A lobsterman’s buoy is a solid wooden cylinder of radius r and mass M. It is weighted at one end so that it floats upright in calm sea water, having density. A passing shark tugs on the slack rope mooring the buoy to a lobster trap, pulling the buoy down a distance x from its equilibrium position and releasing it. Show that the buoy will execute simple harmonic motion if the resistive effects of the water are neglected, and determine the period of the oscillations.
I know that the period is T=2*pie*square root(M/K)
To show that the buoy will execute simple harmonic motion, we need to demonstrate that the restoring force acting on the buoy is proportional to the displacement from its equilibrium position and directed opposite to the displacement.
Let's consider the forces acting on the buoy when it is displaced downward:
1. The weight of the buoy acts downward and can be represented as W = Mg, where M is the mass of the buoy and g is the acceleration due to gravity.
2. The buoyancy force acts upward and is equal to the weight of the water displaced by the buoy. Since the buoy is floating upright, the buoyancy force balances the weight of the buoy when it is in equilibrium. However, when the buoy is displaced downward, the buoyancy force decreases slightly because the volume of water displaced decreases.
3. The tension in the rope attached to the lobster trap also acts upward and increases as the buoy is displaced downward. This tension force provides the restoring force to bring the buoy back to its equilibrium position.
At small displacements, we can approximate the change in buoyancy force and the tension force as linearly dependent on the displacement x. Therefore, we can write the magnitude of the total restoring force as:
F = (M - Δm)g + kx
Where Δm represents the reduction in the buoyancy force due to the displacement and k is the spring constant representing the proportionality between the tension force and displacement.
Now, since Δm is small, we can neglect it and simplify the equation to:
F = Mg + kx
Since F = m*a (mass times acceleration) and the acceleration is the second derivative of displacement with respect to time (a = d²x/dt²), we can rewrite the expression as a differential equation for simple harmonic motion:
Mg + kx = M(d²x/dt²)
Rearranging the terms, we get:
d²x/dt² + (k/M)x = g
This is a second-order linear homogeneous differential equation known as the equation of simple harmonic motion. Its solution will be a sinusoidal function, as is the case for simple harmonic motion.
To find the period of the oscillations, we can use the equation you mentioned: T = 2π√(M/k).