The Wave Equation — Question 6

Section 16-7 · Physics pages 2522–2523

Question 6 Initial data influence only a finite interval at a later event For an infinite string, use the supplied solution y(x, t) = 1 2[h(x −vt) + h(x + vt)] + 1 2v ∫x+vt x−vt g(s)ds, where h and g are initial displacement and velocity. Let v = 3 m/s. x t (x, t) x −vt x + vt Backward characteristics delimit the interval of initial data that can influence an event. (a) Identify the initial region that can affect the event x = 5 m, t = 1 s. (b) If both initial fields vanish outside [−1, 1] m, what is the earliest possible arrival at x = 5 m? Distinguish the boundary arrival from a guaranteed nonzero displacement.
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Question 6 - Solution Initial data influence only a finite interval at a later event x t (x, t) x −vt x + vt Method and mathematical working Read the endpoints of the initial-data interval from the two characteristic directions in the supplied solution. Compare that interval with the support of the initial disturbance. I(x, t) = [x −vt, x + vt], I(5, 1) = [2, 8], tfirst = 5 −1 3 . [5 −3(1), 5 + 3(1)] = [2, 8], 3t ≥5 −1 ⇒t ≥4/3 s. (a) Calculation and conclusion The characteristic endpoints are x −vt = 2 and x + vt = 8 m. Only data within [2, 8] can influence that event; displacement enters at the endpoints and velocity through the interval integral. (b) Interpretation and limits The nearest possible initial disturbance is at x = 1, a distance 4 m away. Earliest possible arrival is 4/3 s. Smooth compact data can vanish at their support edge, so equality need not give nonzero displacement exactly at the first causal boundary. Physical check The earliest possible influence occurs when a characteristic first meets the nearest edge of the support. Actual displacement can still vanish at that edge, so this causal bound need not imply a nonzero reading at the exact first instant.

Question 6

Initial data influence only a finite interval at a later event

For an infinite string, use the supplied solution y(x,t)=12[h(x−vt)+h(x+vt)]+12v∫x−vtx+vtg(s)dsy(x,t)=\tfrac12[h(x-vt)+h(x+vt)]+\tfrac1{2v}\int_{x-vt}^{x+vt}g(s)ds, where hh and gg are initial displacement and velocity. Let v=3m/sv=3\,\mathrm{m/s}.

Diagram shown in the original SVG worksheet.

  1. Identify the initial region that can affect the event x=5m,t=1sx=5\,\mathrm{m},t=1\,\mathrm{s}.

  2. If both initial fields vanish outside [−1,1]m[-1,1]\,\mathrm{m}, what is the earliest possible arrival at x=5mx=5\,\mathrm{m}? Distinguish the boundary arrival from a guaranteed nonzero displacement.

Question 6 - Solution

Initial data influence only a finite interval at a later event

Diagram shown in the original SVG worksheet.

Method and mathematical working

Read the endpoints of the initial-data interval from the two characteristic directions in the supplied solution. Compare that interval with the support of the initial disturbance.I(x,t)=[x−vt,x+vt],I(5,1)=[2,8],tfirst=5−13.[5−3(1),5+3(1)]=[2,8],3t≥5−1⇒t≥4/3s.\begin{aligned}I(x,t)&=[x-vt,x+vt],\\I(5,1)&=[2,8],\qquad t_{\mathrm{first}}=\frac{5-1}{3}.\\[4pt][5-3(1),5+3(1)]&=[2,8],\qquad3t\ge5-1\Rightarrow t\ge4/3\,\mathrm s.\end{aligned}

(a) Calculation and conclusion

The characteristic endpoints are x−vt=2x-vt=2 and x+vt=8mx+vt=8\,\mathrm{m}. Only data within [2,8][2,8] can influence that event; displacement enters at the endpoints and velocity through the interval integral.

(b) Interpretation and limits

The nearest possible initial disturbance is at x=1x=1, a distance 4m4\,\mathrm{m} away. Earliest possible arrival is 4/3s4/3\,\mathrm{s}. Smooth compact data can vanish at their support edge, so equality need not give nonzero displacement exactly at the first causal boundary.

Physical check

The earliest possible influence occurs when a characteristic first meets the nearest edge of the support. Actual displacement can still vanish at that edge, so this causal bound need not imply a nonzero reading at the exact first instant.