JEE Advanced Physics | Max Variation in Weight in an Oscillating Elevator | Frame of Reference скачать в хорошем качестве

JEE Advanced Physics | Max Variation in Weight in an Oscillating Elevator | Frame of Reference

In this video, we solve a classic JEE Advanced Physics problem (2025 Paper 1 Q8) involving Pseudo Forces and Simple Harmonic Motion. We calculate the maximum variation in the observed weight of a 50 kg object inside an elevator with a time-varying height function. Problem Statement: A person inside an elevator performs a weighing experiment with an object of mass $50 \text{ kg}$. The height $y$ of the elevator varies with time $t$ as $y = 8 [1 + \sin(\frac{2\pi t}{T})]$, where $T = 40\pi \text{ s}$. Taking $g = 10 \text{ m/s}^2$, find the maximum variation of the object's weight (in N). JEE Advanced Inertial Frame of Reference Key Concepts Covered: Differentiating displacement to find acceleration ($a = \frac{d^2y}{dt^2}$). Understanding Apparent Weight in an accelerating frame: $W_{app} = m(g \pm a)$. Calculating the "Variation" in weight (Difference between $W_{max}$ and $W_{min}$). Steps to Solve: Identify the displacement equation $y(t)$. Differentiate twice to find the acceleration $a(t)$. Determine the maximum acceleration magnitude. Apply the formula for weight variation: $\Delta W = W_{max} - W_{min} = 2ma_{max}$.