We have, for our infinite wave,

- Z(x,t) = Z
_{0}+ A cos (2π [ν t – x/λ]) = Z_{0}+ A cos (2π [t/T – x/λ])

and we want to show that

- d
^{2}Z/dt^{2}= – (2 π ν)^{2}(Z – Z_{0}) - d
^{2}Z/dx^{2}= – (2 π / λ)^{2}(Z – Z_{0})

A few facts:

- Z – Z
_{0}= A cos (2π [ν t – x/λ]) — just take the red formula and move Z_{0}to the left-hand side. - Since Z
_{0 }is a constant that doesn’t depend on time or space, dZ_{0}/dt = 0 and dZ_{0}/dx = 0. - d(cos t)/dt = – sin t and d(sin t)/dt = + cos t
- 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) .
- 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

d^{2}Z/dt^{2 }= 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 – Z_{0}) .

Since our red formula for the wave would look the same if you exchanged (ν t) with (-x/λ), the computation of d^{2}Z/dx^{2} is the same as for d^{2}Z/dt^{2} 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, d^{2}Z/dt^{2 }= -(2π /λ)^{2 }(Z – Z_{0}) .

## 2 Responses

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

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.