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 superposition of these five waves is

Options

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

Answer

B

Explanation

Here we are asked to find out phase angle which is also called phase constant. It is the phase at `t =0.` Hence let us consider superposition of five given waves at time `t =0.`
Now,
`y_(1) = 2 A sin (omega t ) =0 (because t =0)`
`y _(2) = A/2 sin (omega t + 30^(@)) =A/2 sin 30^(@) =A/4`
`y _(3) = A/2 sin (omega t + 60^(@)) = A/2 sin 60^(@) = (sqrt3A)/(4)`
`y _(2) =A sin (omega t + 90^(@)) = A sin 90^(@) =A`
`y _(5) = A sin (omega t + 180^(@)) = A sin 180^(@) =0`
Now, `y =y _(1) + y _(2) + y _(3) + y _(4) + y_(5)`
`= 0 (A)/(4) + (sqrt3A)/(4) + A + 0`
`= (5A + sqrt3A)/(4) = (A)/(4) (5 + sqrt3) ...(1)`
Now, (at t=0)
`x _(1) = 2 A cos (omega t + 0^(@)) = 2 A cos 0^(@) = 2 A`
`x _(2) = (A)/(2) cos (omega t + 30^(@)) = A/2 cos 30^(@) = (sqrt3A)/(4)`
`x _(3) = A/2 cos (omega t + 60^(@)) = A/2 cos 60^(@) = A/4 `
`x _(4) = A cos (omega t + 90^(@)) = A cos 90^(@) = 0`
`x _(5) = A cos (omega t + 180^(@)) = A cos 180^(@) =-A`
Now `x = x _(1) + x _(2) + x _(3) + x _(4) + x_(5)`
` =2A + (sqrt3A)/(4) + (A)/(4) + 0 -A`
`therefore x =A + (sqrt3A)/(4) + (A)/(4) = (A)/(4) (5 + sqrt3) " "...(2)`
Now, if phase angle of resultant wave at the point of superpositon of above five waves is `phi _(R)` then,
`tan phi _(R) = (y)/(x) =1`
`(because` Here y = x from equations (1) and (2))
`therefore phi _(R) = 45^(@)`