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Question
Show that the distance between two successive nodes or antinodes is λ/2.
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Solution
Condition for node:
Nodes are the points of minimum displacement. This is possible if the amplitude is minimum (zero), i.e.,
`2a cos (2 pi x)/lambda = 0`, or
`cos (2 pi x)/lambda = 0`, or
`(2 pi x)/lambda = pi/2, (3 pi)/2, (5 pi)/2, ....`
∴ `x = lambda/4, (3 lambda)/4, (5 lambda)/4, ....`
i.e., `x = (2 p - 1) lambda/4` where p = 1, 2, 3, ....
The distance between two successive nodes is `lambda/2`.
Condition for antinode:
Antinodes are the points of maximum displacement
i.e., A = ± 2a
∴ `2a cos (2 pi x)/lambda = +-2a`
or, `cos (2 pi x)/lambda = +-1`
∴ `(2 pi x)/lambda = 0, pi, 2pi, 3pi, ....`
or, `x = 0, lambda/2, lambda,(3 lambda)/2, ....`
i.e., `x = (lambda p)/2` where p = 0, 1, 2, 3, ....
The distance between two successive antinodes is `lambda/2`. Nodes and antinodes are formed 2 alternately. Therefore, the distance between a node and an adjacent antinode is `lambda/4`.
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