Consider the one$-$dimensional wave equation: $\frac{de{l}^{2}u}{del{t}^2}=c^2\frac{de{l}^{2}u}{del{x}^2}.$ Applying the Fourier transform with respect to $x,$ the transformed equation becomes:

Oorweeg die eendimensionele golfvergelyking: $\frac{de{l}^{2}u}{del{t}^2}=c^2\frac{de{l}^{2}u}{del{x}^2}.$ Deur die Fourier$-$transform met betrekking tot $x$ toe te pas, word die vergelyking getransformeer na:

$($A$)\frac{de{l}^{2}U(,t)}{del{t}^2}+c^{2}U(,t)=0$

$($B$)\frac{de{l}^{2}U(,t)}{del{t}^2}-c^2{}^{2}U(,t)=0$

$($C$)\frac{de{l}^{2}U(,t)}{del{t}^2}+c^2\frac{de{l}^{2}U(,t)}{\text{d}\text{e}\text{l}{}^2}=0$

$($D$)\frac{de{l}^{2}U(,t)}{del{t}^2}+c^2{}^{2}U(,t)=0$

$($E$)\frac{de{l}^{2}U(,t)}{del{t}^2}-c^{2}U(,t)=0$

Determine the inverse Fourier transform $f(t)$ of the function $F(),$ where $B>0$ and Bepaal die inverse Fourier$-$transform $f(t)$ van die funksie $F(),$ waar $B>0$ en

$F()=\{\begin{array}{ccc}1& \text{v}\text{i}\text{r}& ||B,\\ 0& \text{v}\text{i}\text{r}& ||>B,\end{array}$

$($A$)f(t)=\frac{sin(Bt)}{t}$

$($B$)f(t)=\frac{sin(Bt)}{t}$

$($C$)f(t)=e^{-B|t|}$

$($D$)f(t)=cos(Bt)$

$($E$)f(t)=\frac{1}{}\frac{1}{1+t^2}$