## 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:

Σ F=m g−ρ g A(y o+y)=m a y(12)(12)Σ F=m g−ρ g A(y o+y)=m a y

where A A is the area of the boat’s cross-section at the waterline. y o y o is the equilibrium depth, so m g=ρ g A y o m g=ρ g A y oand

a y=d 2 y d t 2=−ρ g A m y(13)(13)a y=d 2 y d t 2=−ρ g A m y

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

ω=ρ g A m−−−−√(14)(14)ω=ρ g A m

The period of the boat’s oscillation will be

T=2 π m ρ g A−−−−√(15)(15)T=2 π m ρ g A

## Exercise 2: A V-hulled boat

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

V=1 2 β h 2 L=1 2 β L(y o+y)2(16)(16)V=1 2 β h 2 L=1 2 β L(y o+y)2
Σ F=m d 2 y d t 2=m g−1 2 ρ g β L(y 2 o+2 y o y+y 2)(17)(17)Σ F=m d 2 y d t 2=m g−1 2 ρ g β L(y o 2+2 y o y+y 2)

The equilibrium position is given by y o y o, so m g−1 2 ρ g β L y 2 o=0 m g−1 2 ρ g β L y o 2=0 and

d 2 y d t 2=−ρ g β L 2 m(2 y o y+y 2)(18)(18)d 2 y d t 2=−ρ g β L 2 m(2 y o y+y 2)

This can be further simplified as

d 2 y d t 2=−ρ g β L y o m(1+y 2 y o)y(19)(19)d 2 y d t 2=−ρ g β L y o m(1+y 2 y o)y

From that equation we can see that if y 2 y o y 2 y o is small, then the motion is approximately SHO with

ω o=ρ g β L y o m−−−−−−−√(20)(20)ω o=ρ g β L y o m

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

d 2 y d t 2=−ω 2 o(1+y 2 y o)y(21)(21)d 2 y d t 2=−ω o 2(1+y 2 y o)y

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

d 2 γ d t 2=−ω 2 o(1+γ 2)γ(22)(22)d 2 γ d t 2=−ω o 2(1+γ 2)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 y i y i, with v i=0 v i=0, so that the amplitude is y i y i.

Setting different values of amplitude gives different behavior: for small amplitude (γ o=0.1 γ o=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 γ o=0.5 γ o=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.

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

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.