The Wave Equation — Question 4

Question 4 – Solution

Strategy. Displacement amplitudes and energy fractions differ because the strings have different wave impedances.

Step 1: Apply the interface conditions. Every term of the form $F(t\mp x/c_j)$ has $u_{tt}=c_j^2u_{xx}$. At the junction, displacement continuity gives $1+R=D$. The slope condition gives $$\frac{\mathcal T}{c_1}(R-1)F'(t) =-\frac{\mathcal T}{c_2}D F'(t),\qquad \boxed{Z_1(1-R)=Z_2D.}$$ Since $F$ is nonconstant, these amplitude equations follow.

Step 2: Solve and interpret the amplitudes. Solving the two equations gives $$\boxed{R=\frac{Z_1-Z_2}{Z_1+Z_2},\qquad D=\frac{2Z_1}{Z_1+Z_2}.}$$ Here $Z_j=\sqrt{\mathcal T\mu_j}$. A denser second string gives $R<0$, an inverted reflected displacement. A lighter string gives $R>0$. Equal impedances give no reflection and $D=1$.

Step 3: Account for energy flux. The incident rightward power is $Z_1(F')^2$, the reflected leftward power has magnitude $Z_1R^2(F')^2$, and transmitted power is $Z_2D^2(F')^2$. The cross terms in $-\mathcal T u_tu_x$ on the incident side cancel. For a pulse with finite nonzero $\int(F')^2$, integration in time therefore gives $$\boxed{\mathcal R=R^2,\qquad \mathcal D=\frac{Z_2}{Z_1}D^2,\qquad \mathcal R+\mathcal D= \frac{(Z_1-Z_2)^2+4Z_1Z_2}{(Z_1+Z_2)^2}=1.}$$

Step 4: Test examples and extreme impedances. If $\mu_2=9\mu_1$, then $Z_2=3Z_1$, so $R=-1/2$, $D=1/2$, $\mathcal R=1/4$ and $\mathcal D=3/4$. As $Z_2/Z_1\to 0$, $R\to 1$ and $D\to 2$, but $\mathcal D\to 0$. As $Z_2/Z_1\to\infty$, $R\to-1$, $D\to 0$ and again $\mathcal D\to 0$. These are the free-end and fixed-end reflection limits, respectively. Energy depends on impedance as well as squared displacement amplitude.