Question 160: Oscillations

Question

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

Accepted Answer

Here,
$$x = {x_0}\cos \left( {\omega t - \pi /4} ight)$$
$$	herefore$$ Velocity, $$v = {{dx} \over {dt}} =  - {x_0}\omega \sin \left( {\omega t - {\pi  \over 4}} ight)$$
Acceleration,
$$a = {{dv} \over {dt}} =  - {x_0}{\omega ^2}\cos \left( {\omega t - {\pi  \over 4}} ight)$$
$$ = {x_0}{\omega ^2}\cos \left[ {\pi  + \left( {\omega t - {\pi  \over 4}} ight)} ight]$$
$$ = {x_0}{\omega ^2}\cos \left( {\omega t + {{3\pi } \over 4}} ight)$$ $$\,\,\,\,\,\,\,\,\,\,\,...\left( 1 ight)$$
Acceleration, $$a = A\cos \left( {\omega t + \delta } ight)$$ $$\,\,\,\,\,\,\,\,\,\,\,...\left( 2 ight)$$
Comparing the two equations, we get
$$A = {x_0}{\omega ^2}$$ and $$\delta  = {{3\pi } \over 4}.$$

Answer

$$A = {x_0}{\omega ^2},\,\,\delta  = 3\pi /4$$

Answer

$$A = {x_0},\,\,\delta  =  - \pi /4$$

Answer

$$A = {x_0}{\omega ^2},\,\,\delta  = \pi /4$$

Answer

$$A = {x_0}{\omega ^2},\,\,\delta  =  - \pi /4$$

Oscillations

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

Choose the correct answer: