A point mass oscillates along the X-axis acco | vedclass

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(A) The displacement of the particle is given by $x=x_0 \cos \left(\omega t-\frac{\pi}{4}\right)$.Velocity $v$ is the first derivative of displacement

Vedclass · Accepted Answer

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

Vedclass · Answer

(A) The displacement of the particle is given by $x=x_0 \cos \left(\omega t-\frac{\pi}{4}ight)$.Velocity $v$ is the first derivative of displacement with respect to time: $v = \frac{dx}{dt} = -x_0 \omega \sin \left(\omega t-\frac{\pi}{4}ight)$.Acceleration $a$ is the derivative of velocity with respect to time: $a = \frac{dv}{dt} = -x_0 \omega^2 \cos \left(\omega t-\frac{\pi}{4}ight)$.Using the trigonometric identity $\cos(	heta + \pi) = -\cos(	heta)$,we can rewrite the acceleration as:$a = x_0 \omega^2 \cos \left(\omega t - \frac{\pi}{4} + \piight) = x_0 \omega^2 \cos \left(\omega t + \frac{3\pi}{4}ight)$.To match the form $a = A \cos(\omega t - \delta)$,we write the phase as $\omega t - (-\frac{3\pi}{4})$.Thus,$A = x_0 \omega^2$ and $\delta = -\frac{3\pi}{4}$.

$A$ 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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