A320 kgwooden raft floats on a lake. When a 68 kg man stands on the raft, it sinks 3.5 cm deeper into the water. When he steps off, the raft oscillates for a while. (a) What is the frequency of oscillation? (b) What is the total energy of oscillation (ignoring damping)?

In this problem, the total energy of the oscillation is equivalent to the elastic potential energy. The gravitational potential energy is ignored because the displacement is estimated from the equilibrium position.

The mass of the wooden raft is\({\rm{M}} = 320\;{\rm{kg}}\).

The mass of the man is\({\rm{m}} = 68\;{\rm{kg}}\).

The height up to which raft will sink is \({\rm{x}} = 3.5\;{\rm{cm}}\).

When wooden raft is placed on the water then it will be in equilibrium. When man is placed on the raft then it will sink. Now, again, when the man steps off the raft, then it will try to become stable. Thus, mass of the man will act as the restoring force in the oscillation.

The relation to calculate spring constant is given by,

\(\begin{aligned}{F_0} &= {F_{\rm{g}}}\\kx &= mg\\k &= \frac{{mg}}{x}\end{aligned}\)

Here, g is the gravitational acceleration,\({F_0}\)is the spring force and\({F_{\rm{g}}}\)is the force due to gravity.

Substitute the values in the above relation.

\(\begin{aligned}k &= \frac{{\left( {68\;{\rm{kg}}} \right)\left( {9.8\;{\rm{m/}}{{\rm{s}}^2}} \right)}}{{\left( {3.5 \times {{10}^{ - 2}}\;{\rm{m}}} \right)}}\\k &= 1.9 \times {10^4}\;{\rm{N/m}}\end{aligned}\)

The frequencyis calculated as:

\(\begin{aligned}f &= \frac{1}{{2{\rm{\pi }}}}\sqrt {\frac{k}{M}} \\ &= \frac{1}{{2{\rm{\pi}}}}\sqrt{\frac{{1.9\times{{10}^4}\;{\rm{N/m}}}}{{320\;{\rm{kg}}}}} \\ &= 1.2\;{\rm{Hz}}\end{aligned}\)

Hence, the frequency of the oscillations is1.2 Hz.

If vertical gravitational potential energy is ignored, then total energy will be elastic potential energy.

The relation of elastic potential energy is given by,

\(E = \frac{1}{2}k{x^2}\)

Substitute the values in the above relation.

\(\begin{aligned}E &= \frac{1}{2}\left( {1.9 \times {{10}^4}\;{\rm{N/m}}} \right){\left( {3.5 \times {{10}^{ - 2}}\;{\rm{m}}} \right)^2}\\E &= 11.6\;{\rm{J}}\end{aligned}\)

Hence, the total energy of the system is 11.6 J.