An incident wave and a reflected wave are represented by
`xi _(1) = a "sin" (2pi)/(lambda) (vt - x) and xi _(2) = a "sin"(2pi)/(lambda) (vt + x) `
Derive the equation of the stationary wave and calcu-late the position of the nodes and antinodes.

Explanation

According to the principle of superposition
`xi = xi_(1) + xi _(2) = a "sin" (2pi)/(lambda)(vt - x) + a "sin"(2pi)/(lambda) (vt - x) `
` = 2 a "cos"(2pix)/(lambda) "sin"(2pi)/(lambda) vt`
` = 2 a "cos (2pi)/(lambda) x sin 2 pi n t [:. v = n lambda] `
` A sin 2 pi n t , ` where amplitude, ` A = 2 a "cos" (2pi)/(lambda) x `
This equation represents a stationary wave .
For nodes, `xi = 0 ` for all values of t,i.e.,
` "cos"(2pi)/(lambda)x = 0 [ :. A = 0 ] `
or, `"cos"(2pix)/(lambda)= cos(2n+1)(pi)/(2),n=0,1,2...`
or, `x = (2n + 1) (lambda)/(4)`
The position of nodes at which particles do not vibrate, are ` x (lambda)/(4),(3lambda)/(4),(5lambda)/(4)` ect. Distance between two successive nodes ` = (lambda)/(2)` .
For amtinodes, ` "cos"(2pi)/(lambda) = pm 1 ` , so that the particles vibrate with maximum amplitude `pm2a` . Hence, antinodes occur at a position for which
`"cos"(2pix)/(lambda) = cos npi , n = 0 , 1 , 2 ...`
or, ` (2pix)/(lambda) = n pi or, x = (nlambda)/(2)`
So, the position of antinodes are x ` = 0,(lambda)/(2),lambda,(3lambda)/(2)` ect.
The distance between two antinodes ` = (lambda)/(2)` .