Answer
The proof is below.
Work Step by Step
Equation 13.3 is:
$m\frac{d^2x}{dt^2}=-kx$
If the given equation is a solution to equation 13.3, it will result in a true statement when plugged into equation 13.3. Thus, we find:
$m\frac{d^2x}{dt^2}=-kx$
$m\frac{d^2(Asin\omega t)}{dt^2}=-kAsin\omega t$
$-mA\omega^2=-kA$
We know that $\omega^2 = \frac{k}{m}$, so we find:
$-mA \frac{k}{m}=-kA$
$-kA=-kA$
This is true, so we see that the given equation is a solution to equation 13.3.
