cos(kx+\omega t)$
We plug these into equation 14.5 to find:
$-A \omega^2cos(kx+\omega t) =\frac{1}{v^2}(-Ak^2cos(kx+\omega t))$
$-A \omega^2=\frac{1}{v^2}(-Ak^2)$
We substitute the given value of v to find:
$-A\omega^2 = -A\omega^2$
Thus, the proof is complete.
