Answer
32 watts. (Your answer may vary, because it depends on the assumed area of the chest.)
Work Step by Step
From the given data, the amplitude A = 0.50 m, and the wave speed $v=\frac{2.5\;m}{4.0\;s}=0.625\;m/s$.
Use equation 11–17b for the average power. The density of seawater is found in Table 10–1. Estimate the adult’s chest as a square that is 0.40 meters on a side. (Your answer may vary, because it depends on the area of the chest.)
$$P_{av}=2\pi^2\rho Svf^2A^2$$
$$P_{av}=2\pi^2(1025\;kg/m^3)(0.40\;m)^2(0.625\;m/s)(0.25\;Hz)^2(0.50\;m)^2$$
$$=32\;W$$
