Question

Estimate the average power of a water wave when it hits the chest of an adult male standing in the water

at the seashore. Assume that the amplitude of the wave is 0.50 m, the wavelength is 2.5 m, and the

period is 4.0 s.

Answer 1

The power of the wave is given as follows (see http://large.stanford.edu/courses/2010/ph240/bonifacio1/):

$$P = \dfrac{\rho g^2 T H^2}{32\pi}$$

where $\rho \approx 1000kg/m^3$ is the density of water, $g= 9.81m/s^2$ is the gravitational acceleration, $T = 4.0s$ is the period, and $H = 0.50m$ is the amplitude. Thus, obtain:

$$P = \dfrac{1000\cdot 9.81^2 \cdot 4\cdot 0.5^2}{32\pi} \approx 9.6\times 10^2W$$

Answer. $9.6\times 10^2W$.