The minimum phase difference between two simple harmonic oscillations, $y_{1} = \frac{1}{2} s i n ω t + \frac{\sqrt{3}}{2} c o s ω t, y_{2} = s i n ω t + c o s ω t$ , is:

\frac{π}{6}

rad

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\frac{7 π}{12}

rad

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\frac{π}{3}

rad

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\frac{π}{12}

rad

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Solution

The correct option is D $\frac{π}{12}$ rad
$y_{1} = \frac{1}{2} sin ω t + \frac{\sqrt{3}}{2} cos ω t$
$= A_{1} sin (ω t + ϕ_{1})$
$y_{2} = sin ω t + cos ω t$
$= A_{2} sin (ω t + ϕ_{2})$ where ,
$ϕ_{1} = {tan}^{- 1} (\frac{\sqrt{3} / 2}{1 / 2} = {tan}^{- 1} \sqrt{3} = \frac{π}{3}$
$ϕ_{2} = {tan}^{- 1} (1 / 1) = \frac{π}{4}$
Phase difference $= ϕ = ϕ_{1} - ϕ_{2} = \frac{π}{3} - \frac{π}{4} = \frac{π}{12}$

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Similar questions

y_{1} = \frac{1}{2} sin ω t + \frac{\sqrt{3}}{2} cos ω t

y_{2} = sin ω t + cos ω t

y_{1} = \frac{1}{2} s i m ω t + \frac{\sqrt{3}}{2} c o s ω t

y_{2} = s i m ω t + c o s ω t

π

0.4

π / 2

y_{1} = 8 cos ω t, y_{2} = 4 cos (ω t + \frac{π}{2}), y_{3} = 2 cos (ω t + π), y_{4} = cos (ω t + \frac{3 π}{2}),

y_{1} = \frac{1}{2} sin ω t + \frac{\sqrt{3}}{2} cos ω t

y_{2} = \frac{\sqrt{3}}{2} sin ω t + \frac{1}{2} cos ω t

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