A person in an elevator weighs a 50 kg object. The height of the elevator is y = 8[1 + sin(2πt/T)] m, where T = 40π s. Taking g = 10 m s⁻², find the maximum variation of the object's observed weight in newtons.
Physics · Circular Motion
JEE Advanced 2025 Paper 1

Question

A person in an elevator weighs a 50 kg object. The height of the elevator is y = 8[1 + sin(2πt/T)] m, where T = 40π s. Taking g = 10 m s⁻², find the maximum variation of the object's observed weight in newtons.

Physics · Circular Motion

JEE Advanced 2025 Paper 1

G1 Physics · Accepted Answer

Angular frequency: Ω = 2π/T = 1/20 s⁻¹. Acceleration: a = d²y/dt² = −8Ω² sin(Ωt). Hence aₘₐₓ = 8(1/20)² = 0.02 m s⁻². Apparent weight N = m(g+a). The peak-to-peak variation is ΔNₘₐₓ = m[(g+aₘₐₓ) − (g−aₘₐₓ)] = 2maₘₐₓ = 2(50)(0.02) = 2 N.