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Stumbled upon this problem lately. Maybe someone could help me clarify some subtleties I do not see?
1. Consider the propagation speed ##c## of periodic surface of gravity waves with wavelength ##\lambda## and amplitude ##a## in water of depth ##H##. Let ##\rho_{a}## and ##\rho_{w}## be the density of air and water. Ignoring viscosity and surface tension, use diminutional analysis to find an expression for the propagation speed ##c##. Since ##\frac{\rho_{a}}{\rho_{w}} \approx 0.001## the density of air is usually ignored. How does your expression simplify in this case? What about in the small amplitude limit ##a \rightarrow 0##?
1. I expressed ##c## as follows:
##c=f(H, \rho_{a}, \rho_{w}, t, a, \lambda)##, where ##H## is the depth of water, ##t## is time.
I used the ##Buckingham-\pi## Theorem.
2. I non-dimensionalized ##\pi_1## as follows: ##\pi_1 = \frac{\lambda}{a}##.
3. Now, ##c^* = f_1(\rho_{a}^*, \rho_{w}^*, t, \frac{\lambda}{a})## - I divided by ##a## which has units of length. So now we have the following dimensions in the variables of ##f_1##:
##[\rho_{a}^*] = ML^{-4}##
##[\rho_{w}^*] = ML^{-4}##
##[t] = T##
And ##\frac{\lambda}{a}## is dimensionless.
Where M is in kilograms, L is in meters, T is in seconds.
So, ##\pi_2 = c^{*\alpha}\rho_a^{*\beta}\rho_w^{*\gamma} t^{\delta}## (is this correct?)
4. Now, as per the ##Buckingham-\pi## Theorem, we have
##[\pi_2]=T^{-\alpha+\delta} M^{\beta + \gamma} L^{-4\beta - 4\gamma}##
So,
##-\alpha + \delta = 0##
##\beta + \gamma = 0##
##-4\beta - 4\gamma = 0##
Thus ##\alpha = \delta##, ##\beta = -\gamma##. And now we have:
##\pi_2 = (ct)^{\alpha}(\frac{\rho_{a}}{\rho_{w}})^{\beta}##.
So we could express the dimensionless equation as
##F(\frac{\lambda}{a}) = ct(\frac{\rho_{a}}{\rho_{w}})^{\beta}##. (I omitted the *'s).
I think there is or are some errors in my deduction because if we use the same method that I described but eliminate the density of air or the amplitude of waves, we get ##{\pi_2=1}##.
Thank you for your attention!
1. Consider the propagation speed ##c## of periodic surface of gravity waves with wavelength ##\lambda## and amplitude ##a## in water of depth ##H##. Let ##\rho_{a}## and ##\rho_{w}## be the density of air and water. Ignoring viscosity and surface tension, use diminutional analysis to find an expression for the propagation speed ##c##. Since ##\frac{\rho_{a}}{\rho_{w}} \approx 0.001## the density of air is usually ignored. How does your expression simplify in this case? What about in the small amplitude limit ##a \rightarrow 0##?
Homework Equations
: [/B]noneThe Attempt at a Solution
:[/B]1. I expressed ##c## as follows:
##c=f(H, \rho_{a}, \rho_{w}, t, a, \lambda)##, where ##H## is the depth of water, ##t## is time.
I used the ##Buckingham-\pi## Theorem.
2. I non-dimensionalized ##\pi_1## as follows: ##\pi_1 = \frac{\lambda}{a}##.
3. Now, ##c^* = f_1(\rho_{a}^*, \rho_{w}^*, t, \frac{\lambda}{a})## - I divided by ##a## which has units of length. So now we have the following dimensions in the variables of ##f_1##:
##[\rho_{a}^*] = ML^{-4}##
##[\rho_{w}^*] = ML^{-4}##
##[t] = T##
And ##\frac{\lambda}{a}## is dimensionless.
Where M is in kilograms, L is in meters, T is in seconds.
So, ##\pi_2 = c^{*\alpha}\rho_a^{*\beta}\rho_w^{*\gamma} t^{\delta}## (is this correct?)
4. Now, as per the ##Buckingham-\pi## Theorem, we have
##[\pi_2]=T^{-\alpha+\delta} M^{\beta + \gamma} L^{-4\beta - 4\gamma}##
So,
##-\alpha + \delta = 0##
##\beta + \gamma = 0##
##-4\beta - 4\gamma = 0##
Thus ##\alpha = \delta##, ##\beta = -\gamma##. And now we have:
##\pi_2 = (ct)^{\alpha}(\frac{\rho_{a}}{\rho_{w}})^{\beta}##.
So we could express the dimensionless equation as
##F(\frac{\lambda}{a}) = ct(\frac{\rho_{a}}{\rho_{w}})^{\beta}##. (I omitted the *'s).
I think there is or are some errors in my deduction because if we use the same method that I described but eliminate the density of air or the amplitude of waves, we get ##{\pi_2=1}##.
Thank you for your attention!
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