Question
A point mass oscillates along the $$x$$-axis according to the law $$x = {x_0}\cos \left( {\omega t - \frac{\pi }{4}} \right).$$ If the acceleration of the particle is written as $$a = A\cos \left( {\omega t + \delta } \right),$$ then
A.
$$A = {x_0}{\omega ^2},\delta = \frac{{3\pi }}{4}$$
B.
$$A = {x_0},\delta = \frac{{ - \pi }}{4}$$
C.
$$A = {x_0}{\omega ^2},\delta = \frac{\pi }{4}$$
D.
$$A = {x_0}{\omega ^2},\delta = \frac{{ - \pi }}{4}$$
Answer :
$$A = {x_0}{\omega ^2},\delta = \frac{{3\pi }}{4}$$
Solution :
$$\eqalign{
& {\text{Here,}} \cr
& \,x = {x_0}\cos \left( {\omega t - \frac{\pi }{4}} \right) \cr
& \therefore {\text{Velocity,}}\,v = \frac{{dx}}{{dt}} = - {x_0}\omega \sin \left( {\omega t - \frac{\pi }{4}} \right) \cr
& {\text{Acceleration,}} \cr
& a = \frac{{dv}}{{dt}} = - {x_0}{\omega ^2}\cos \left( {\omega t - \frac{\pi }{4}} \right) \cr
& = {x_0}{\omega ^2}\cos \left[ {\pi + \left( {\omega t - \frac{\pi }{4}} \right)} \right] \cr
& = {x_0}{\omega ^2}\cos \left( {\omega t + \frac{{3\pi }}{4}} \right)\,......\left( 1 \right) \cr
& {\text{Acceleration,}}\,a = A\cos \left( {\omega t + \delta } \right)......\left( 2 \right) \cr} $$
Comparing the two equations, we get
$$A = {x_0}{\omega ^2}\,{\text{and}}\,\delta = \frac{{3\pi }}{4}$$