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Hence show that the distance between node and adjacent antinode is λ/4

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#### Solution

Amplitude of anitodes is maximum, A=±2a

A=2acos`(2pix)/lambda`

∴±2a=2acos`(2pix)/lambda`

∴cos`(2pix)/lambda`=±1

∴`(2pix)/lambda=0,pi,2pi.....`

or`(2pix)/lambda=Ppi`

∴`x=(Ppi)/2=P(lambda/2)........(where P=0,1,2......)`

For x=

`0,lambda/2,lambda,(3lambda)/2,.......`antinodes are produced.

Thus, distance between any two successive antinodes is `lambda/2`

Amplitude of nodes is zero, A=0

∴A=2acos`(2pix)/lambda`

∴0=2acos`(2pix)/lambda`

∴cos`(2pix)/lambda`=0

∴`(2pix)/lambda=pi/2, (3pi)/2,(5pi)/2........`

∴x=(2P-1)`lambda/4`...........(where P=1,2......)

For x=

`lambda/4,(3lambda)/4, (5lambda)/4 ...... `

Thus, distance between any two successive nodes is`lambda/2.`

The distance between node and adjacent anitnodes is`lambda/4`

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