Question

Two simple harmonic motions are represented by the equations $y_1=0.1sin\left(100\pi t+\frac{\pi }{3}\right)$ and $y_2=0.1cos\left(100\pi t\right)$ . The phase difference of the velocity of a particle $1$ with respect to the velocity of the particle $2$ is

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

Answer 1

Answer: (C)

The velocity is given by the rate of change of displacement, i.e.,
$v_1=\frac{d{y}_1}{dt}=0.1\times 100\pi cos\left(100\pi t+\frac{\pi }{3}\right)$
$v_2=\frac{d{y}_2}{dt}=-0.1\pi sin\pi t=0.1\pi cos\left(\pi t+\frac{\pi }{2}\right)$
The phase difference of velocity of the first particle with respect to the velocity of $2^{nd}$ particle at $t=0$ will be,
$\Delta \phi ={\phi }_1-{\phi }_2=\frac{\pi }{3}-\frac{\pi }{2}=-\frac{\pi }{6}$