Plane waves and phase velocity

What "plane wave" means in 1D. At any instant, every point sharing the same $x$-coordinate shares the same phase — there is no dependence on any transverse direction. In 3D these surfaces of constant phase become literal planes perpendicular to the travel direction; in the 1D string/sound picture of this chapter it simply means the disturbance is a function of $x$ and $t$ alone.

Reading off $k$ and $\omega$. A harmonic plane wave is written

where $A$ is the amplitude, $k$ is the wavenumber, $\omega$ is the angular frequency, and $\phi$ is a constant phase offset. The wavenumber and angular frequency connect directly to the more familiar wavelength and period through

where: $\lambda$ is the wavelength (distance over which the pattern repeats at a fixed instant), $f$ is the frequency, and $T_{\text{period}}=1/f$ is the time over which the oscillation at a fixed point repeats.

Phase velocity. The phase of the wave is $\Theta(x,t)=kx-\omega t+\phi$. Following one particular crest means following a point of constant phase, $d\Theta = 0$, which fixes how fast that crest travels — this is worked out fully below.

Energy scales as $A^2$. Every element of the medium oscillates like an SHM oscillator, so its energy goes as $\omega^2A^2$ (recall $E=\tfrac12 m\omega_0^2A^2$). A travelling wave carries this oscillatory energy along at speed $v_p$, so the intensity — power transported per unit time — obeys $I\propto A^2$: doubling the amplitude quadruples the transported power, at fixed frequency and medium.

A harmonic wave on a string is given by $y(x,t) = 0.05\cos(4\pi x - 800\pi t)$ (SI units).

Step 1. Read off the constants by comparing with $A\cos(kx-\omega t)$: $A = 0.05\ \text{m}$, $k = 4\pi\ \text{rad/m}$, $\omega = 800\pi\ \text{rad/s}$.

Step 2. Find wavelength, frequency, and period:

Step 3. Compute the phase velocity two independent ways as a check:

Step 4. Now suppose a second wave travels on the identical string with double the amplitude, $A' = 0.10\ \text{m}$, at the same frequency. Since $I\propto A^2$, the transported power scales as $(A'/A)^2 = 4$, so the intensity quadruples — but the phase velocity stays at $200\ \text{m/s}$, because $v_p$ is set by the string's tension and mass per length, not by how large the oscillation is. The wave's speed is unchanged at $200\ \text{m/s}$; only the transported power, which quadruples, depends on amplitude.

Competitive stretch

The same wave now crosses into a second identical-looking string segment where the phase velocity is measured to be $v_p' = 250\ \text{m/s}$ instead. Since the source still drives the string at the same $f=400\ \text{Hz}$, the wavelength in the new segment must adjust: $\lambda' = v_p'/f = 250/400 = 0.625\ \text{m}$, and correspondingly $k' = 2\pi/\lambda' \approx 10.05\ \text{rad/m}$ — larger $v_p$ means a longer wavelength and smaller wavenumber at fixed frequency.

• Confusing the wavenumber $k=2\pi/\lambda$ (rad/m) with plain spatial frequency $1/\lambda$ (cycles/m) — check for the missing factor of $2\pi$ before using $v_p=\omega/k$.
• Believing phase velocity depends on amplitude — for the media in this chapter $v_p$ is fixed purely by the medium ($T/\mu$ or $B/\rho$); amplitude only changes energy and intensity, never speed.
• Assuming wavelength stays fixed across a medium boundary — it is the frequency that stays fixed (set by the driving source), while $\lambda$ and $k$ adjust to satisfy $v_p=f\lambda$ with the new speed.
• Writing $I\propto A$ instead of $I\propto A^2$ — wave intensity goes as amplitude squared, exactly like SHM energy $E=\tfrac12 m\omega_0^2A^2$.

Ask the tutor: Want more practice converting between $k$, $\omega$, $\lambda$, and $f$, or a worked example on waves crossing between two different strings? Ask the AI tutor for extra problems.