Q. The motion of a particle executing SHM is given by$x = 0.01\sin{100p(t + 0.05)}$, where $x$ is in metre and time t is in second. The time period is
A) 0.2 s
B) 0.1 s
C) 0.02 s
D) 0.01 s
Explanation
(c) $x = 0.01\sin{100\pi}(t + 0.05)$
$= 0.01\sin{(100\pi t + 5\pi)}$
$\therefore$ Angular frequency $\omega = 100\pi = \frac{2\pi}{T}$
or $T = \frac{2}{100} = 0.02$s
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.