Two simple harmonic motions are represented by the equations, $y_{1} = 0.1 \sin (100 πt + \frac{π}{3})$ and $y_{2} = 0.1 cosπt$ .The phase difference of velocity of particle 1 with respect to the velocity of particle 2 is

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$\frac{- π}{3}$

$\frac{π}{6}$

$\frac{π}{3}$

$\frac{- π}{6}$

answer is C.

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Detailed Solution

$\begin{array}{l} v_{1} = \frac{{dy}_{1}}{dt} = {Aω}_{1} \cos (ω_{1} t + ϕ_{1}) \\ v_{2} = \frac{{dy}_{2}}{dt} = A_{2} ω_{2} \sin (ω_{2} t + ϕ_{2}) Δϕ = ϕ_{1} - ϕ_{2} \end{array}$

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