Two simple harmonic motions are represented by the equations y1=0.1 sin100πt+π3 and y2=0.1 cos100πt. The phase difference of the velocity of particle 1, with respect to the velocity of particle 2 is-

y1=0.1 sin100πt+π/3 ⇒ V1=0.1×100π cos100πt+π/3 and y2=0.1 cos100πt ⇒ V2=-0.1×100πsin 100πt or V2=10π sin100πt+π Now ϕ12=ϕ1-ϕ2=100πt+π3+π2-100πt+π =-π6rad.

Two simple harmonic motions are represented by the equations y1=0.1 sin100πt+π3 and y2=0.1 cos100πt. The phase difference of the velocity of particle 1, with respect to the velocity of particle 2 is-

y1=0.1 sin100πt+π/3 ⇒ V1=0.1×100π cos100πt+π/3 and y2=0.1 cos100πt ⇒ V2=-0.1×100πsin 100πt or V2=10π sin100πt+π Now ϕ12=ϕ1-ϕ2=100πt+π3+π2-100πt+π =-π6rad.

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

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$\frac{π}{3}$

$\frac{π}{6}$

$- \frac{π}{6}$

$- \frac{π}{3}$

answer is A.

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Detailed Solution

$y_{1} = 0.1 \sin (100 πt + π / 3) \Rightarrow V_{1} = 0.1 \times 100 π \cos (100 πt + π / 3) and y_{2} = 0.1 \cos 100 πt \Rightarrow V_{2} = - 0.1 \times 100 πsin 100 πt or V_{2} = 10 π \sin (100 πt + π) Now ϕ_{12} = ϕ_{1} - ϕ_{2} = [100 πt + \frac{π}{3} + \frac{π}{2}] - [100 πt + π] = - \frac{π}{6} rad .$