How does $\zeta=A\cos(\omega\cdot t-kx)$ satisfy this equation:

Question

How does $\zeta=A\cos(\omega\cdot t-kx)$ satisfy this equation:

$$\frac{\partial^2\zeta}{\partial x^2}=\frac{1}{c^2}\frac{\partial^2\zeta}{\partial t ^2}$$

Answer 1

As stated, it doesn't obviously. However, if there is a relationship between $\omega$ and $k$, then it can. Note that $$ \frac{\partial^2 \zeta}{\partial x^2} = -Ak^2\cos(\omega t - kx) $$ and $$ \frac{\partial^2 \zeta}{\partial t^2} = -A\omega^2\cos(\omega t - kx) $$ and so the equation $$ -Ak^2\cos(\omega t - kx) = \frac{\partial^2 \zeta}{\partial x^2} = \frac{1}{c^2}\frac{\partial^2 \zeta}{\partial t^2} =-\frac{1}{c^2}A\omega^2\cos(\omega t - kx) $$ is satisfied if $c^2 = \omega^2/k^2$.

Answer 2

Differentiate $\zeta $ twice relative to $x$ and relative to $t$ we get

$$\frac{\partial^2\zeta}{\partial x^2}=k^2\zeta$$ and $$\frac{\partial^2\zeta}{\partial t^2}=\omega^2 \zeta$$ so in order to have the given equation we should have $$\frac{\omega}{c}=k$$