Five sinusoidal waves have the same frequency 500 Hz but their amplitudes are in the ratio `2 : (1)/(2) : (1)/(2) : 1 : 1` and their phase angles `0, (pi)/(6), (pi)/(3), (pi)/(2)` and `pi` respectively. The phase angle of resultant wave obtained by the supersposition of these five waves is

Options

A. `30^(@)`
B. `45^(@)`
C. `60^(@)`
D. `90^(@)`

Answer

B

Explanation

`y _(1) = 2 A sin omega t`
`y_(2) = (A)/(2) sin ( omega t + (pi)/(6))`
`y_(3) = (A)/(2) sin ( omega t + (pi)/( 3))`
`y _(4) = A sin ( omega t (pi)/(2))`
`y_(5) = A sin ( omega t + pi)`
By phaser diagram ,
`tan phi = (PQ)/(OQ) = 1`
`phi = 45^(@)`
Alternatively : ` y = 2 A sin omega t + (A)/(2) ( sin omega cos 30^(@) + cos omega t sin 30^(@)) + (A)/(2) ( sin omega t cos 60^(@) + cos omega t sin 60^(@)) + A cos omega t - A sin omega t`
` = A' cos phi sin omega t + A' sin phi cos omega t`
where `A' cos phi = [ A + (A)/(4) ( sqrt (3) + 1) ]`
where `A' sin phi = [ A + (A)/(4) ( sqrt (3) + 1) ]`
`tan phi = 1`
`phi = 45 ^(@)`