## Essential University Physics: Volume 1 (3rd Edition)

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.