Answer
See answer.
Work Step by Step
The force of the man’s weight causes the raft to sink by 3.5 cm, and the extra buoyant force is a restoring force. This is just like pulling down on a mass that is hanging from a spring. The effective spring constant is found like this:
$$k=\frac{F}{x}=\frac{(68kg)(9.80m/s^2)}{0.035m}=1.904\times10^4\;N/m$$
a. Find the frequency of oscillation from the effective spring constant and the raft’s mass.
$$f=\frac{1}{2\pi}\sqrt{\frac{k}{m}}=\frac{1}{2\pi}\sqrt{\frac{1.904\times10^4\;N/m }{320kg}}\approx 1.2\;Hz$$
b. For a vertical spring, when the displacement is measured from the oscillator’s equilibrium position, the total energy is all elastic potential energy.
$$E=\frac{1}{2}kA^2=\frac{1}{2}(1.904\times10^4\;N/m)(0.035\;m)^2=12\;J$$
