Energy and Power of a Wave Traveling Along a String — Question 9

Question 9

Amplitude attenuation doubles its exponent in the power law

In a weakly lossy uniform string, suppose a slowly varying sinusoidal amplitude is $A(x)=A_0e^{-\alpha x}$, with $\alpha>0$. Use the local traveling-wave relation $\overline P=C A^2$, where $C$ is constant; neglect reflections.

1. Find the power profile and power lost per unit length.

2. If amplitude halves over distance $L$, what fraction of initial power remains and what mean power is deposited between $0$ and $L$?

Question 9 - Solution

Amplitude attenuation doubles its exponent in the power law

Method and mathematical working

Power is quadratic in amplitude. The energy deposited per unit length per unit time equals the decrease of forward power with position.$$\begin{aligned}\overline P(x)&=\overline P_0e^{-2\alpha x},\\q(x)&=-\frac{d\overline P}{dx}=2\alpha\overline P(x).\\[4pt]A(L)/A_0=1/2&\Rightarrow\overline P(L)/\overline P_0=1/4,\\\int_0^L2\alpha\overline P(x)\,dx&=\overline P_0-\overline P(L)=3\overline P_0/4.\end{aligned}$$

(a) Calculation and conclusion

$\overline P(x)=P_0e^{-2\alpha x}$. The positive deposited power per unit length is $-d\overline P/dx=2\alpha P_0e^{-2\alpha x}$, with units $\mathrm{W/m}$. This follows from energy balance on a short interval.

(b) Interpretation and limits

At $A(L)=A_0/2$, $P(L)=P_0/4$. The segment receives net mean power $P_0-P(L)=3P_0/4$. Halving amplitude does not halve power because energy transport is quadratic in the oscillatory motion.

Physical check

For positive attenuation $q$ is positive and has units $\mathrm{W/m}$. Integrating $q$ between two positions gives exactly the incoming power minus the outgoing power.