A physics laboratory records the transverse displacement (in metres) of a point on a…

PHYSICS · Subjective · Fermi University Physics · Mathematical Methods for Physics · Numerical and computational mathematics · Numerical and computational mathematics

Computational Physics, Ch Differentiating and integrating numbers

Problem

A physics laboratory records the transverse displacement x(t)x(t) (in metres) of a point on a vibrating string every Δt=0.1 s\Delta t = 0.1\,\mathrm{s}. Over one interval the displacement obeys, to a good approximation, x(t)=t3−tx(t) = t^3 - t, and seven readings are taken at t=0,0.1,0.2,0.3,0.4,0.5,0.6 st = 0, 0.1, 0.2, 0.3, 0.4, 0.5, 0.6\,\mathrm{s}: x=[0,−0.099,−0.192,−0.273,−0.336,−0.375,−0.384] mx = [0, -0.099, -0.192, -0.273, -0.336, -0.375, -0.384]\,\mathrm{m}. (a) Form the second difference [xj+1−2xj+xj−1]/Δt2[x_{j+1} - 2x_j + x_{j-1}]/\Delta t^2 at every interior point by offset slicing on this array of 77 values, and state how many entries the resulting array has. Give its value at t=0.3 st = 0.3\,\mathrm{s} and compare it with the exact acceleration x¨(t)=6t\ddot x(t) = 6t at that instant. (b) The central-difference estimate of the velocity x˙(0.3 s)\dot x(0.3\,\mathrm{s}) built from this same record gives D0.2=−0.69 m s−1D_{0.2} = -0.69\,\mathrm{m\,s^{-1}} at step h=0.2 sh=0.2\,\mathrm{s} and D0.1=−0.72 m s−1D_{0.1} = -0.72\,\mathrm{m\,s^{-1}} at step h=0.1 sh=0.1\,\mathrm{s}. Combine these two estimates by Richardson extrapolation and compare the result with the exact velocity x˙(0.3 s)=3t2−1\dot x(0.3\,\mathrm{s}) = 3t^2-1.

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