Maximum Weight Variation in Sinusoidally Moving Elevator

To find the maximum variation of the object's weight observed during the experiment, we need to analyze the acceleration of the elevator and how it affects the apparent weight of the object.

1. Position and Acceleration of the Elevator: The position $y(t)$ of the elevator as a function of time $t$ is given by: $y(t) = 8 \left[1 + \sin\left(\frac{2\pi t}{T}\right)\right]$

The angular frequency $\omega$ of the motion is: $\omega = \frac{2\pi}{T} = \frac{2\pi}{40\pi} = \frac{1}{20}\text{ rad/s} = 0.05\text{ rad/s}$

Differentiating $y(t)$ with respect to $t$ gives the velocity $v(t)$: $v(t) = \frac{dy}{dt} = 8\omega \cos(\omega t)$

Differentiating $v(t)$ with respect to $t$ gives the vertical acceleration $a(t)$ of the elevator: $a(t) = \frac{d^2y}{dt^2} = -8\omega^2 \sin(\omega t)$

2. Maximum Acceleration: The magnitude of the maximum acceleration $a_{\max}$ is: $a_{\max} = 8\omega^2 = 8 \times \left(\frac{1}{20}\right)^2 = \frac{8}{400} = 0.02\text{ m/s}^2$

3. Apparent Weight and Maximum Variation: The apparent weight $W(t)$ of the object measured by a weighing machine inside the elevator is given by: $W(t) = m(g + a(t))$

The maximum apparent weight observed is: $W_{\max} = m(g + a_{\max})$

The minimum apparent weight observed is: $W_{\min} = m(g - a_{\max})$

The maximum variation in the object's weight $\Delta W_{\max}$ is defined as the difference between the maximum and minimum values of the apparent weight: $\Delta W_{\max} = W_{\max} - W_{\min} = 2 m a_{\max}$

4. Calculation: Given $m = 50\text{ kg}$ and $a_{\max} = 0.02\text{ m/s}^2$: $\Delta W_{\max} = 2 \times 50 \times 0.02 = 2\text{ N}$

The maximum variation of the object's weight as observed in the experiment is 2.