30 May 2025

Standing Waves, Phase and Group Velocities, and Energy Transfer

Fixed and free string boundaries, phase and group velocities, stationary-wave kinematics, and vibrating-string energy transfer.

waves-and-optics standing-waves vibrating-string energy-transfer phase-velocity group-velocity

Formation and boundary phase

Superpose equal waves travelling in opposite directions:

\[y_+=A\sin(kx-\omega t), \qquad y_-=A\sin(kx+\omega t).\]

The identity $\sin p+\sin q=2\sin[(p+q)/2]\cos[(p-q)/2]$ gives

\[\boxed{y(x,t)=2A\sin kx\cos\omega t}.\]

There is no factor of the form $kx\mp\omega t$, so the pattern does not travel. Reflection at an ideal fixed end reverses displacement and adds phase $\pi$, producing a node. At an ideal free end the reflected displacement has no phase reversal, producing an antinode.

At a fixed end, $y=0$. At a free end, the transverse force must vanish. For a small slope that force is $-T\,\partial y/\partial x$, so the boundary condition is

\[\boxed{y=0\quad\text{(fixed end)}}, \qquad \boxed{\frac{\partial y}{\partial x}=0\quad\text{(free end)}}.\]

Fixed-fixed, free-free, and fixed-free strings

Let the string occupy $0\le x\le L$.

For two fixed ends, use $y=X(x)\cos\omega t$. The condition $X(0)=0$ selects $X=C\sin kx$, while $X(L)=0$ requires

\[\sin kL=0 \quad\Longrightarrow\quad k_n=\frac{n\pi}{L}.\]

Since $\omega_n=vk_n$,

\[\boxed{y_n=C_n\sin\left(\frac{n\pi x}{L}\right)\cos(\omega_nt+\phi_n)}, \qquad \boxed{f_n=\frac{nv}{2L}}, \quad n=1,2,3,\ldots\]

For two free ends, $X’(0)=0$ selects $X=C\cos kx$, and $X’(L)=0$ again gives $k_n=n\pi/L$. The $n=0$ solution is a rigid translation with zero frequency; the vibrational modes have $n\ge1$ and the same frequencies $nv/(2L)$.

For a fixed end at $x=0$ and a free end at $x=L$, $X=C\sin kx$ and

\[X'(L)=Ck\cos kL=0.\]

Therefore

\[k_nL=\frac{(2n-1)\pi}{2}, \qquad \boxed{f_n=\frac{(2n-1)v}{4L}}, \quad n=1,2,3,\ldots\]

Only odd multiples of the fundamental occur in the fixed-free case.

Changes with position and time

For $y=2A\sin kx\cos\omega t$, fixing $t$ gives the spatial sinusoid $2A\cos\omega t\sin kx$. Its nodes and antinodes are

\[\boxed{x_{\rm node}=\frac{n\pi}{k}=\frac{n\lambda}{2}}, \qquad \boxed{x_{\rm antinode}=\frac{(2n+1)\pi}{2k} =\frac{(2n+1)\lambda}{4}}.\]

Fixing $x$ instead gives simple harmonic motion of signed amplitude $2A\sin kx$:

\[u_y=\frac{\partial y}{\partial t} =-2A\omega\sin kx\sin\omega t, \qquad \frac{\partial^2y}{\partial t^2}=-\omega^2y.\]

All points within one loop are in phase. Since $\sin kx$ changes sign across a node, neighboring loops differ in phase by $\pi$. When $\cos\omega t=0$, the whole string passes through equilibrium and particle speeds are greatest; at $\cos\omega t=\pm1$, displacement is extremal and all particle speeds vanish.

Energy density and transfer on a string

For linear density $\mu$, tension $T$, and small slope, the kinetic and elastic potential energies per unit length are

\[\boxed{u_K=\frac12\mu y_t^2}, \qquad \boxed{u_U=\frac12T y_x^2}.\]

For the progressive wave $y=A\cos(kx-\omega t)$,

\[y_t=A\omega\sin(kx-\omega t), \qquad y_x=-Ak\sin(kx-\omega t).\]

Because $v^2=T/\mu$ and $\omega=vk$, $Tk^2=\mu\omega^2$. Hence

\[u_K=u_U=\frac12\mu A^2\omega^2\sin^2(kx-\omega t),\]

and the time-averaged total energy density is

\[\boxed{\langle u\rangle =\frac12\mu A^2\omega^2}\qquad({\rm J\,m^{-1}}).\]

The transverse force does work across a section of string at the rate

\[\boxed{P=-T y_x y_t}.\]

For the right-moving wave this becomes $P=Tk\omega A^2\sin^2(kx-\omega t)$, so

\[\boxed{\langle P\rangle =\frac12Tk\omega A^2 =\frac12\mu v\omega^2A^2}\qquad({\rm W}).\]

For the standing wave,

\[u_K=2\mu A^2\omega^2\sin^2kx\sin^2\omega t,\] \[u_U=2Tk^2A^2\cos^2kx\cos^2\omega t,\]

and

\[P=4TA^2k\omega\sin kx\cos kx\sin\omega t\cos\omega t.\]

Thus energy alternates locally between kinetic and elastic forms, but

\[\boxed{\langle P\rangle_t=0}.\]

A perfect standing wave therefore has no net time-averaged energy transfer.

Phase and group velocities

For a component $\cos(kx-\omega t)$, constant phase gives

\[\boxed{v_p=\frac{\omega}{k}}.\]

Now add two nearby components:

\[y=\cos(k_1x-\omega_1t)+\cos(k_2x-\omega_2t).\]

With $\bar k=(k_1+k_2)/2$, $\Delta k=k_1-k_2$, $\bar\omega=(\omega_1+\omega_2)/2$, and $\Delta\omega=\omega_1-\omega_2$,

\[y=2\cos\left(\frac{\Delta k\,x-\Delta\omega\,t}{2}\right) \cos(\bar kx-\bar\omega t).\]

The carrier phase travels at approximately $\bar\omega/\bar k$. A point of constant envelope phase satisfies $\Delta k\,x-\Delta\omega\,t={\rm constant}$, so

\[v_{\rm env}=\frac{\Delta\omega}{\Delta k}.\]

For a narrow packet, take the limit:

\[\boxed{v_g=\frac{d\omega}{dk}}.\]

In a nondispersive medium $\omega=vk$, hence $v_p=v_g=v$. In a dispersive medium the two velocities need not be equal.

The string panels satisfy the displayed boundary conditions; the packet separates carrier phase from envelope motion.

The boundary spectra, energy identities, and carrier-envelope algebra are checked in the Unit I Maxima worksheet.