JEE MAIN - Physics (2022 - 27th June Evening Shift - No. 10)
The equation of a particle executing simple harmonic motion is given by $$x = \sin \pi \left( {t + {1 \over 3}} \right)m$$. At t = 1s, the speed of particle will be
(Given : $$\pi$$ = 3.14)
0 cm s$$-$$1
157 cm s$$-$$1
272 cm s$$-$$1
314 cm s$$-$$1
Explanation
$$x = \sin \left( {\pi t + {\pi \over 3}} \right)m$$
$$ \Rightarrow {{dx} \over {dt}} = \pi \cos \left( {\pi t + {\pi \over 3}} \right)$$
$$ = \pi \cos \left( {\pi + {\pi \over 3}} \right)$$ at $$t = 1\,s$$
$$ = - {\pi \over 2}$$ m/s
or $$\left| {{{dx} \over {dt}}} \right| = 157$$ cm/s
Comments (0)
