A person sitting inside an elevator performs a weighing experiment with an object of mass $50\text{kg}$.

Suppose that the variation of the height $y\left(m\right)$ of the elevator from the ground, with time $t\left(s\right)$, is given b y=8[1+\sin (\frac{2}{\pi }\left)\right]

where $T=40\pi \text{ s}$. Taking acceleration due to gravity $g=10\text{m/s²}$, the maximum variation of the object's weight (in N) observed in the experiment is ______.

The correct answer is 2.

Step-by-step Explanation:

We are given the variation of the height of the elevator from the ground as:

$y=8\left[1+\sin \left(\frac{2\pi t}{T}\right)\right]$

where the time period is given by:

$T=40\pi \text{s}$

Let the angular frequency of the motion be represented by $\omega$:

$\omega =\frac{2\pi }{T}=\frac{2\pi }{40\pi }=\frac{1}{20}{\text{s}}^{-1}$

Substituting this back into the expression for height, we get:

$y=8+8\sin \left(\omega t\right)$

To find the acceleration of the elevator, we differentiate the height $y$ twice with respect to time $t$.
First, the velocity $v$ is:

$v=\frac{dy}{dt}=8\omega \cos \left(\omega t\right)$

Second, the acceleration $a$ is:

$a=\frac{dv}{dt}=-8{\omega }^2\sin \left(\omega t\right)$

The maximum magnitude of the acceleration is:

$a_{\text{max}}=8{\omega }^2=8\times {\left(\frac{1}{20}\right)}^2=\frac{8}{400}=0.02{\text{m/s}}^2$

The apparent weight of the object inside the elevator varies depending on the acceleration. The apparent weight $W$ is given by:

$W=m(g+a)$

Therefore, the maximum apparent weight $W_{\text{max}}$ and the minimum apparent weight $W_{\text{min}}$ are:

$W_{\text{max}}=m(g+a_{\text{max}})$

$W_{\text{min}}=m(g-a_{\text{max}})$

The maximum variation of the object's weight observed during the experiment is the difference between the maximum and minimum apparent weights:

$\Delta W=W_{\text{max}}-W_{\text{min}}=2m{a}_{\text{max}}$

Substitute the mass of the object $m=50\text{kg}$ and the maximum acceleration $a_{\text{max}}=0.02{\text{m/s}}^2$ into the equation:

$\Delta W=2\times 50\times 0.02=2\text{N}$

Thus, the maximum variation of the object's weight observed in the experiment is 2 N.