Answer to Question #244255 in Physics for Josh

Question

A 32Okg wooden raft floats on a lake. When a 75kg man stands on the raft, it sinks 3.5cm deeper into the water. When he steps off, the raft oscillates for a while.

i. What is the frequency of oscillation? 

ii. What is the total energy of oscillation (ignoring damping)? 

Answer 1

The additional buoyancy force when the man is standing:

$$F_a=\rho g\Delta V=\rho g(A\Delta x).$$

The angular frequency of oscillations can be expressed as

$$a=\omega^2\Delta x,\\\space\\ \omega=\sqrt\frac a{\Delta x}.$$

The restoring force that appears when the man is standing is, according to Newton's second law:

$$F_r=F_a,\\ F_r=Ma,\\\space\\ a=\frac{F_r}{M}=\frac{F_a}{M},\\\space\\ \omega=\sqrt{\frac{F_a}{M\Delta x}}=\sqrt{\frac{\rho gA\Delta x}{M\Delta x}}=\sqrt{\frac{\rho gA}{M}}.$$

Find the area from the equilibrium of forces for the wood floating alone and with the man standing on it:

$$Mg=\rho gAh,\\ (M+m)g=\rho gA(h+\Delta x),$$

divide one by another:

$$\frac{Mg}{(M+m)g-\rho gA\Delta x}=1,\\\space\\ \rho gA\Delta x=mg,\\\space\\ \rho gA=\frac{mg}{\Delta x}.$$

Finally,

$$f=\frac\omega{2\pi}=\frac{1}{2\pi}\sqrt{\frac{\rho gA}{M}}=\frac{1}{2\pi}\sqrt{\frac{mg}{M\Delta x}}\\\space\\ f=\frac{1}{2\pi}\sqrt{\frac{75·9.8}{320·0.035}}=1.29\text{ Hz}.$$

The total energy is

$$E=\frac 12Mv^2=\frac12 M(\omega\Delta x)^2=\frac12 M(2\pi f\Delta x)^2=\\=12.9\text{ J}.$$