Q. The SHM of a particle is given by $x(t) = 5\ \cos\ \left( 2\pi t + \frac{\pi}{4} \right)$ in MKS units. Calculate the displacement and the magnitude of acceleration of the particle at $t = 1.5\ $s.
A) $- 3.0\ m,100\ \frac{m}{s^{2}}$
B) $+ 2.54\ m,\ 200\ \frac{m}{s^{2}}$
C) $- 3.54\ m,\ 140\ \frac{m}{s^{2}}$
D) $+ 3.55\ m,\ 120\ \frac{m}{s^{2}}$
Explanation
(c) Displacement $x(t) = 5\cos\left( 2\pi \times \frac{3}{2} + \frac{\pi}{4} \right)\ $
= $5\cos\left( \frac{13\pi}{4} \right) = \ - 3.5$ m
$y = 5\cos\left( 2\pi t + \frac{\pi}{4} \right)$
$\therefore$Velocity $(v) = \frac{dy}{dt} = \ - 10\pi\sin\left( 2\pi t + \frac{\pi}{4} \right)$
$\therefore$Acceleration $a = \frac{dv}{dt} = - 20\ \pi^{2}\cos\left( 2\pi t + \frac{\pi}{4} \right)$
$= \ - 20\ \pi^{2}\cos\left( 2\pi t\frac{3}{2} + \frac{\pi}{4} \right)$
$= 20\ \pi^{2}\cos{\frac{13\pi}{4} = 140\ \frac{m}{s^{2}}}\ $
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.