JEE MAIN - Physics (2005 - No. 27)
Two simple harmonic motions are represented by the equations $${y_1} = 0.1\,\sin \left( {100\pi t + {\pi \over 3}} \right)$$ 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
$${\pi \over 3}$$
$${{ - \pi } \over 6}$$
$${\pi \over 6}$$
$${{ - \pi } \over 3}$$
Explanation
$${v_1} = {{d{y_1}} \over {dt}} = 0.1 \times 100\pi \cos \left( {100\pi t + {\pi \over 3}} \right)$$
$${v_2} = {{d{y_2}} \over {dt}} = - 0.1\pi sin\pi t = 0.1\pi cos\left( {\pi t + {\pi \over 2}} \right)$$
$$\therefore$$ Phase diff. $$ = {\phi _1} - {\phi _2} = {\pi \over 3} - {\pi \over 2} = {{2\pi - 3\pi } \over 6} = {\pi \over 6}$$
$${v_2} = {{d{y_2}} \over {dt}} = - 0.1\pi sin\pi t = 0.1\pi cos\left( {\pi t + {\pi \over 2}} \right)$$
$$\therefore$$ Phase diff. $$ = {\phi _1} - {\phi _2} = {\pi \over 3} - {\pi \over 2} = {{2\pi - 3\pi } \over 6} = {\pi \over 6}$$
Comments (0)
