Show that the distance between any two consecutive nodes or antinodes in a stationary wave is `lambda//2`.

Explanation

We know that `y = 2 A cos kx sin omega t `.The resultant amplitude `A(x)= 2Acoskx`
For node `A ( x) = 0= sin kx `
This implies `kx = 1 pi //2, 3pi //2, 5pi // 2,"…." ( 2n + 1 ) pi //2`
i.e., `kx = ( 2x + 1) pi //2, n = 0,1,2,"…..."`
i.e., `(cancel(2) pi )/( lambda).x = ( 2x+1)( pi)/(cancel(2))`
i.e.,` x = ( 2x+1) (lambda)/( 4)`
for `n = 0,1,"....."x = 1 lambda //4, 3 lambda//4, 5 lambda //4,"....." ( 2x + 1) lambda //4`
Distance between two consecutive nodes `= ( 3lambda )/( 4) - (1 lambda)/( 4) = ( lambda)/( 2)`
Similarly, for antinode `A_("max") =2A rArr kx = 0,pi, 2 pi ,".."n pi`
so that `x=0,pi//k, 2pi//k, 3pi//k,"......"`
i.e., `x =0 , ( lambda)/(2) , lambda, ( 3)/( 2) lambda,"..............."`
Here distance between two consecutive antinodes,
rArr`=( lambda)/( 2) - 0 = ( lambda)/( 2) ` and `( 3)/(2)lambda - 1 lambda = (lambda)/(2)`