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