The Principle of Superposition for Waves — Question 7

Question 7

A junction can transmit a larger displacement than the incident amplitude

Two ideal strings meet at $x=0$ with common tension and a massless junction. For harmonic displacement amplitudes, continuity gives $1+r=t$ and force balance gives $Z_1(1-r)=Z_2t$, where $Z_i=\sqrt{F\mu_i}$, $r=A_r/A_i$ and $t=A_t/A_i$.

1. Solve for $r,t$ and evaluate them for $Z_2=Z_1/3$.

2. Use the power ratio $\mathcal T=(Z_2/Z_1)t^2$ and $\mathcal R=r^2$ to check conservation. Does $t>1$ violate it?

Question 7 - Solution

A junction can transmit a larger displacement than the incident amplitude

Method and mathematical working

Solve the two boundary equations for displacement-amplitude ratios. Convert amplitudes to power ratios with the impedance factor on the transmitted side.$$\begin{aligned}r&=\frac{Z_1-Z_2}{Z_1+Z_2},\qquad t=\frac{2Z_1}{Z_1+Z_2},\\R&=r^2,\qquad T=\frac{Z_2}{Z_1}t^2.\\[4pt]r&=\frac{1-1/3}{1+1/3}=\frac12,\quad t=\frac32,\\R+T&=\frac14+\frac13\left(\frac32\right)^2=1.\end{aligned}$$

(a) Calculation and conclusion

Substitution gives $r=(Z_1-Z_2)/(Z_1+Z_2)$ and $t=2Z_1/(Z_1+Z_2)$. With $Z_2=Z_1/3$, $r=1/2$ and $t=3/2$. The reflection is upright.

(b) Interpretation and limits

$\mathcal R=1/4$ and $\mathcal T=(1/3)(9/4)=3/4$, summing to one. The transmitted string has lower mechanical impedance, so a larger displacement amplitude can carry less than the incident power. Amplitude ratios are not power fractions.

Physical check

For the given impedances, $R+T=1$. A transmitted amplitude above unity is allowed because its power is weighted by the smaller impedance.