JEE MAIN - Physics (2007 - No. 25)
A point mass oscillates along the $$x$$-axis according to the law $$x = {x_0}\,\cos \left( {\omega t - \pi /4} \right).$$ If the acceleration of the particle is written as $$a = A\,\cos \left( {\omega t + \delta } \right),$$ then
$$A = {x_0}{\omega ^2},\,\,\delta = 3\pi /4$$
$$A = {x_0},\,\,\delta = - \pi /4$$
$$A = {x_0}{\omega ^2},\,\,\delta = \pi /4$$
$$A = {x_0}{\omega ^2},\,\,\delta = - \pi /4$$
Explanation
Here,
$$x = {x_0}\cos \left( {\omega t - \pi /4} \right)$$
$$\therefore$$ Velocity, $$v = {{dx} \over {dt}} = - {x_0}\omega \sin \left( {\omega t - {\pi \over 4}} \right)$$
Acceleration,
$$a = {{dv} \over {dt}} = - {x_0}{\omega ^2}\cos \left( {\omega t - {\pi \over 4}} \right)$$
$$ = {x_0}{\omega ^2}\cos \left[ {\pi + \left( {\omega t - {\pi \over 4}} \right)} \right]$$
$$ = {x_0}{\omega ^2}\cos \left( {\omega t + {{3\pi } \over 4}} \right)$$ $$\,\,\,\,\,\,\,\,\,\,\,...\left( 1 \right)$$
Acceleration, $$a = A\cos \left( {\omega t + \delta } \right)$$ $$\,\,\,\,\,\,\,\,\,\,\,...\left( 2 \right)$$
Comparing the two equations, we get
$$A = {x_0}{\omega ^2}$$ and $$\delta = {{3\pi } \over 4}.$$
$$x = {x_0}\cos \left( {\omega t - \pi /4} \right)$$
$$\therefore$$ Velocity, $$v = {{dx} \over {dt}} = - {x_0}\omega \sin \left( {\omega t - {\pi \over 4}} \right)$$
Acceleration,
$$a = {{dv} \over {dt}} = - {x_0}{\omega ^2}\cos \left( {\omega t - {\pi \over 4}} \right)$$
$$ = {x_0}{\omega ^2}\cos \left[ {\pi + \left( {\omega t - {\pi \over 4}} \right)} \right]$$
$$ = {x_0}{\omega ^2}\cos \left( {\omega t + {{3\pi } \over 4}} \right)$$ $$\,\,\,\,\,\,\,\,\,\,\,...\left( 1 \right)$$
Acceleration, $$a = A\cos \left( {\omega t + \delta } \right)$$ $$\,\,\,\,\,\,\,\,\,\,\,...\left( 2 \right)$$
Comparing the two equations, we get
$$A = {x_0}{\omega ^2}$$ and $$\delta = {{3\pi } \over 4}.$$
Comments (0)
