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

Options

A. A
B. `sqrt2A`
C. `sqrt3A`
D. `2A`

Answer

C

Explanation

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`