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Q.
Two travelling waves, $y_1 = A \sin [ k ( x + ct) ]$ and $y_2 = A \sin [ k (x - ct) ]$ are superposed on a string. The distance between adjacent antinodes is
Given,
$Y_{1}=A \sin [k(x +c t)]\,...(i)$
and $Y_{2}=A \sin [k(x -c t)]\,...(ii)$
By the principle of superposition, the resultant displacement of the particle is given by
$Y=Y_{1}+Y_{2} $
$Y=A[\sin \{k(x+ c t)\}+\sin \{k(x- c t)\}]$
By the formula
$\sin C+\sin D=2 \sin \frac{C+D}{2} \cdot \cos \frac{C-D}{2}$
We have
$y=2 A \sin \frac{k x+ k c t+k x-k c t}{2} $
$\cdot \cos \frac{k x +k c t-k x +k c t}{2}$
$y=2 A \sin k x \cdot \cos k c t$
For first antinode
$\sin k x_{1}=1$
$\sin k x_{1}=\sin \frac{\pi}{2}$
$k \,x_{1}=\frac{\pi}{2}\,...(iii)$
For second antinode
$ \sin k x_{2} =-1 $
$\sin k x_{2} =\sin \frac{3 \pi}{2}$
$k x_{2} =\frac{3 \pi}{2}\,...(iv) $
$\therefore $ The distance between adjacent antinodes
$k x_{2}-k x_{1}=\frac{3 \pi}{2}-\frac{\pi}{2}$
$k\left(x_{2}-x_{1}\right)=\pi$
$\Delta x=\frac{\pi}{k}$