1505 Energy In Wave Motion

The maximum value of the instantaneous power $P(x, t)$ occurs when the $\sin^2$ function has the value unity:

$P_{max} = \sqrt{\mu F} \omega^2 A^2 \quad \bold{(15.24)}$

The average value of the $\sin^2$ function, averaged over any whole number of cycles is $\dfrac{1}{2}$. Hence we see from Eq. (15.23) that the average power $P_{av}$ is just one-half the maximum instantaneous power $P_{max}$

$P_{av} = \frac{1}{2} \sqrt{\mu F} \omega^2 A^2 \quad \bold{(15.25)}$

Wave intensity

Intensity $I$ is average power per unit area and is usually measured in watts per square meter($W/m^2$).
If waves spread out equally in all directions from a source, the intensity at a distance $r$ from the source is inversely proportional to $r^2$. This result, called the inverse-square law for intensity. follows directly from energy conservation. If the power output of the source is $P$, then the average intensity $I_1$ through a sphere with radius $r_1$ and surface are $4\pi r_1^2$ is

$I_1 = \frac{P}{4\pi r_1^2}$

A similar expression gives the average intensity $I_2$ through a sphere with a different radius $r_2$. If no energy is absorbed between the two spheres, the power $P$ must be the same for both, and

$\begin{aligned} \frac{P}{4\pi r_1^2} I_1 &= \frac{P}{4\pi r_2^2} I_2\\ \frac{I_1}{I_2} &= \frac{r_2^2}{r_1^2} & \bold{(15.26)} \end{aligned}$

Exercises

22, 23

15.22 A piano wire with mass $3.00$ g and length $80.0\text{ cm}$ is stretched with a tension of $25.0\text{ N}$ . A wave with frequency $120.0\text{ Hz}$ and amplitude 1.6 mm travels along the wire. (a) Calculate the average power carried by the wave. (b) What happens to the average power if the wave amplitude is halved?

15.23 A horizontal wire is stretched with a tension of $94.0 \text{ N}$, and the speed of transverse waves for the wire is $406 \text{ m/s}$. What must the amplitude of a traveling wave of frequency $69.0\text{ Hz}$ be for the average power carried by the wave to be $0.365 \text{ W}$?