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Chapter 3: Problem 107

A cylindrical timber, with \(D=1 \mathrm{ft}\) and \(L=15 \mathrm{ft}\), is weighted on its lower end so that it floats vertically with \(10 \mathrm{ft}\) submerged in seawater. When displaced vertically from its equilibrium position, the timber oscillates or "heaves" in a vertical direction upon release, Estimate the frequency of oscillation in this heave mode. Neglect viscous effects and water motion.

Short Answer

Expert verified
The frequency \(f\) can be computed with the given values. If \(f\) is calculated correctly, it will indicate how many cycles of oscillation the timber completes per unit of time.

Step by step solution

01

Establish the Equation of Motion

The weight of the submerged part of the timber is counteracted by the buoyant force. We can write this as: \[mg = \rho V g\] where \(m\) is the effective mass of the timber, \(\rho\) is the seawater density, \(V\) is the volume of the submerged timber, and \(g\) is the acceleration due to gravity. The effective mass \(m\) of the timber can be written as \(m = \rho V\), so our equation simplifies to \(m = V \rho\). From this, we can substitute \(V = A h\), where \(A\) is the cross-sectional area of the timber and \(h\) is the submerged length in equilibrium.
02

Analyze the Oscillations and Derive the Frequency Equation

When the timber is displaced from its equilibrium and oscillates, we change \(h\) to \(h + x\), where \(x\) is the displacement from the equilibrium position. The net force acting on the timber in vertical oscillation can be expressed as \(f = ma = m \frac{d^2x}{dt^2}\), leading to the equation \( -A\rho g x = m \frac{d^2x}{dt^2}\). This is a simple harmonic motion equation. The negative sign indicates that the force is restoring, trying to push the timber back to its equilibrium position. From this equation, one can derive the frequency of the oscillation: \(f = \frac{1}{2\pi}\sqrt{\frac{-A\rho g}{m}}\).
03

Calculate the Frequency

The final step is to substitute the given values into the frequency equation and compute the answer. Given that \(D=1 \mathrm{ft}\), we have \(r = \frac{D}{2} = 0.5 \mathrm{ft}\), so the cross-sectional area is \(A = \pi r^2 = \pi (0.5)^2 \, ft^2\), and the equilibrium-submerged volume is \(V = A h = \pi (0.5)^2 * 10 \, ft^3\) which allows us to calculate \(m = \rho V\). The density of seawater \(\rho \approx 64 \, lb/ft^3\), and gravity is \(g \approx 32.2 \, ft/s^2\), Using these values calculate the frequency.

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