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Incident wave y= A sin (ax + bt+ pi/2) i...

Incident wave `y= A sin (ax + bt+ pi/2)` is reflected by an obstacle at x = 0 which reduces intensity of reflected wave by 36%. Due to superposition, the resulting wave consists of a standing wave and a travelling wave given by
`y= -1.6 sin ax sin bt + cA cos (bt + ax)`
where A, a, b and c are positive constants.
3. Position of second antinode is

A

`x = pi/(3a)`

B

`x = (3pi)/a`

C

`x = (3pi)/(2a)`

D

`x = (2pi)/(3a)`

Text Solution

Verified by Experts

The correct Answer is:
C
`A_x = -1.6 sin ax `
`:. X = 0` is a node `
` Second antinode is at a distance. `
`x = lambda/4 + lambda/2 = (3lambda)/4 = 3/4 (2pi)/k`
But, k=a
`:. x = (3pi)/(2a)`.
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