Q 1 :

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(2πtT)]. where T=40π s. Taking acceleration due to gravity, g=10 m/s2, the maximum variation of the object's weight (in N) as observed in the experiment is ______                [2025]



(2)

Given y=8[1+sin(2πtT)]

ω=2πT=2π40π=120

With respect to elevator, variation in weight will be

ΔW=m(Δa)max

ΔW=m×2ω2A    [ amax=Aω2]

Here elevator is performing SHM

ΔW=2m(2πT)2×A

or,  ΔW=2×50(2π40π)2×8

ΔW=2×50×1400×8

  ΔW=800400=2N