Of Particular Significance

4a. Waves (The Quick Calculation)

We have, for our infinite wave,

  • Z(x,t) = Z0 + A cos (2π [ν t – x/λ]) = Z0 + A cos (2π [t/T – x/λ])

and we want to show that

  • d2Z/dt2 = – (2 π ν)2 (Z – Z0)
  • d2Z/dx2 = – (2 π / λ)2 (Z – Z0)

A few facts:

  1. Z – Z0 = A cos (2π [ν t – x/λ])  —  just take the red formula and move Z0 to the left-hand side.
  2. Since Z0 is a constant that doesn’t depend on time or space, dZ0/dt = 0 and dZ0/dx = 0.
  3. d(cos t)/dt = – sin t  and  d(sin t)/dt = + cos t
  4. d(F[a t +b x])/dt =  a d(F[a t +b x])/d(a t+ b x), where a and b are constants and F is any function of (a t + b x) .
  5. d[A f(t)]/dt = A  d[f(t)]/dt , where f(t) is any function of t and A is a constant

Altogether this means

dZ/dt = d[ A cos (2π [ν t – x/λ]) ] /dt= A d[ cos (2π [ν t – x/λ]) ] /dt

= A(2π ν) d[ cos (2π [ν t – x/λ]) ]/d(2π [ν t – x/λ]) = -(2π ν) A sin (2π [ν t – x/λ])


d2Z/dt2 = d[-(2π ν)A sin (2π [ν t – x/λ])]/dt = -(2π ν)A d[sin (2π [ν t – x/λ])]/dt

= -(2π ν)2 A cos (2π [ν t – x/λ]) = -(2π ν)2 (Z – Z0) .

Since our red formula for the wave would look the same if you exchanged (ν t) with (-x/λ), the computation of d2Z/dx2 is the same as for d2Z/dt2 except that instead of d/dt giving a factor of (2π ν) we have d/dx giving a factor of (- 2π/λ); but since there are two such factors in the answer, we just replace (2π ν)2 with (- 2π/λ)2 = (+2π/λ)2; the minus sign doesn’t matter.  (There’s still an overall minus sign in addition.)  So, as we needed to show, d2Z/dt2 =  -(2π /λ)2 (Z – Z0) .

2 Responses

  1. Typo on 1. …the left-hand side*
    Feel free to delete this.

    Great site, far more helpful than everything else I’ve been struggling with for the past year.

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