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The string speed is $c=\sqrt{T/\rho }$. The characteristic impedance is the ratio of the forward-wave transverse force to transverse velocity, $Z=T/c=\sqrt{T\rho }$. With peak displacement amplitude $A$, the mean power isDisplacement and transverse force are continuous at a massless junction. Since $Z_2=2Z_1$, the displacement amplitudes are $r=(Z_1-Z_2)/(Z_1+Z_2)=-1/3$ and $t=2Z_1/(Z_1+Z_2)=2/3$. Taking the incident amplitude real, the light-string motion isIn this standing-and-travelling decomposition of a reflected string wave, the first term travels and the second stands; the decomposition is not a separation into independent incident-power fractions because these fields interfere. The reflected wave has a phase reversal. power coefficients and standing-wave ratio are

For a uniform inserted matching section, unequal exterior impedances require a quarter-wave impedance transformer: $Z_M=\sqrt{Z_1{Z}_2}$ and $k_{M}a=(2j+1)\pi /2$. At equal tension this means middle density $2{\rho }_0$ and an odd-quarter-wavelength length in that section. Reflections at its two ends then cancel at the design frequency. An arbitrary density or an arbitrary half-wavelength segment cannot perfectly match these unequal exterior strings; the ideal match is frequency-selective.

amplitude scaling along a slowly tapered string instead follows from matching over an adiabatic frequency range. The source fixes frequency, while conserved power and the final characteristic impedance determine amplitude. In the heavy string,Its wave speed halves. The no-reflection approximation needs small fractional characteristic impedance change over each local wavelength, rather than length alone for a taper containing abrupt features.

Put $\delta =k_{M}l$. Eliminating the middle-segment complex amplitudes using the right-hand boundary conditions givesHence the displacement reflection coefficient isFor finite positive real mechanical impedances, its denominator cannot vanish at real $\delta$. The real and imaginary parts of the numerator must therefore vanish separately:This gives all three possibilities.

If the exterior strings differ, $Z_L\ne Z_R$, then $\cos \delta =0$, so $\sin \delta \ne 0$ andThe middle segment is a quarter-wave impedance transformer. Its two reflected contributions cancel at these odd quarter-wavelength thicknesses. Here ${\lambda }_M=2\pi /k_M$ is specifically the wavelength in the middle string; the matching is not generally valid at every frequency.

If $Z_L=Z_R$ but $Z_M$ differs, the second condition requires $\sin \delta =0$, givingA whole number of half-wavelengths reproduces the exterior mechanical impedance at the input. The formal $n=0$ limit is zero frequency or zero segment length, rather than a finite-wavelength resonance.

Finally, if $Z_L=Z_M=Z_R$, both conditions hold identically: $\boxed{r=0\text{ at every wavelength}}$. Since the system is lossless, each of these cases also gives unit transmitted power.

For equal-mass one-dimensional scattering between regions with real wavenumbers $k_0>k_2>0$, insert a layer of width $a$ and wavenumber $k_1$. For a purely right-moving incident wave of amplitude $A$ and no reflection, continuity at the first edge gives interior amplitudes $C=A(1+k_0/k_1)/2$ and $D=A(1-k_0/k_1)/2$. If $k_{1}a=\pi /2$, continuity at the second edge gives a transmitted edge amplitude $F=i(k_0/k_1)A$ and the matching condition $k_1^2=k_0{k}_2$. Thus $F/A=i\sqrt{k_0/k_2}$ and the transmitted probability current equals the incident current, despite their unequal amplitudes.

For potentials $0,V_1,V_2$ with $E>V_2>V_1>0$, the condition is $(E-V_1{)}^2=E(E-V_2)$, or $E=V_1^2/(2V_1-V_2)$, requiring $V_2/2<V_1<V_2$. It is the quantum-wave counterpart of a quarter-wave impedance transformer: the intermediate wavenumber is the geometric mean and the two reflected amplitudes cancel.