A 0.10, kg block oscillates back and forth al | vedclass

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Question

(A) The displacement equation is given by $x = A \cos(\omega t + \phi)$,where $A = 10\, cm = 0.10\, m$ and $\omega = 10\, rad/s$.The acceleration $a$

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$10\, m/s^2$

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(A) The displacement equation is given by $x = A \cos(\omega t + \phi)$,where $A = 10\, cm = 0.10\, m$ and $\omega = 10\, rad/s$.The acceleration $a$ of a particle in simple harmonic motion is given by $a = -\omega^2 x$.The maximum acceleration $a_{\max}$ is given by the formula $a_{\max} = \omega^2 A$.Substituting the given values: $a_{\max} = (10\, rad/s)^2 	imes (0.10\, m) = 100 	imes 0.10 = 10\, m/s^2$.Therefore,the maximum acceleration experienced by the block is $10\, m/s^2$.

$A$ $0.10\, kg$ block oscillates back and forth along a horizontal surface. Its displacement from the origin is given by: $x = (10\,cm)\cos [(10\,rad/s)\,t + \pi /2\,rad]$. What is the maximum acceleration experienced by the block?

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