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 to “4a. Waves (The Quick Calculation)

  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.

  2. Ronald R. Bisson

    Your notation is confusing. Your missing a few dt in line 4. I do believe

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