A person sitting inside an elevator performs a weighing experiment with an object of mass 50 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 \(T=40 \pi \mathrm{~s}\). Taking acceleration due to gravity, \(g=10 \mathrm{~m} / \mathrm{s}^{2}\), the maximum variation of the object's weight (in N ) as observed in the experiment is ____.

JEE Advanced · 2025 · 18 May 2025 · Paper 1 · Laws of Motion

A person sitting inside an elevator performs a weighing experiment with an object of mass 50 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 [1 + sin (\frac{2 π t}{T})]

, where

T = 40 π s

. Taking acceleration due to gravity,

g = 10 m / s^{2}

, the maximum variation of the object's weight (in N ) as observed in the experiment is ____.

Answer: 2

The elevator undergoes SHM:

y = 8 + 8 sin \frac{2 π t}{T}

Amplitude A = 8 m, angular frequency

ω = \frac{2 π}{T} = \frac{2 π}{40 π} = \frac{1}{20}

rad/s

Maximum acceleration:

a_{ma x} = ω^{2} A = (\frac{1}{20})^{2} \times 8 = \frac{8}{400} = 0.02

m/s²

The weight variation is due to the vertical acceleration of the elevator:

Δ W = m \times 2 a_{ma x} = 50 \times 2 \times 0.02 = 2 N

The factor of 2 accounts for the full range from maximum upward to maximum downward acceleration.

JEE Advanced · 2025 · 18 May 2025 · Paper 1 · Numerical answer