Two simple harmonic motions are represented by the equation $y_{1}=0.1 \sin \left(100 \pi t+\frac{\pi}{3}\right)$ and $y_{2}=0.1 \cos \pi t$. The phase difference of the velocity of particle 1 w.r.t. the velocity of the particle 2 is

JEE Main · 2005 · 08 May 2005 · Offline · Elasticity & SHM

Two simple harmonic motions are represented by the equation

y_{1} = 0.1 sin (100 π t + \frac{π}{3})

and

y_{2} = 0.1 cos π t

. The phase difference of the velocity of particle 1 w.r.t. the velocity of the particle 2 is

Answer: (A)

Differentiate each displacement to get the velocity and compare the phase constants. From

y_{1} = 0.1 sin (100 π t + \frac{π}{3})

the velocity goes as

cos (100 π t + \frac{π}{3})

, with phase

100 π t + \frac{π}{3}

. From

y_{2} = 0.1 cos (π t)

the velocity is

- 0.1 π sin (π t)

, which goes as

cos (π t + \frac{π}{2})

, with phase

π t + \frac{π}{2}

. Comparing the constant parts, the first velocity carries

\frac{π}{3}

and the second

\frac{π}{2}

, so particle

1

's velocity leads particle

2

's by

\frac{π}{3} - \frac{π}{2} = - \frac{π}{6}

JEE Main · 2005 · 08 May 2005 · Offline · Single correct option