The Wave Equation — Question 5

Section 16-7 · Physics pages 2520–2521

Question 5 A one-direction initial state requires a derivative relation An infinite uniform string has initial displacement h(x) and initial velocity g(x). The desired motion is a pulse traveling only to the right at speed v. x v Pulse geometry and direction; the profile is schematic unless a function is given. (a) Find the necessary initial-velocity field in terms of h. (b) For h = Ae−x2/a2, find the required velocity on each side of the peak. Would releasing this shape from rest produce the desired motion?
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Question 5 - Solution A one-direction initial state requires a derivative relation x v Method and mathematical working A pure rightward solution imposes a specific relation between the initial displacement derivative and the initial velocity. Differentiate the Gaussian before applying this relation. h′(x) = −2Ax a2 e−x2/a2, g(x) = −vh′(x) = 2vAx a2 e−x2/a2. x < 0 ⇒g(x) < 0, x > 0 ⇒g(x) > 0, g(0) = 0. (a) Calculation and conclusion Pure rightward translation is y = h(x −vt), so g(x) = yt(x, 0) = −vh′(x). Displacement and velocity must be prepared consistently. (b) Interpretation and limits h′ = −(2Ax/a2)e−x2/a2, hence g = (2vAx/a2)e−x2/a2. The left side moves down, the right side up, and the peak initially has zero velocity. Release from rest fails this relation except at isolated points and produces components traveling in both directions. Physical check The required velocity is odd while the displacement is even. On the right flank the element moves upward, exactly as expected when a positive pulse moves toward increasing x.

Question 5

A one-direction initial state requires a derivative relation

An infinite uniform string has initial displacement h(x)h(x) and initial velocity g(x)g(x). The desired motion is a pulse traveling only to the right at speed vv.

Diagram shown in the original SVG worksheet.

  1. Find the necessary initial-velocity field in terms of hh.

  2. For h=Ae−x2/a2h=Ae^{-x^2/a^2}, find the required velocity on each side of the peak. Would releasing this shape from rest produce the desired motion?

Question 5 - Solution

A one-direction initial state requires a derivative relation

Diagram shown in the original SVG worksheet.

Method and mathematical working

A pure rightward solution imposes a specific relation between the initial displacement derivative and the initial velocity. Differentiate the Gaussian before applying this relation.h′(x)=−2Axa2e−x2/a2,g(x)=−vh′(x)=2vAxa2e−x2/a2.x<0⇒g(x)<0,x>0⇒g(x)>0,g(0)=0.\begin{aligned}h^{\prime}(x)&=-\frac{2Ax}{a^2}e^{-x^2/a^2},\\g(x)&=-v h^{\prime}(x)=\frac{2vAx}{a^2}e^{-x^2/a^2}.\\[4pt]x<0&\Rightarrow g(x)<0,\quad x>0\Rightarrow g(x)>0,\qquad g(0)=0.\end{aligned}

(a) Calculation and conclusion

Pure rightward translation is y=h(x−vt)y=h(x-vt), so g(x)=yt(x,0)=−vh′(x)g(x)=y_t(x,0)=-vh^{\prime}(x). Displacement and velocity must be prepared consistently.

(b) Interpretation and limits

h′=−(2Ax/a2)e−x2/a2h^{\prime}=-(2Ax/a^2)e^{-x^2/a^2}, hence g=(2vAx/a2)e−x2/a2g=(2vAx/a^2)e^{-x^2/a^2}. The left side moves down, the right side up, and the peak initially has zero velocity. Release from rest fails this relation except at isolated points and produces components traveling in both directions.

Physical check

The required velocity is odd while the displacement is even. On the right flank the element moves upward, exactly as expected when a positive pulse moves toward increasing xx.