# Hence Show that the Distance Between Node and Adjacent Antinode Is λ/4 - Physics

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 islambda/2.

The distance between node and adjacent anitnodes islambda/4

Concept: Formation of Stationary Waves on String
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