Q. 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
NTA AbhyasNTA Abhyas 2022
Solution:
The velocity is given by the rate of change of displacement, i.e.,
$v_{1}=\frac{d y_{1}}{d t}=0.1\times 100\pi cos\left(100 \pi t + \frac{\pi }{3}\right)$
$v_{2}=\frac{d y_{2}}{d t}=-0.1\pi sin\pi t=0.1\pi cos\left(\pi t + \frac{\pi }{2}\right)$
The phase difference of velocity of the first particle with respect to the velocity of $2^{nd }$ particle at $t=0$ will be,
$\Delta \varphi=\varphi_{1}-\varphi_{2}=\frac{\pi }{3}-\frac{\pi }{2}=-\frac{\pi }{6}$
$v_{1}=\frac{d y_{1}}{d t}=0.1\times 100\pi cos\left(100 \pi t + \frac{\pi }{3}\right)$
$v_{2}=\frac{d y_{2}}{d t}=-0.1\pi sin\pi t=0.1\pi cos\left(\pi t + \frac{\pi }{2}\right)$
The phase difference of velocity of the first particle with respect to the velocity of $2^{nd }$ particle at $t=0$ will be,
$\Delta \varphi=\varphi_{1}-\varphi_{2}=\frac{\pi }{3}-\frac{\pi }{2}=-\frac{\pi }{6}$