Let an object of mass $m$ and volume $V$ be dropped in water from height $h$, and $a$ be the amplitude of the wave generated. What is the relation between $a$ and $h$. How many waves are generated? What is the relation between the amplitudes of successive waves? Does it depend on the shape of the particle?
Assume the particle is spherical. What would be the shape of the water that rises creating first wave?
[Physics] Waves on water generated by a falling object
fluid dynamicshomework-and-exercisesnewtonian-mechanicswaves
Best Answer
This is crude.
Maybe there can be an energy approach. Initially the mass has potential energy $T=m g h$. At the point of peak splash-back lets assume all the energy has been transferred to the water with peak potential energy related to the radial wave height function $y(r)=?$. A small volume of water a distance $r$ from impact has differential volume ${\rm d}V = y(r) 2\pi r {\rm d}r$ .
The potential energy of the small volume of water is ${\rm d}T = \rho g \frac{y}{2} {\rm d}V$ where $\rho$ is density of water. The total energy is thus:
$$ T = \int_0^\infty \rho g \frac{y(r)^2}{2} 2\pi r {\rm d} r $$
Putting a nice smooth wave height function of $$y(r) = Y \exp(-\beta\, r) \left(\cos(\kappa\, r) +\frac{\beta}{\kappa} \sin(\kappa\, r)\right)$$ with $Y$ a height coefficient. This has the properties of ${\rm d}y/{\rm d}r=0$ at $r=0$ with $y(0)=Y$.
$$ T = \frac{\pi Y^2 g \rho \left(9 \beta^4+2 \beta^2 \kappa^2+\kappa^4\right)}{8 \beta^2 \left( \beta^2+\kappa^2 \right)^2 } = m g h $$
So wave height should be $$ Y = \propto \sqrt{ \frac{h m}{\rho \pi }} $$