Q. A point mass oscillates along the $x$-axis according to the law $x = x_{0}\ cos(\omega t - \frac{\pi}{4)}$. If the acceleration of the particle is written as:$a = A\ cos(\omega t + \delta)$, then
A) $A = x_{0},\delta = - \frac{\pi}{4}$
B) $A = x_{0}\omega^{2},\delta = \frac{\pi}{4}$
C) $A = x_{0}\omega^{2},\delta = - \frac{\pi}{4}$
D) $A = x_{0}\omega^{2},\delta = \frac{3\pi}{4}$
Explanation
(d) $x = x_{0}\cos\left( \omega t - \frac{\pi}{4} \right)$
Acceleration, $a = \frac{d^{2}x}{dt^{2}}$
$= - \omega^{2}x_{0}\cos\left( \omega t - \frac{\pi}{4} \right)$
$= - \omega^{2}x_{0}\cos\left( \omega t + \frac{3\pi}{4} \right)$
So, $A = \omega^{2}x_{0}$
and $\delta = \frac{3\pi}{4}$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.