Exercise 1: A straight-sided boat

For a boat with vertical sides, the equation of motion can be derived by starting with Newton’s second law:

(12)ΣF=mgρgA(yo+y)=may

where A is the area of the boat’s cross-section at the waterline. yo is the equilibrium depth, so mg=ρgAyoand

(13)ay=d2ydt2=ρgAmy

This is in the form of the equation for SHO, with

(14)ω=ρgAm

The period of the boat’s oscillation will be

(15)T=2πmρgA

Exercise 2: A V-hulled boat

The derivation for the V-hulled boat follows the same general procedure.

(16)V=12βh2L=12βL(yo+y)2
(17)ΣF=md2ydt2=mg12ρgβL(yo2+2yoy+y2)

The equilibrium position is given by yo, so mg12ρgβLyo2=0 and

(18)d2ydt2=ρgβL2m(2yoy+y2)

This can be further simplified as

(19)d2ydt2=ρgβLyom(1+y2yo)y

From that equation we can see that if y2yo is small, then the motion is approximately SHO with

(20)ωo=ρgβLyom

There are an awful lot of constants in that equation, so to simplify our numeric calculations let’s just redefine things.

(21)d2ydt2=ωo2(1+y2yo)y

We can also define the unitless variable γyyo, which allows us to rewrite the equation in (almost) unit-independent form:

(22)d2γdt2=ωo2(1+γ2)γ

That’s the equation we need to send through the ODE solver.

The equation is approximately SHO if the amplitude is small. The easist way to control the amplitude is to set initial position to yi, with vi=0, so that the amplitude is yi.

Setting different values of amplitude gives different behavior: for small amplitude (γo=0.1) the result is nearly indistinguishable from SHO, but for larger values both the period and symmetry change. The figure below shows that at γo=0.5, the boat spends more time higher (the graph is inverted, since down is positive in our initial setup) and the period lengthens relative to the SHO approximation.


Solution to the V-hull boat oscillation

These effects are even more prominent at larger values of γo, as shown in the next figure.

Solution to the V-hull boat oscillation at large amplitude

There’s an opportunity here for the model to break down. If the boat gets completely out of the water, then the width of the boat becomes negative (ok, it doesn’t really but it does in the model) so then the volume becomes negative and the buoyant force becomes negative and the boat leaps away from the water exactly the way that real boats don’t. Be prepared to discuss this interesting result should the students get the amplitude too high.

Adding damping is relatively easy: just add another force term in the definition of the ODE. Results are shown below.

Solution to the V-hull boat oscillation with damping