Energy and Power of a Wave Traveling Along a String — Question 8

Question 8

Signed energy flux distinguishes the two propagation directions

For a string under constant tension, define rightward instantaneous power as $P=-F y_x y_t$. Compare $y_R=f(x-vt)$ and $y_L=g(x+vt)$, with $v>0$.

1. Find $P$ for each wave and state its sign.

2. Why can leftward power be negative while the mechanical energy density remains nonnegative?

Question 8 - Solution

Signed energy flux distinguishes the two propagation directions

Method and mathematical working

Use the signed power crossing a fixed point. The chain rule gives opposite signs for the relation between time derivative and spatial slope in the two propagation directions.$$\begin{aligned}y=f(x-vt)&:\quad P=Fv(f^{\prime})^2,\\y=f(x+vt)&:\quad P=-Fv(f^{\prime})^2.\\[4pt]P_R&\ge0,\quad P_L\le0,\qquad u=\tfrac12\mu y_t^2+\tfrac12Fy_x^2\ge0.\end{aligned}$$

(a) Calculation and conclusion

For the rightward wave, $y_t=-vf^{\prime}$ and $y_x=f^{\prime}$, giving $P_R=Fv(f^{\prime})^2\ge0$. For the leftward wave, $y_t=vg^{\prime}$, yielding $P_L=-Fv(g^{\prime})^2\le0$.

(b) Interpretation and limits

Power is a signed flux through an oriented section; negative means energy crosses toward decreasing $x$. Energy density is a sum of kinetic and elastic squares and remains nonnegative. A direction convention cannot make stored mechanical energy negative.

Physical check

The sign reports energy-flow direction, while the energy density stays nonnegative. Squaring a displacement amplitude alone cannot retain this directional information.