Q133 / 160 · TS EAMCET 2020 (10 Sep Shift 2) · Physics

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 $a=A \cos (\omega t-\delta)$, then

  • Correct answer
    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
    $A=x_0 \omega^2, \delta=\frac{3 \pi}{4}$

Solution

Displacement of particle is $x=x_0 \cos \left(\omega t-\frac{\pi}{4}\right)$ Velocity, $v=\frac{d x}{d t}$ $\Rightarrow v=-x_0 \omega \sin \left(\omega t-\frac{\pi}{4}\right)$ Acceleration, $a=\frac{d v}{d t}=-x_0 \omega^2 \cos \left(\omega t-\frac{\pi}{4}\right)$ $=x_0 \omega^2 \cdot \cos \left(\pi+\left(\omega t-\frac{\pi}{4}\right)\right)$ $=x_0 \omega^2 \cdot \cos \left(\omega t+\frac{3 \pi}{4}\right)$ $a=x_0 \omega^2 \cos \left(\omega t-\left(-\frac{3 \pi}{4}\right)\right)$ Comparing with $a=A \cos (\omega t-\delta)$, we have $A=x_0 \omega^2, \delta=-\frac{3 \pi}{4}$

Hence, the correct answer is option: 'A'.

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