Question 35: Oscillations

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

Accepted Answer

Here, we will use the formula for apparent weight in elevator as the elevator has some acceleration.
The normal force (apparent weight) adjusts for the elevations acceleration
relative to gravity.
Apparent weight: $w=m(g+a)$, where $a$ is the elevator's acceleration.
Here, elevator's position, $y$ is given so we can find elevator's acceleration using
$$ a=\frac{d^2 y}{d t^2} $$
Given, $m=50 \mathrm{~kg}$
$$ \begin{aligned} & 	ext { and } y=8\left[1+\sin \left(\frac{2 \pi t}{T}ight)ight] \ & T=40 \pi \mathrm{~s} . \ & \Rightarrow y=8(1+\sin (\omega t)) \quad\left(as, \omega=\frac{2 \pi}{T}ight) \ & \omega=\frac{2 \pi}{40 \pi}=\frac{1}{20} \ & a=\frac{d^2 y}{d t^2}=-8 \omega^2 \sin (\omega t) \ & \left|a_{	ext {max }}ight|=8 \omega^2=8\left(\frac{1}{20}ight)^2=8 	imes \frac{1}{400}  \ & \left|a_{	ext {max }}ight|=\frac{1}{50} \end{aligned} $$
$W_{	ext {max }}=m\left(g+a_{	ext {max }}ight)$ (when lift goes up)
and. $W_{	ext {min }}=m\left(g-a_{	ext {max }}ight)$ (when lift goes down)
so
$$ \begin{aligned} \Delta w & =w_{	ext {max }}-w_{	ext {min }}=2 m a_{	ext {max }} \ & =2 	imes 50 	imes \frac{1}{50} \Rightarrow \Delta w=2 \mathrm{~N} \end{aligned} $$

Oscillations

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