Two simple harmonic motions are represented by the equations y1=0.1 sin 100πt+π3 and y2=0.1 cos πt.The phase difference of the velocity of particle l, with respect to the velocity of particle 2 is :

v1=dy1dt=0.1×100 π cos 100πt+π3 =10 π cos 100 πt+π3 v2=dy2dt=0.1×100π sin (100 πt)=10 πcosπ2+100πt ∴ Phase difference =π3-π2=-π6

Two simple harmonic motions are represented by the equations y1=0.1 sin 100πt+π3 and y2=0.1 cos πt.The phase difference of the velocity of particle l, with respect to the velocity of particle 2 is :

v1=dy1dt=0.1×100 π cos 100πt+π3 =10 π cos 100 πt+π3 v2=dy2dt=0.1×100π sin (100 πt)=10 πcosπ2+100πt ∴ Phase difference =π3-π2=-π6

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

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$\frac{- π}{6}$

$\frac{π}{3}$

$\frac{- π}{3}$

$\frac{π}{6}$

answer is A.

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Detailed Solution

$v_{1} = \frac{{dy}_{1}}{dt} = 0.1 \times 100 π \cos (100 πt + \frac{π}{3}) = 10 π \cos (100 πt + \frac{π}{3}) v_{2} = \frac{{dy}_{2}}{dt} = 0.1 \times 100 π \sin (100 πt) = 10 πcos (\frac{π}{2} + 100 πt) ∴ Phase difference = \frac{π}{3} - \frac{π}{2} = - \frac{π}{6}$