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Phase difference between two waves y (1)...

Phase difference between two waves `y _(1) = A sin (omega t - kx) and y _(2)= A cos (omega t - kx) ` is ……

A

`pi/4`

B

`pi`

C

`pi/8`

D

`pi/2`

Text Solution

Verified by Experts

The correct Answer is:
D
`y _(1) =A sin (omega t - kx )`
` y _(2) = A cos (omega t - kx )`
`therefore y _(2) = A sin ((pi)/(2) + omega t - kx)`
`(because cos theta = sin ((pi )/(2) + theta ))`
Phase difference `= ((pi)/(2) + omega t - kx) - (omega t - kx)` br> `= (pi)/(2) + omega t - kx - omega t + kx`
`=(pi)/(2) rad`
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