A person sitting inside an elevator performs a weighing experiment with an object of mass 50 kg. Sup

Q: Why is the variation $2ma_{\text{max}}$ and not just $ma_{\text{max}}$?

The variation is the difference between the maximum weight (when accelerating upwards) and the minimum weight (when accelerating downwards). So, $m(g+a) - m(g-a) = 2ma$.

Q: What does the $8$ outside the bracket in the position equation represent?

It shifts the equilibrium position of the elevator to $y = 8$ meters above the ground. It doesn't affect the acceleration or the variation in weight.

Q: Why did we differentiate the position equation twice?

Weight variation depends on the normal force, which depends on acceleration. Acceleration is the second derivative of position with respect to time.

Q: Does the mass of the elevator matter in this calculation?

No, only the mass of the object on the weighing scale matters for its apparent weight. The elevator's mass would only matter if we were calculating the tension in the elevator cable.

Q: What if the elevator was moving with a constant velocity?

If velocity is constant, acceleration is zero. The apparent weight would just be $mg$ at all times, and the variation would be exactly zero.

Question

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 $y = 8\left[1 + \sin\left(\frac{2\pi t}{T}ight)ight]$, where $T = 40\pi$ s. Taking acceleration due to gravity, $g = 10 	ext{ m/s}^2$, the maximum variation of the object's weight (in N) as observed in the experiment is _______.

Sigma Percentile · Accepted Answer

The problem of an oscillating elevator is a classic test of your understanding of non-inertial frames and apparent weight. It beautifully combines the kinematics of Simple Harmonic Motion (SHM) with the dynamics of Newton's Laws. Let's break down the physics step-by-step.

### The Oscillating Elevator

Imagine you are standing on a weighing scale inside an elevator. If the elevator is stationary or moving with a constant velocity, the scale simply reads your true weight, $mg$. However, this elevator is performing Simple Harmonic Motion.

The position of the elevator is given by the equation:
$$y(t) = 8\left[1 + \sin\left(\frac{2\pi t}{T}ight)ight]$$

This equation tells us that the elevator is oscillating around an equilibrium position of $y = 8$ meters, with an amplitude $A = 8$ meters. The time period $T$ is given as $40\pi$ seconds.

### Decoding Apparent Weight

What exactly does a weighing scale measure? It does not measure gravity directly; it measures the **normal force** ($N$) it exerts to support you. This normal force is what we call your **apparent weight** ($W_{	ext{app}}$).

When the elevator accelerates upwards with an acceleration $a$, the floor pushes harder against your feet. By Newton's Second Law, the net force is $N - mg = ma$, which gives:
$$W_{	ext{app}} = N = m(g + a)$$

Conversely, when the elevator accelerates downwards, the floor falls away slightly, and the normal force decreases:
$$W_{	ext{app}} = m(g - a)$$

### The Calculus of Motion

To find the variation in weight, we first need to determine the maximum acceleration of the elevator. Acceleration is the second derivative of position with respect to time.

Differentiating $y(t)$ twice, we get the acceleration function $a(t)$. For any Simple Harmonic Motion, the maximum acceleration is directly related to the amplitude and the angular frequency ($\omega$):
$$a_{	ext{max}} = A\omega^2$$

Here, the angular frequency is $\omega = \frac{2\pi}{T} = \frac{2\pi}{40\pi} = \frac{1}{20} 	ext{ rad/s}$.

### The Final Calculation

The question asks for the **maximum variation** of the object's weight. This is the difference between the maximum apparent weight (at the lowest point of oscillation, accelerating upwards) and the minimum apparent weight (at the highest point, accelerating downwards).

$$\Delta W = W_{	ext{max}} - W_{	ext{min}}$$
$$\Delta W = m(g + a_{	ext{max}}) - m(g - a_{	ext{max}})$$
$$\Delta W = 2m a_{	ext{max}}$$

Notice how elegantly the acceleration due to gravity ($g$) cancels out! The variation depends entirely on the elevator's acceleration.

Now, we substitute our known values:
$$\Delta W = 2(50) 	imes \left[8 \left(\frac{1}{20}ight)^2ight]$$
$$\Delta W = 100 	imes 8 	imes \frac{1}{400}$$
$$\Delta W = \frac{800}{400} = 2 	ext{ N}$$

The maximum variation in the object's weight is exactly **$2 	ext{ N}$**.

Always remember to check the constraints in such problems. If the downward acceleration ever exceeded $g$, the object would lose contact with the scale, and the apparent weight would drop to zero!

Enter Numerical Value:

Visualized Solution

\text{Visualizing the Elevator}

\text{Apparent Weight}

\text{Acceleration of the Elevator}

\text{Maximum Variation in Weight}

\text{Substituting the Values}

\text{Final Calculation}

\text{The Way Forward}

The Sigma Insight: Pseudo Force

The Oscillating Elevator

Decoding Apparent Weight

The Calculus of Motion

The Final Calculation

Similar Questions

A person standing on a spring balance inside a stationary lift measures $60 kg$ . The weight of that person, if the lift descends with uniform downward acceleration of $1.8 m/s^{2}$ will be ......... $N$ . $[g = 10 m/s^{2}]$

A lift is moving down with acceleration $a$ . A man in the lift drops a ball inside the lift. The acceleration of the ball as observed by the man in the lift and a man standing stationary on the ground are respectively

A car is moving on a plane inclined at $3 0^{\circ}$ to the horizontal with an acceleration of $10 ms^{- 2}$ parallel to the plane upward. A bob is suspended by a string from the roof of the car. The angle in degrees which the string makes with the vertical is ......... (Take, $g = 10 ms^{- 2}$ )

A block is kept on a frictionless inclined surface with angle of inclination $α$ . The incline is given an acceleration $a$ to keep the block stationary. Then, $a$ is equal to

A block of mass $m$ slides on the wooden wedge, which in turn slides backward on the horizontal surface. The acceleration of the block with respect to the wedge is [Given, $m = 8 kg, M = 16 kg$ ] Assume all the surfaces shown in the figure to be frictionless.

System shown in figure is in equilibrium and at rest. The spring and string are massless, now the string is cut. The acceleration of mass $2 m$ and $m$ just after the string is cut will be (2006)

Comprehension Passage

Comprehension Passage

The distance $r$ of the block at time $t$ is :

The net reaction of the disc on the block is :

Enter Numerical Value:

Visualized Solution

\text{Visualizing the Elevator}

\text{Apparent Weight}

\text{Acceleration of the Elevator}

\text{Maximum Variation in Weight}

\text{Substituting the Values}

\text{Final Calculation}

\text{The Way Forward}

The Sigma Insight: Pseudo Force

The Oscillating Elevator

Decoding Apparent Weight

The Calculus of Motion

The Final Calculation

Similar Questions

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } A block is kept on a frictionless inclined surface with angle of inclination α. The incline is given an acceleration a to keep the block stationary. Then, a is equal to

Comprehension Passage

Comprehension Passage

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } The distance r of the block at time t is :

The net reaction of the disc on the block is :

A block is kept on a frictionless inclined surface with angle of inclination $α$ . The incline is given an acceleration $a$ to keep the block stationary. Then, $a$ is equal to

The distance $r$ of the block at time $t$ is :