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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
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The correct answer is:
$A=x_0 \omega^2, \delta=\frac{-3 \pi}{4}$
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
$A=x_0 \omega^2, \delta=-\frac{3 \pi}{4}$
$x=x_0 \cos \left(\omega t-\frac{\pi}{4}\right)$
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
$A=x_0 \omega^2, \delta=-\frac{3 \pi}{4}$
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