The user wants to find the maximum variation in the apparent weight of an object inside an elevator whose motion is described by a sinusoidal function.

Step-by-step Solution:

1. Analyze the given motion of the elevator. The height of the elevator from the ground is given as a function of time $t$: $y(t) = 8\left[1 + \sin\left(\frac{2\pi t}{T}\right)\right]$ where $T = 40\pi$ s. This can be rewritten as: $y(t) = 8 + 8\sin(\omega t)$ with angular frequency $\omega = \frac{2\pi}{T} = \frac{2\pi}{40\pi} = \frac{1}{20} \text{ rad/s}$.

2. Calculate the acceleration of the elevator. The apparent weight depends on the acceleration of the elevator. We find the acceleration, $a(t)$, by differentiating the position function $y(t)$ twice with respect to time.

   First, find the velocity $v(t)$: $v(t) = \frac{dy}{dt} = \frac{d}{dt}\left(8 + 8\sin(\omega t)\right) = 8\omega\cos(\omega t)$

   Next, find the acceleration $a(t)$: $a(t) = \frac{dv}{dt} = \frac{d}{dt}\left(8\omega\cos(\omega t)\right) = -8\omega^2\sin(\omega t)$

3. Find the maximum and minimum acceleration. The acceleration $a(t)$ varies sinusoidally. Its amplitude is $a_{amp} = 8\omega^2$. Let's calculate this value: $\omega^2 = \left(\frac{1}{20}\right)^2 = \frac{1}{400} \text{ (rad/s)}^2$ $a_{amp} = 8 \times \frac{1}{400} = \frac{8}{400} = \frac{1}{50} \text{ m/s}^2$

   The acceleration function is $a(t) = -\frac{1}{50}\sin(\omega t)$. The value of $\sin(\omega t)$ ranges from -1 to +1.

   • The maximum upward acceleration, $a_{max}$, occurs when $\sin(\omega t) = -1$: $a_{max} = -\frac{1}{50}(-1) = +\frac{1}{50} \text{ m/s}^2$
   • The minimum acceleration (maximum downward acceleration), $a_{min}$, occurs when $\sin(\omega t) = +1$: $a_{min} = -\frac{1}{50}(+1) = -\frac{1}{50} \text{ m/s}^2$

4. Calculate the apparent weight. The apparent weight, $W_{app}$, is the normal force exerted by the weighing scale on the object. From Newton's second law, considering the upward direction as positive: $W_{app} - mg = ma$ $W_{app} = m(g + a)$ where $m = 50$ kg and $g = 10$ m/s².

5. Calculate the maximum and minimum apparent weights.

   • The maximum apparent weight, $W_{max}$, occurs when the acceleration is maximum ($a_{max}$): $W_{max} = m(g + a_{max}) = 50\left(10 + \frac{1}{50}\right) = 50 \times 10 + 50 \times \frac{1}{50} = 500 + 1 = 501 \text{ N}$
   • The minimum apparent weight, $W_{min}$, occurs when the acceleration is minimum ($a_{min}$): $W_{min} = m(g + a_{min}) = 50\left(10 - \frac{1}{50}\right) = 50 \times 10 - 50 \times \frac{1}{50} = 500 - 1 = 499 \text{ N}$

6. Find the maximum variation of the weight. The maximum variation is the difference between the maximum and minimum apparent weights. $\Delta W = W_{max} - W_{min} = 501 - 499 = 2 \text{ N}$

   Alternatively, the variation can be calculated as: $\Delta W = W_{max} - W_{min} = m(g+a_{max}) - m(g+a_{min}) = m(a_{max} - a_{min})$ $\Delta W = 50\left(\frac{1}{50} - \left(-\frac{1}{50}\right)\right) = 50\left(\frac{2}{50}\right) = 2 \text{ N}$

Final Answer: The maximum variation of the object's weight is 2 N.