When two waves y _(1) =A sin (omega t + (pi)/(6)) and y _(2) = A cos (omega t) superpose, find amplitude of resultant wave.

When two waves y _(1) =A sin (omega t + (pi)/(6)) and y _(2) = A cos (

1.	When two waves y _(1) =A sin (omega t + (pi)/(6)) and y _(2) = A cos (omega t) superpose, find amplitude of resultant wave.
Answer» A `sqrt2A` `sqrt3A` `2A` Solution :Here `y _(1)= A sin (omega t + (pi)/(6))` `y _(2) =A cos (omega t)` `= A sin ((pi )/(2) + omega t ) ( because cos theta =sin ((pi)/(2) + theta ))` Phase different between above WAVES is, `RHO = ((pi)/(2) + omega t ) - (omega t + (pi )/(6)) = (pi)/(3)` Now, amplitude of resultant wave is, `A . = sqrt (A _(1) ^(2) + A _(2) ^(2) + 2 A _(1) A _(2) cos rho )` `therefore A.= sqrt (A ^(2) + A ^(2) + 2 A A cos ((pi)/(3)))` `therefore A.= sqrt (2A ^(2) + 2 A ^(2) ((1)/(2))) = sqrt ( 3 A ^(2)) = sqrt3 A`