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(D) The displacement of the particle is given by $x = x_0 \sin \left(\omega t - \frac{\pi}{6}\right)$.Acceleration $a$ is given by $a = \frac{d^2x}{dt

Vedclass · Accepted Answer

$A = x_0 \omega^2, \delta = \frac{5\pi}{6}$

Vedclass · Answer

(D) The displacement of the particle is given by $x = x_0 \sin \left(\omega t - \frac{\pi}{6}ight)$.Acceleration $a$ is given by $a = \frac{d^2x}{dt^2} = -\omega^2 x$.Substituting the expression for $x$:$a = -\omega^2 x_0 \sin \left(\omega t - \frac{\pi}{6}ight)$.Using the trigonometric identity $-\sin(	heta) = \sin(	heta + \pi)$:$a = \omega^2 x_0 \sin \left(\omega t - \frac{\pi}{6} + \piight)$.$a = \omega^2 x_0 \sin \left(\omega t + \frac{5\pi}{6}ight)$.Comparing this with $a = A \sin(\omega t + \delta)$,we get $A = x_0 \omega^2$ and $\delta = \frac{5\pi}{6}$.

$A$ point mass oscillates along the $x$-axis according to $x = x_0 \sin \left(\omega t - \frac{\pi}{6}\right)$. If the acceleration of the point mass is written as $a = A \sin (\omega t + \delta)$,then:

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