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

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Question

(A) Given the displacement equation: $x = x_0 \cos(\omega t - \frac{\pi}{4})$.Velocity $v = \frac{dx}{dt} = -x_0 \omega \sin(\omega t - \frac{\pi}{4})

Vedclass · Accepted Answer

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

Vedclass · Answer

(A) Given the displacement equation: $x = x_0 \cos(\omega t - \frac{\pi}{4})$.Velocity $v = \frac{dx}{dt} = -x_0 \omega \sin(\omega t - \frac{\pi}{4})$.Acceleration $a = \frac{dv}{dt} = -x_0 \omega^2 \cos(\omega t - \frac{\pi}{4})$.Using the trigonometric identity $\cos(	heta + \pi) = -\cos(	heta)$,we can rewrite the acceleration as:$a = x_0 \omega^2 \cos(\omega t - \frac{\pi}{4} + \pi) = x_0 \omega^2 \cos(\omega t + \frac{3\pi}{4})$.Comparing this with the given form $a = A \cos(\omega t + \delta)$,we find:$A = x_0 \omega^2$ and $\delta = \frac{3\pi}{4}$.

$A$ point mass oscillates along the $x$-axis according to the law $x=x_0 \cos(\omega t - \frac{\pi}{4})$. If the acceleration of the particle is written as $a=A \cos(\omega t + \delta)$,then:

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