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])

and

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) .