Wave Boundary Conditions: Reflection and Transmission
Let us consider a situation where two different strings are tied together, and a wave propagates from left to right along string 1. Since the propagation speed of the wave is related to the mass, the speed changes as the wave passes through the point where the strings are tied. For convenience, let’s denote the location of the knot as $x=0$ and assume that the wave enters from the left. Then, the incident wave can be represented by a complex wave function as follows.
$$ \tilde {f} _{\text{I}}(x,t) = \tilde{A}_{\text{I}} e^{i(k_{1}x-wt)},\quad x\lt 0 $$
The incident wave creates a reflected wave that turns back along string 1 and a transmitted wave that continues along string 2.
$$ \tilde {f} _{\text{R}}(x,t) = \tilde{A}_{\text{R}} e^{i(-k_{1}x-wt)},\quad x\lt 0 $$
$$ \tilde {f} _{\text{T}}(x,t) = \tilde{A}_{\text{T}} e^{i(k_{2}x-wt)},\quad x\gt 0 $$
An important point to note here is that although the speed of the wave changes depending on the mass of the string, the frequency $\omega$ remains unchanged. Since the source of the wave is the same, the frequency at all parts of the string remains $\omega$ and does not change. However, since the wave speeds differ in the two strings, both the wavelength and wavenumber change.
$$ \dfrac{\lambda_{1}}{\lambda_{2}}=\dfrac{\frac{2\pi}{k_{1}}}{\frac{2\pi}{k_{2}}}=\dfrac{k_{2}}{k_{1}}=\dfrac{v_{1}}{v_{2}} $$
In the region of $x\lt 0$, where both the incident and reflected waves are present, the net displacement of the string can be expressed as follows.
$$ \tilde{f}(x,t) = \begin{cases} \tilde{A}_{\text{I}} e^{i(k_{1}x-wt)} + \tilde{A}_{\text{R}} e^{i(-k_{1}x-wt)} & x\lt 0 \\ \tilde{A}_{\text{T}} e^{i(k_{2}x-wt)} & x\gt 0 \end{cases} $$
Naturally, if the string is not severed at the knot$(x=0)$, the following equation must hold true. $f$ must be continuous at $x=0$.
$$ \lim \limits_{x \rightarrow 0^-} f(x,t) = \lim \limits_{x \rightarrow 0^+} f(x,t) $$
If the mass of the knot can be neglected, then the derivative of $f$ must also be continuous at $x=0$.
$$ \dfrac{\partial f}{\partial x} \Bigg|_{0^-} = \dfrac{\partial f}{\partial x} \Bigg|_{0^+} $$
Otherwise, as shown in the figure below, the tension will not cancel out, causing the knot to experience a net force and continuously increasing acceleration. The above boundary condition applies to real wave functions $f(x,t)$ but also equally to complex wave functions $\tilde{f}$. The imaginary part of $\tilde{f}$ differs from the real part only in that the cosine is replaced by a sine.
$$ \lim \limits_{x \rightarrow 0^-} \tilde{f}(x,t) = \lim \limits_{x \rightarrow 0^+} \tilde{f}(x,t) \\ $$
$$ \dfrac{\partial \tilde{f}}{\partial x} \Bigg|_{0^-} = \dfrac{\partial \tilde{f}}{\partial x} \Bigg|_{0^+} $$
Applying the above condition to $(2)$ yields the following two equations.
