Question 8

The height of the elevator is given by $y = 8 \left[ 1 + \sin \left( \frac{2\pi t}{T} \right) \right]$.

The velocity of the elevator is $v = \frac{dy}{dt} = 8 \left( \frac{2\pi}{T} \right) \cos \left( \frac{2\pi t}{T} \right)$.

The acceleration of the elevator is $a = \frac{dv}{dt} = -8 \left( \frac{2\pi}{T} \right)^2 \sin \left( \frac{2\pi t}{T} \right)$.

Given $T = 40\pi$ s, the angular frequency is $\omega = \frac{2\pi}{T} = \frac{2\pi}{40\pi} = 0.05 \text{ rad/s}$.

The maximum magnitude of acceleration is $a_{max} = 8 \omega^2 = 8 \times (0.05)^2 = 8 \times 0.0025 = 0.02 \text{ m/s}^2$.

The weight of the object as observed in the elevator (apparent weight) is $W = m(g + a)$.

The maximum weight observed is $W_{max} = m(g + a_{max})$ and the minimum weight observed is $W_{min} = m(g - a_{max})$.

The maximum variation in weight is the difference between these extremes: $\Delta W = W_{max} - W_{min} = 2 m a_{max}$ $\Delta W = 2 \times 50 \times 0.02 = 2 \text{ N}$