## Exercise 1: A straight-sided boat

Imagine a boat with vertical sides, floating. There will be two forces on this boat: its weight m g m g, and the buoyant force −ρ g V−ρ g V. (I’ve defined down to be the positive direction, here, so the buoyant force is negative since it’s upwards.) The volume V V in the buoyant force component is the volume of water displaced by the boat: in other words, the volume of boat that is below water level. In equilibrium, the bottom of the boat will be some distance x o x o below the water, so V=A x o V=A x o, where A A is the area of the boat at the waterline.

If we then push the boat down some additional distance x x and let go, the boat will bob up and down.

1. Show that the boat’s vertical motion is SHO. It will probably be helpful to follow the same pattern as for the introduction example: the key is to get the equation in the form of the equation for SHO.
2. What is the period of the boat’s motion?

## Exercise 2: A V-hulled boat

Instead of a straight-sided boat, imagine a boat with a V-shaped hull profile, as shown below.

The width w w of the hull at the waterline depends on the depth below waterline h h:

w=β h(7)(7)w=β h

The volume of water displaced by this boat is the area of the triangle below water level, times the length of the boat:

V=1 2 β h 2 L(8)(8)V=1 2 β h 2 L

1. Show that the vertical motion of this boat is not SHO. Replace h h with (y o+y)(y o+y), where y o y o is the equilibrium depth, and follow the example in the introduction.

2. Rearrange the equation you got in the previous problem to be as close to SHO as possible by putting it in this form:

   d 2 y d t 2=−ω 2 o(1+y 2 y o)y(9)(9)d 2 y d t 2=−ω o 2(1+y 2 y o)y

   In this form, you can see that if y 2 y o y 2 y o is small, the motion is approximately SHO with angular frequency ω o ω o. What is that ω o ω o? What must be small for this approximation to be valid?

3. If y 2 y o y 2 y o is not small, would the period of the boat’s oscillation be larger or smaller than T o=2 π ω o T o=2 π ω o? Use a numeric solution of the equation of motion for the boat to verify your answer.

4. Plot the motion of the boat for various amplitudes. In addition to effects on T T, how else does the motion differ from SHO? It will be helpful to plot both the solution to the differential equation and the SHO approximation,

   y(t)=y m a x cos(ω o t)(10)(10)y(t)=y m a x cos⁡(ω o t)

5. We’ve neglected viscous damping, which is a bad idea in liquids! Redo your calculations, and plots, adding a viscous damping force

   F v=−δ d y d t(11)(11)F v=−δ d y d t

   to your equations.