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The path difference between the two wave...

The path difference between the two waves
`y_(1)=a_(1) sin(omega t-(2pi x)/(lambda)) and y(2)=a_(2) cos(omega t-(2pi x)/(lambda)+phi)` is

A

`lambda/(2pi) (phi)`

B

`lambda/(2pi) (phi+pi/2)`

C

`(2pi)/lambda (phi-pi/2)`

D

`(2pi)/lambda (phi)`

Text Solution

Verified by Experts

The correct Answer is:
B

As `y_(2)=a_(2) cos (omega t-(2pi x)/lambda +phi)`
`=a_(2) sin [pi/2+(omega t-(2pi x)/lambda +phi)]`
Compare it with `y_(1)=a_(1) sin (omega t-(2pi x)/lambda)`
Phase difference `=(pi/2 +phi)`
`:.` Path difference `=lambda/(2pi)(pi/2 +phi)`.
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MHTCET PREVIOUS YEAR PAPERS AND PRACTICE PAPERS-WAVE MOTION-Exercise 1
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