Question

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

• A. $\frac{- \pi}{3}$
• B. $\frac{\pi}{6}$
• C. $\frac{- \pi}{6}$
• D. $\frac{\pi}{3}$

Answer 1

Answer: (C)

$\frac{- \pi}{6}$

Answer 2

$v_{1} = \frac{dy_{1}}{dt} = 0.1 \times 100\pi\cos\left( 100\pi t + \frac{\pi}{3} \right)$

$v_{2} = \frac{dy_{2}}{dt} = - 0.1\pi\sin\pi t = 0.1\pi\cos\left( \pi t + \frac{\pi}{2} \right)$

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

$\Delta\varphi = \varphi_{1} - \varphi_{2} = \frac{\pi}{3} - \frac{\pi}{2} = - \frac{\pi}{6}.$