If in a stationary wave the amplitude corresponding to antinode is 4cm then the amplitude corresponding to a particle of medium located exactly at midway between node and antinode is
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The amplitude corresponding to particle is 21 cm
Explanation:
A'=2ASin kx
K = 2π/λ
x= λ/8 [Since the distance between node and anti-node = λ/2. Therefore distance between amplitude and 0 = λ/4 and distance between mid points of these two points= λ/8]
2A=4 [given]
4Sin [2π/λ × λ/8] 4 sin[π/4]= 4/21/2 = 2 x 21/2 cm
= 42/2 cm = 21 cm
The amplitude corresponding to particle is 21 cm
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