The Wave Equation — Question 4

Section 16-7 · Physics pages 2518–2519

Question 4 A snapshot determines acceleration but not initial velocity At t = 0, a uniform string of speed v = 4 m/s has profile y = A cos(kx) with A = .010 m and k = 3 rad/m. No velocity data are provided. x y v A displacement snapshot supplies spatial curvature, but does not specify the initial material velocities. (a) Find the acceleration field at that instant. (b) Give two different future motions with this same initial profile, and identify the missing initial data.
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Question 4 - Solution A snapshot determines acceleration but not initial velocity x y v Method and mathematical working Use the wave equation directly on the measured snapshot to find acceleration. Initial velocity remains independent data for a second-order time-evolution equation. yxx(x, 0) = −Ak2 cos kx, ytt(x, 0) = −v2Ak2 cos kx. ytt(x, 0) = −(4)2(0.010)(3)2 cos 3x = −1.44 cos 3x m/s2, y±(x, t) = 0.010 cos(3x ∓12t). (a) Calculation and conclusion The wave equation gives ytt(x, 0) = v2yxx = −v2k2A cos kx = −1.44 cos(3x) m/s2. (b) Interpretation and limits Both A cos(3x −12t) and A cos(3x + 12t) share the snapshot and acceleration but have opposite initial velocities ±.12 sin(3x) m/s. A second-order time equation requires both displacement and velocity fields to select the evolution. Physical check The acceleration points back toward equilibrium for this sinusoidal snapshot. Two waves traveling in opposite directions have the same displacement and acceleration at the chosen instant but opposite velocities.

Question 4

A snapshot determines acceleration but not initial velocity

At t=0t=0, a uniform string of speed v=4m/sv=4\,\mathrm{m/s} has profile y=Acos⁡(kx)y=A\cos(kx) with A=.010mA=.010\,\mathrm{m} and k=3rad/mk=3\,\mathrm{rad/m}. No velocity data are provided.

Diagram shown in the original SVG worksheet.

  1. Find the acceleration field at that instant.

  2. Give two different future motions with this same initial profile, and identify the missing initial data.

Question 4 - Solution

A snapshot determines acceleration but not initial velocity

Diagram shown in the original SVG worksheet.

Method and mathematical working

Use the wave equation directly on the measured snapshot to find acceleration. Initial velocity remains independent data for a second-order time-evolution equation.yxx(x,0)=−Ak2cos⁡kx,ytt(x,0)=−v2Ak2cos⁡kx.ytt(x,0)=−(4)2(0.010)(3)2cos⁡3x=−1.44cos⁡3xm/s2,y±(x,t)=0.010cos⁡(3x∓12t).\begin{aligned}y_{xx}(x,0)&=-Ak^2\cos kx,\\y_{tt}(x,0)&=-v^2Ak^2\cos kx.\\[4pt]y_{tt}(x,0)&=-(4)^2(0.010)(3)^2\cos3x=-1.44\cos3x\,\mathrm{m/s^2},\\y_{\pm}(x,t)&=0.010\cos(3x\mp12t).\end{aligned}

(a) Calculation and conclusion

The wave equation gives ytt(x,0)=v2yxx=−v2k2Acos⁡kx=−1.44cos⁡(3x)m/s2y_{tt}(x,0)=v^2y_{xx}=-v^2k^2A\cos kx=-1.44\cos(3x)\,\mathrm{m/s^2}.

(b) Interpretation and limits

Both Acos⁡(3x−12t)A\cos(3x-12t) and Acos⁡(3x+12t)A\cos(3x+12t) share the snapshot and acceleration but have opposite initial velocities ±.12sin⁡(3x)m/s\pm.12\sin(3x)\,\mathrm{m/s}. A second-order time equation requires both displacement and velocity fields to select the evolution.

Physical check

The acceleration points back toward equilibrium for this sinusoidal snapshot. Two waves traveling in opposite directions have the same displacement and acceleration at the chosen instant but opposite velocities.