Question

A person sitting inside an elevator performs a weighing experiment with an object of mass $ 50 \, \text{kg} $. Suppose that the variation of the height $ y $ (in m) of the elevator, from the ground, with time $ t $ (in s) is given by $$ y = 8\left[1 + \sin\left(\frac{2\pi t}{T}\right)\right], $$ where $ T = 40\pi \, \text{s} $. Taking acceleration due to gravity, $ g = 10 \, \text{m/s}^2 $, the maximum variation of the object's weight (in N) as observed in the experiment is ____.

Hint:

To compute the apparent weight in a non-inertial frame such as an elevator, use the effective acceleration \( g \pm a \), where \( a \) is the vertical acceleration of the frame. Always differentiate the position function twice to get acceleration.

Answer 1

Correct Answer: 1

Step 1: Understand the motion of the elevator The vertical position of the elevator is given by: \[ y(t) = 8\left[1 + \sin\left(\frac{2\pi t}{T}\right)\right] \] This is a sinusoidal function, indicating vertical oscillatory motion. 
Step 2: Find the acceleration of the elevator Acceleration is the second derivative of position with respect to time: \[ a(t) = \frac{d^2 y}{dt^2} \] First derivative: \[ \frac{dy}{dt} = 8 \cdot \cos\left(\frac{2\pi t}{T}\right) \cdot \left(\frac{2\pi}{T}\right) \] Second derivative: \[ \frac{d^2 y}{dt^2} = -8 \cdot \sin\left(\frac{2\pi t}{T}\right) \cdot \left(\frac{2\pi}{T}\right)^2 \] Substitute \( T = 40\pi \): \[ a(t) = -8 \cdot \left(\frac{2\pi}{40\pi}\right)^2 \cdot \sin\left(\frac{2\pi t}{40\pi}\right) = -8 \cdot \left(\frac{1}{20}\right)^2 \cdot \sin\left(\frac{t}{20}\right) = -\frac{8}{400} \cdot \sin\left(\frac{t}{20}\right) = -0.02 \cdot \sin\left(\frac{t}{20}\right) \, \text{m/s}^2 \] 
Step 3: Maximum acceleration The sine function has maximum absolute value 1, so: \[ a_{\text{max}} = 0.02 \, \text{m/s}^2 \] 
Step 4: Maximum variation in apparent weight Apparent weight in an accelerating elevator is given by: \[ W_{\text{apparent}} = m(g + a) \] Thus, the variation in apparent weight is: \[ \Delta W = m \cdot a_{\text{max}} = 50 \cdot 0.02 = 1 \, \text{N} \]