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(A) Given the displacement equation: $x = x_0 \cos(\omega t + \pi/4)$.The velocity $v$ is the derivative of displacement with respect to time: $v = \f

Vedclass · Accepted Answer

$A = x_0 \omega^2, \delta = 3\pi/4$

Vedclass · Answer

(A) Given the displacement equation: $x = x_0 \cos(\omega t + \pi/4)$.The velocity $v$ is the derivative of displacement with respect to time: $v = \frac{dx}{dt} = -x_0 \omega \sin(\omega t + \pi/4)$.The acceleration $a$ is the derivative of velocity with respect to time: $a = \frac{dv}{dt} = -x_0 \omega^2 \cos(\omega t + \pi/4)$.Using the trigonometric identity $-\cos(	heta) = \cos(	heta + \pi)$,we can rewrite the acceleration as:$a = x_0 \omega^2 \cos(\omega t + \pi/4 + \pi) = x_0 \omega^2 \cos(\omega t + 5\pi/4)$.Since $\cos(	heta)$ has a period of $2\pi$,$\cos(\omega t + 5\pi/4)$ is equivalent to $\cos(\omega t + 5\pi/4 - 2\pi) = \cos(\omega t - 3\pi/4)$.Comparing $a = x_0 \omega^2 \cos(\omega t - 3\pi/4)$ with $a = A \cos(\omega t + \delta)$,we get $A = x_0 \omega^2$ and $\delta = -3\pi/4$ or $5\pi/4$. Note: Checking the provided options,the correct phase shift corresponds to $3\pi/4$ if the initial phase was $-\pi/4$. Given the input equation $x = x_0 \cos(\omega t + \pi/4)$,the acceleration is $a = -x_0 \omega^2 \cos(\omega t + \pi/4) = x_0 \omega^2 \cos(\omega t + \pi/4 + \pi) = x_0 \omega^2 \cos(\omega t + 5\pi/4)$. This is equivalent to $x_0 \omega^2 \cos(\omega t - 3\pi/4)$. If the question intended $x = x_0 \cos(\omega t - \pi/4)$,then $a = x_0 \omega^2 \cos(\omega t - \pi/4 + \pi) = x_0 \omega^2 \cos(\omega t + 3\pi/4)$. Thus,$A = x_0 \omega^2$ and $\delta = 3\pi/4$.

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

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