The phase shift of the resultant $y = 3 sin ω t + 4 cos ω t$ is:

Question

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Answer 1

The equations can be split up as $y_{1} = 3 s i n (ω t)$ and $y_{2} = 4 s i n (\frac{π}{2} - ω t)$ .
The phase shift is given by, $δ = tan^{- 1} (\frac{a _{1} s i n α _{1} + a _{2} s i n α _{2}}{a _{1} c o s α _{1} + a _{2} c o s α _{2}})$ . Substituting the given values, we get answer as $53.1 3^{0}$ . Hence, option B is correct.

Answer 2

The equations can be split up as

y_{1} = 3 s i n (ω t)

and

y_{2} = 4 s i n (\frac{π}{2} - ω t)

.
The phase shift is given by,

δ = tan^{- 1} (\frac{a _{1} s i n α _{1} + a _{2} s i n α _{2}}{a _{1} c o s α _{1} + a _{2} c o s α _{2}})

. Substituting the given values, we get answer as

53.1 3^{0}

. Hence, option B is correct.