A body oscillates with S.H.M. according to th | vedclass

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(B) The equation of motion is $x = A \cos(\omega t + \phi)$,where $A = 5.0 \, m$ and $\omega = 2 \pi \, rad \, s^{-1}$.The acceleration $a$ in $S.H.M.

Vedclass · Accepted Answer

$139.56$

Vedclass · Answer

(B) The equation of motion is $x = A \cos(\omega t + \phi)$,where $A = 5.0 \, m$ and $\omega = 2 \pi \, rad \, s^{-1}$.The acceleration $a$ in $S.H.M.$ is given by $a = -\omega^2 x$.First,find the displacement $x$ at $t = 1.5 \, s$:$x = 5 \cos(2 \pi 	imes 1.5 + \pi / 4) = 5 \cos(3 \pi + \pi / 4)$.Since $\cos(3 \pi + 	heta) = -\cos(	heta)$:$x = -5 \cos(\pi / 4) = -5 / \sqrt{2} \, m$.Now,calculate the acceleration:$a = -\omega^2 x = -(2 \pi)^2 	imes (-5 / \sqrt{2})$.$a = 4 \pi^2 	imes (5 / \sqrt{2}) = 4 	imes 9.8696 	imes 3.5355 \approx 139.56 \, m / s^2$.

$A$ body oscillates with $S.H.M.$ according to the equation $x = (5.0 \, m) \cos [(2 \pi \, rad \, s^{-1}) t + \pi / 4]$. At $t = 1.5 \, s$,its acceleration is ....... $m / s^2$.

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