HSC Science (Electronics) 12th Board ExamMaharashtra State Board
Account
It's free!

Share

Books Shortlist

# Solution - Hence Show that the Distance Between Node and Adjacent Antinode Is λ/4 - HSC Science (Electronics) 12th Board Exam - Physics

ConceptFormation of Stationary Waves on String

#### 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 islambda/2.

The distance between node and adjacent anitnodes islambda/4

Is there an error in this question or solution?

#### Video TutorialsVIEW ALL [1]

Solution for question: Hence Show that the Distance Between Node and Adjacent Antinode Is λ/4 concept: Formation of Stationary Waves on String. For the courses HSC Science (Electronics), HSC Science (Computer Science), HSC Science (General)
S