Question

a stationary wave is y=12 sin 300t cos2x . What is the distance between two nearest nodes

Answer 1

node :- point on wave where velocity is velocity of particle is maxi maximum. eg
mean position of a wave is known as node .

y =12sin300t.cos2x

use trigonometry formula
2sinA.cosB =sin(A+B)+sin(A-B)

now,

y =6{sin(300t+2x) +sin(300t -2x)}

=6sin(300t+2x) +6sin(300t-2x)

at mean position node will be formed

so,
y =0

sin(300t+2x) = -sin(300t -2x)
sin(300t+2x) =sin(π +300t-2x)

300t +2x = 300t -2x +π

4x =π

x =π/4

hence,
node form at x =π/4

now ,
again
sin(300t +2x) =sin(-π + 300t-2x)
300t +2x = -π +300t -2x
4x =-π
x =-π/4

hence, node formed at near to x=π/4 is -π/4

so , distance between node =π/4 +π/4 =π/2

Answer 2

Nodes means the displacement from mean position is always zero. The points are not vibrating.
 
    y = 12  Sin(300*t)  Cos(2x)
 
    Find points where y = 0 always.
    Cos 2x = 0    for  2x = π/2,  3π/2, 5π/2 .....  for any time t.
    So nodes are at  x = π/4,  3π/4 ,  5π/4 ..... units

     Hence the distance between two nearest nodes = 3π/4 - π/4 = π/2  units