# Standing Wave problem-In deep water limit $h\to\infty$ show that $\omega^2=gk$

The equations I have are

$\phi=(-ag/\omega)\cos(kx)\sin(\omega t)e^{kz}$

and

$\eta=a\cos(kx)\cos(\omega t)$

I know that $d\phi/dz=d\eta/dt$

but when I partially differentiate and rearrange I get

$gk e^{kz}= \omega^2$ and I don't know how to get rid of the exponential function.

-
 How does $h$ enter your equations? Do the parameters depend on $h$? – Johan Nov 26 '12 at 13:23 The original equation is $\phi=-ag/\omega cos(kx)sin(\omega t) (cosh(k[z+h])/ cosh(kH))$ but as h tends to inifity $\phi$ simplifies to the equation in the question. – Adam Nov 26 '12 at 13:37 Do you know why the original equation simplifies as h tends to infinity? How do I show that the original equations simplifies as h tends to infinity? – Becka Dec 8 '12 at 23:41