Two strings attached together

We have considered the behavior of a single uniform massive string. Now we will look at systems with two different strings. We will first consider two strings of densities $\rho_1$ and $\rho_2$ tied together.

We saw that one solution to the wave equation was a traveling wave of the form $f(x + vt)$ . We also saw that when a traveling wave reached a wall or ring on a pole, it turned back (with negative amplitude in the case of a wall). We will make the assumption that when a traveling wave from the left reaches $x=0$ it will split into two waves, one reflected back on string 1 in the $-x$ direction, and one transmitted to string 2 in the $+x$ direction.

The initial wave is called the incoming or incident wave, and the resulting waves are called the reflected and transmitted waves. We can write the vertical displacement along the string as a superposition of these waves.

\begin{align*} y_1 &= f_i + f_r = \text{incident pulse + reflected pulse} \\ y_2 &= f_t = \text{transmitted pulse}. \end{align*}

Consider the boundary conditions. Clearly since the strings are attached $y_1(x=0, t) = y_2(x=0,t)$ .

Now let’s look at the infinitesimal mass element at $x=0$ . There are two tension forces acting on it. Since we want the knot to remain at $x=0$ , the horizontal components must cancel, so $T_1 \cos \theta_1 = T_2 \cos \theta_2$ . Taking the limit as $dx \to 0$ , $dm \to 0$ the angles become very small, so we get $T_1 = T_2$ .

Now consider the vertical components of force $F_y = dm\, \ddot y(x=0,t) = T \sin \theta_2 - T \sin \theta_1$ . As $dm \to 0$ , the force $F \to 0$ as well. Since we have shown the tensions are equal, $\sin \theta_1 = \sin \theta_2$ . Since the angle is very small, $\sin \theta \approx \tan \theta = \frac{\partial y}{\partial x}$ . This gives us the second boundary condition $\left. \frac{\partial y_1}{\partial x} \right|_{x=0} = \left. \frac{\partial y_2}{\partial x} \right|_{x=0}$ .

Now we use these two boundary conditions. Condition 1 gives us

\begin{align*} f_i(\omega t) + f_r(\omega t) &= f_t(\omega t). \tag{1} \end{align*}

Condition 2 gives us

\begin{align*} \frac{\partial f_i}{\partial x} + \frac{\partial f_r}{\partial x} &= \frac{\partial f_t}{\partial x}. \end{align*}

Now note that for $f(x-vt)$ ,

\begin{align*} \frac{\partial f}{\partial x} &= \frac{\partial x}{\partial (x-v t)} \\ \frac{\partial f}{\partial t} &= -v\frac{\partial x}{\partial x-v t} = -v \frac{\partial f}{\partial x}. \end{align*}

We replace the spacial derivatives with time derivatives to find

\begin{align*} -\frac{1}{v_l}\frac{\partial f_i}{\partial t} + \frac{1}{v_l}\frac{\partial f_r}{\partial t} &= -\frac{1}{v_r} \frac{\partial f_t}{\partial t}. \end{align*}

Integrate from $t=-\infty$ to $\infty$ (evaluating at $x=0$ ) and multiply by $v_l v_r$

-v_r f_i + v_r f_r = -v_l f_t. \tag{2}

We now solve equation (1) and (2) to find

\begin{align*} f_r &= R f_i &\quad R &= \frac{v_r-v_l}{v_r+v_l} \\ f_t &= T f_i &\quad T &= \frac{2v_r}{v_r+v_l}. \end{align*}

$R$ and $T$ are the reflection and transmission coefficients for this system.

Note some interesting properties. If $v_l = v_r$ then $R=0$ and the system behaves as if it was just a single string. $T$ can be zero when $v_r=\sqrt{\frac{T}{\rho_2}}=0$ , when the mass on the right side is infinite, basically a wall.

When the wave goes from one string to the other, the velocity changes. $\omega$ is the same, so $\mathcal K$ must chance. Since $\mathcal K=\frac{2\pi}{\text{wave length}}$ is the wave number, the wave length changes.

TODO: massless ring (the above is not general)

Boundary condition $T_1 \psi_1' = T_2 \psi_2'$ – forces need to balance.