$$ \tilde{A}_{I}+\tilde{A}_{R}=\tilde{A}_{T} \\ k_{1} (\tilde{A}_{I} -\tilde{A}_{R})=k_{2}\tilde{A}_{T} $$
Solving these equations simultaneously to express the reflected and transmitted waves in terms of the incident wave gives $$ \tilde{A}_{R}=\dfrac{k_{1}-k_{2}}{k_{1}+k_{2}}\tilde{A}_{I}, \quad \tilde{A}_{T}=\dfrac{2k_{1}}{k_{1}+k_{2}}\tilde{A}_{I} $$ The details of the calculation are omitted due to the length of the text; if curious, refer to the bottom of the article$^{\ast}$. Furthermore, applying the relationship between wavenumber and speed $(1)$ to the above equation yields $$ \tilde{A}_{R}=\dfrac{v_{2}-v_{1}}{v_{2}+v_{1}}\tilde{A}_{I}, \quad \tilde{A}_{T}=\dfrac{2v_{2}}{v_{2}+v_{1}}\tilde{A}_{I} $$ Likewise, if you’re curious about the calculation process, refer to the bottom of the article$^{**}$. Therefore, the relationship between the real amplitude and phase is as follows. $$ A_{R}e^{i\delta_{R}}=\dfrac{v_{2}-v_{1}}{v_{2}+v_{1}}A_{I}e^{i\delta_{I}}, \quad A_{T}e^{i\delta_{T}}=\dfrac{2v_{2}}{v_{2}+v_{1}}A_{I}e^{i\delta_{I}} $$ If string 2 is lighter than string 1, then $\mu_{2} <\mu_{1} \ \rightarrow v_{1} < v_{2} \left( \because v=\sqrt{\frac{T}{\mu}}\right) $, and the phase angles of all three waves are the same. $\delta_{I}=\delta_{R}=\delta_{T}$ Therefore, the amplitudes of the reflected and transmitted waves are $$ A_{R}=\dfrac{v_{2}-v_{1}}{v_{2}+v_{1}}A_{I}, \quad A_{T}=\dfrac{2v_{2}}{v_{2}+v_{1}}A_{I} $$ If string 2 is heavier than string 1, then $\mu_{1} <\mu_{2} \ \rightarrow v_{2} < v_{1} \left( \because v=\sqrt{\frac{T}{\mu}}\right) $, and the phase of the reflected wave is displaced by $\pi$. $\delta_{R}+\pi=\delta_{I}=\delta_{T}$. And since $\cos(-k_{1}x-\omega t +\delta_{I} - \pi) =-\cos(-k_{1}x-\omega t +\delta_{I})$, the amplitudes of the reflected and transmitted waves are $$ A_{R}=\dfrac{v_{1}-v_{2}}{v_{2}+v_{1}}A_{I}, \quad A_{T}=\dfrac{2v_{2}}{v_{2}+v_{1}}A_{I} $$ Especially, if string 2 is very heavy compared to string 1 (or if the end of 1 is fixed), then $v_{2} « v_{1}$, so the amplitudes of the reflected and transmitted waves are $$ A_{R}=A_{I}, \quad A_{T}=0 $$ That is, there is no transmitted wave, and it is all reflected.
* $$ \tilde{A}_{I}+\tilde{A}_{R}=\tilde{A}_{T} \\ k_{1} (\tilde{A}_{I} -\tilde{A}_{R})=k_{2}\tilde{A}_{T} $$ Substituting the equation above into this equation gives $$ \begin{array}{rc} & k_{2}(\tilde{A}_{I} + \tilde{A}_{R}) = k_{1}(\tilde{A}_{I} -\tilde{A}_{R}) \\ \implies & (k_{1}-k_{2})\tilde{A}_{I} = (k_{1}+k_{2})\tilde{A}_{R} \\ \implies &\tilde{A}_{R} = \dfrac{k_{1}-k_{2}}{k_{1}+k_{2}}\tilde{A}_{I} \end{array} $$ Substituting the equation above into this equation gives $$ \begin{array}{rc} & k_{1}(\tilde{A}_{I} + \tilde{A}_{I} -\tilde{A}_{T}) = k_{2}\tilde{A}_{T} \\ \implies & 2k_{1}\tilde{A}_{I} = (k_{1}+k_{2})\tilde{A}_{T} \\ \implies &\tilde{A}_{T} = \dfrac{2k_{1}}{k_{1}+k_{2}}\tilde{A}_{I} \end{array} $$ ** $$ \dfrac{k_{1}-k_{2}}{k_{1}+k_{2}}=\dfrac{\dfrac{k_{1}}{k_{2}}-1}{\dfrac{k_{1}}{k_{2}}+1}=\dfrac{\dfrac{v_{2}}{v_{1}}-1}{\dfrac{v_{2}}{v_{1}}+1}=\dfrac{v_{2}-v_{1}}{v_{2}+v_{1}} $$
$$ \dfrac{2k_{1}}{k_{1}+k_{2}}=\dfrac{2\dfrac{k_{1}}{k_{2}}}{\dfrac{k_{1}}{k_{2}}+1}=\dfrac{2\dfrac{v_{2}}{v_{1}}}{\dfrac{v_{2}}{v_{1}}+1}=\dfrac{2v_{2}}{v_{2}+v_{1}} $$