Homework Help: Dispersion relation for a surface wave of a pool of water

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1. Dec 23, 2017

Pushoam

1. The problem statement, all variables and given/known data

2. Relevant equations

3. The attempt at a solution

$v = \frac { \omega } k$

$\omega = \sqrt{ kg \tanh (k) }$

I have no idea to guess the graph.

I put g = 9.8 and tried to calculate $\omega$ for different values of k.

$\omega (0 ) = 0, \omega (30) = 17.15 ~~\omega (20) = 14$

But, these values don’t match any of the graph.

So, what to do now?

Last edited: Dec 23, 2017
2. Dec 23, 2017

You missed a factor of $h$ in $\tanh (kh)$. (Editing: Oh I missed the part $h=1$). One problem with trying to rely on numbers on the axes here is they don't label their units, so you don't know what number to use for $g$. $\\$ A hint that should help: What is the functional behavior of $\tanh(k)$ for large $k$? Thereby, what does $\omega$ vs. $k$ look like for large $k$? $\\$ Also, what is the slope for very small $k$? Is it some constant or is it zero? (Hint: Expand $\sinh(x)$ and $\cosh(x)$ in Taylor series for small $x$). $\\$ You could also try using $g=980$ cm/sec^2 , but for this problem, I think that leads you to an incorrect answer.

Last edited: Dec 23, 2017
3. Dec 23, 2017

Pushoam

I didn’t know the graph of $\tanh ( k )$ and I don’t want to look for it right now. 1st I want to solve this question with the knowledge I have and then I will look to the graph.

$\tanh ( k) = \frac { e^k - e^{-k} } { e^k + e^{-k} }$ .....(.1)

As k tends to infinity, $\tanh ( k)$ tends to 1.

So, for very large k, $\omega \approx \sqrt{ kg}$ .....(2)

So, as k goes to infinity, $\omega$ should go to infinity . .....(3)

(b) and (d) fulfils condition (3).

$\tanh ( k ) = k \frac { 1 + \frac { k^2}{3!} +…}{ 1 + \frac { k^2}{2!} + …}$ .....(4)

So, for small k, $\tanh ( k) \approx k$ .....(5)

$\omega \approx k\sqrt{ g}$ .....(6)

So, $\omega$ goes from $\sqrt{ g}$ k for small k to $\sqrt {gk}$ for large k. This is shown by (b).

So, the correct option is (b).

Is this correct?

4. Dec 23, 2017

Correct=I would pick (b) also. A couple of things to observe: For large $k$, $\omega=\sqrt{g} \sqrt{k}$ so instead of looking like a straight line, the second derivative will be negative. $\\$ Also for small $k$, $d \omega /dk=\sqrt{g}$, rather than 0. $\\$ (d) comes close to working in some ways, but the second derivative in (d) is slightly positive for large $k$.

5. Dec 23, 2017

Pushoam

Thanks.
Is there any insight article in PF, which helps in understanding how to guess a graph without plotting it exactly?

6. Dec 23, 2017

Not that I know of. I thought of one additional feature though that helps to show (b) is correct: For large $k$ (in this case even 18 is large), $\omega(36)/\omega(18)$ needs to be approximately $\sqrt{36/18}=\sqrt{2}$. This would rule out (c).

7. Dec 24, 2017

Pushoam

That's a nice idea to get rid of g.
Then I can have one ratio for small k and another ratio for large k and see which graph meets this ratio. If I got one meeting this ratio, then surely this will be the required graph.
I see that tanh (k) ≈ k is a good approximation for k < 0.5,
tanh (k) ≈ 1 is a good approximation for k > 10.

Now, small k approximation is not useful here due to scale of the graphs. So, one can use large k approximation to eliminate some options.
Thanks for it.

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