A wave carries energy without carrying matter. Shake one end of a long rope and, a moment later, a bit of rope far away starts moving: energy has traveled down the rope. How much power does a wave deliver, and how does it depend on how hard you shake? You know the wave function $y = A\sin(kx - \omega t)$; the answer follows from it.
🔑 Key idea
A sinusoidal wave on a string with linear mass density $\mu$ carries the average power $$P_{\text{avg}} = \tfrac{1}{2}\,\mu\,\omega^2 A^2\,v.$$ Power grows with the square of the amplitude and the square of the frequency: double $A$ and the power is 4 times as large.
💡 Why it works
Each bit of string of mass $\Delta m = \mu\,\Delta x$ moves up and down in simple harmonic motion with amplitude $A$ and angular frequency $\omega$. A simple harmonic oscillator's total energy equals its maximum kinetic energy, $\tfrac{1}{2}\Delta m\,(A\omega)^2$. So one wavelength of string holds energy $E_\lambda = \tfrac{1}{2}\mu\lambda\,\omega^2A^2$. One wavelength passes any point every period $T$, so the power is $$P = \frac{E_\lambda}{T} = \tfrac{1}{2}\mu\omega^2A^2\,\frac{\lambda}{T} = \tfrac{1}{2}\mu\omega^2A^2\,v.$$ The square comes from kinetic energy: speed squared, and the bit's top speed $A\omega$ is proportional to $A$.
Energy per meter of string is $\tfrac{1}{2}\mu\omega^2A^2$, and that energy moves at speed $v$: power is energy per length times speed, the same idea as a flow rate. For sound and light the same reasoning gives intensity proportional to the amplitude squared.
Worked examples
✏️ Worked example: Power on a rope
A rope with $\mu = 0.25$ kg/m carries waves at 12 m/s. You shake it at 2.0 Hz with amplitude 4.0 cm. What average power do you supply?
Solve for $A$: $$A = \sqrt{\frac{2P}{\mu\omega^2 v}} = \sqrt{\frac{2(2.0)}{(0.0050)(377)^2(126.5)}}.$$
$= \sqrt{4.45 \times 10^{-5}\ \text{m}^2} = 6.7 \times 10^{-3}$ m.
Answer: $A \approx 6.7$ mm
⚠️ Common mistake
Assuming the power grows in proportion to the amplitude. It grows with the square: twice the amplitude means twice the top speed of each bit, so four times the kinetic energy. A quick check: if $A$ goes up by 10%, $P$ goes up by about 21%, since $1.1^2 = 1.21$.
⚠️ Common mistake
Changing the wave speed when the frequency changes. On a given string under a given tension, $v = \sqrt{F_T/\mu}$ does not depend on $f$. A higher frequency gives a shorter wavelength, not a faster wave.
Raise the amplitude and watch the red particle: it covers more distance in the same period, so its top speed and its kinetic energy rise. Doubling $A$ doubles the speed and quadruples the energy.
Practice
Pass with 3 right answers, the last 2 in a row. Each right answer earns 2 XP.