Q.8
A person sitting inside an elevator performs a weighing experiment with an object of mass 50 kg. 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=40πs.

Taking the acceleration due to gravity g=10m/s2, the maximum variation of the object’s weight (in N) as observed in the experiment is ______.

Answer: 2


Solution:

1. Understanding the Problem

  • An elevator moves vertically with its height y described by a sinusoidal function:

    y(t)=8[1+sin(2πtT)],T=40πs.
  • The object has a mass m=50kg, and g=10m/s2.

  • We need to find the maximum variation in the object's apparent weight as observed on a scale inside the elevator.

2. Key Physics Concepts

  • Apparent Weight: The normal force N exerted by the scale, which equals the object's apparent weight.

    • When the elevator accelerates upward, N=m(g+a).

    • When the elevator accelerates downward, N=m(ga).

  • The variation in weight is due to the elevator's acceleration a, derived from the second time derivative of y(t).

3. Derive the Elevator's Acceleration

Given:

y(t)=8+8sin(2πtT).

  • First Derivative (Velocity):

    v(t)=dydt=82πTcos(2πtT).
  • Second Derivative (Acceleration):

    a(t)=dvdt=8(2πT)2sin(2πtT).

Substitute T=40π:

a(t)=8(2π40π)2sin(2πt40π)=8(120)2sin(t20).a(t)=81400sin(t20)=150sin(t20)m/s2.

4. Determine Apparent Weight

The apparent weight W varies with the acceleration a(t):

W=m(g+a(t))=50(10150sin(t20)).

The maximum variation occurs when sin(t20) reaches its extreme values (±1):

  • Maximum Weight:

    Wmax=50(10+150)=501N.
  • Minimum Weight:

    Wmin=50(10150)=499N.
  • Variation in Weight:

    ΔW=WmaxWmin=501499=2N.

5. Final Answer

The maximum variation of the object’s weight is:

\boxed{2}

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