Question

A person normally weighing $60\, kg$ stands on a platform which oscillates up and down harmonically at a frequency $2.0\, s ^{-1}$ and an amplitude $5.0\, cm$. If a machine on the platform gives the person's weight against time, find the ratio of maximum and minimum reading it will show, take $g=10\, m / s ^{2}$.

Answer 1

As shown in figure platform is executing SHM with amplitude and angular frequency given as
$A=5.0 \,cm$
and $\omega=2 \pi n=4 \pi rad / s \,\,\,\left[\right.$ As $\left.n=2 s ^{-1}\right]$
Here weighing machine will show weight more then that of man when it is below its equilibrium position when the acceleration of platform is in upward direction. In this situation the free body diagram of man relative to platform is shown in figure. Here $m a$ is the pseudo force on man in downward direction relative to platform (or weighing machine). As weighing machine will read the normal reaction on it thus for equilibrium of man relative to platform, we have $N=m g+m a$
or $ N=m g+m\left(\omega^{2} y\right) \left[\right.$ As $\left.|a|=\omega^{2} y\right]$
where $v$ is the displacement of platform from its mean position. We wish to find the maximum weight shown by the weighing machine, which is possible when platform is at its lowest extreme position as shown in figure thus maximum reading of weighing machine will be
$N_{\max }=m g+m \omega^{2} A$
$=60 \times 10+60 \times(4 \pi)^{2} \times 0.05=600+480$
$=1080 N =108\, kg\, wt$
Similarly the machine will show minimum reading when it is at its upper extreme position when pseudo force on man will be in upward direction, thus minimum reading of weighing machine will be
$N_{\min }=m g-m \omega^{2} A$
$=600-480=120\, N=12\, kg \,w t$
Hence required ratio, $\frac{N_{\max }}{N_{\min }}=\frac{108}{12}=9$