Question

If a water wave with length L moves with a velocity 𝑣 across a body of water with depth d, as in figure below,

then

𝑣2 = 𝑔𝐿 tanh 2𝜋𝑑, 2𝜋 𝐿

a) If the water is deep, show that 𝑣 ≈ √𝑔𝐿. 2𝜋

b) If the water is shallow, use the Maclaurin series for tanh to show that 𝑣 ≈ √𝑔𝑑. (Thus, in shallow water the velocity of a wave tends to be independent of the length of the wave.)

Answer 1

a)by definition we know that

$tanh(\frac{2\pi d}{L})=\frac{e^{(\frac{2\pi d}{L})}-e^{-(\frac{2\pi d}{L})}}{e^{\frac{2\pi d}{L}}+e^{\frac{2\pi d}{L}}}$

${lim}_{\ d \to\infin}\ \ tanh(\frac{2\pi d}{L})=lim_{\ d \to\infin}\ \frac{e^{(\frac{2\pi d}{L})}-e^{-(\frac{2\pi d}{L})}}{e^{\frac{2\pi d}{L}}+e^{\frac{2\pi d}{L}}}$

since we have a situation of $\frac{\infin}{\infin}$ we use L 'hopital's rule, which yield;

$lim_{d\to\infin} \frac{4\pi e^{4\pi d}}{4\pi e^{4\pi d}}$ but this is just $lim_{d\to\infin}\ 1=1$

thus, we have shown that $lim_{d\to\infin}\ tanh(\frac{2\pi d}{L})=1$

now we have that , $v=\sqrt{(\frac{gL}{2\pi}).tanh(\frac{2\pi d}{L})}=\sqrt{(\frac{gL}{2\pi}).1}$

therefore, $v\approx \sqrt{(\frac{gL}{2\pi}).}$

b)Find the maclaurin series for tanh x

$f(x)=tanh x\ \ \ \ \ f(0)=0$

$f'(x)=sech^{2}x\ \ \ \ \ \ f'(0)=1$

$f''(x)=-2tanhx\ sech^{2}x\ \ \ \ f''(0)=0$

$f'''(x)=(cosh\ 2x-2)sech^{4}x\ \ \ f'''(0)=-2$

$f^{4}(x)=-2(sin(3x)-11sinh\ x)$

$tanh\ x \approx0+x+\frac{0}{2!}x^{2}-\frac{2}{3!}+0$

$\approx x-\frac{1}{3}x^{2}$

letting $x=\frac{2\pi d}{L}$

$tanh \frac{2\pi d}{L}\approx \frac{2\pi d}{L}-\frac{1}{3}(\frac{2\pi d}{L})$

now put it into the equation

$v^{2}=\frac{gL}{2\pi}tanh\frac{2\pi d}{L}$

$\approx\frac{gL}{2\pi}(\frac{2\pi}{L}-\frac{1}{3}(\frac{2\pi d}{L})^{3})$

$\approx g(d-\frac{1}{3}(\frac{2\pi d}{L})^{2})$

in shallow water d and it simplifies to:

$v^2\approx g(d-0)$

$\approx gd$

$v\approx \sqrt{gd}$