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(D) The height of the elevator is given by $y = 8 + 8 \sin(\frac{2 \pi t}{T})$.Taking the second derivative with respect to time $t$,we find the accel

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$2$

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(D) The height of the elevator is given by $y = 8 + 8 \sin(\frac{2 \pi t}{T})$.Taking the second derivative with respect to time $t$,we find the acceleration of the elevator: $a = \frac{d^2y}{dt^2} = -8(\frac{2 \pi}{T})^2 \sin(\frac{2 \pi t}{T})$.The apparent weight of the object is $W' = m(g + a)$.The variation in weight is $\Delta W = m \cdot |a|$.The maximum acceleration is $a_{\max} = A \omega^2$,where $A = 8 \ m$ and $\omega = \frac{2 \pi}{T} = \frac{2 \pi}{40 \pi} = 0.05 \ rad/s$.Thus,$a_{\max} = 8 	imes (0.05)^2 = 8 	imes 0.0025 = 0.02 \ m/s^2$.The variation in weight is $\Delta W = m \cdot a_{\max} = 50 	imes 0.02 = 1 \ N$.However,considering the full range of oscillation (from $+a_{\max}$ to $-a_{\max}$),the total variation is $2 	imes m 	imes a_{\max} = 2 	imes 50 	imes 0.02 = 2 \ N$.

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