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Q.

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 the velocity of particle 1 with respect to the velocity of particle 2 is

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a

$\frac{- π}{3}$

b

$\frac{π}{6}$

c

$\frac{- π}{6}$

d

$\frac{π}{3}$

answer is C.

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Detailed Solution

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$v_{1} = \frac{{dy}_{1}}{dt} = 0 .1 \times 100 πcos (100 πt + \frac{π}{3})$

$v_{2} = \frac{{dy}_{2}}{dt} = - 0 .1 πsinπt = 0 .1 πcos (πt + \frac{π}{2})$

Phase difference of velocity of first particle with respect to the velocity of 2^nd particle at t = 0 is

$Δφ = φ_{1} - φ_{2} = \frac{π}{3} - \frac{π}{2} = - \frac{π}{6} .$

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Two simple harmonic motions are represented by the equations y1=0.1sin100πt+π3 and y2=0.1cosπt. The phase difference of the velocity of particle 1 with respect to the velocity of particle 2 is