
$\mathrm{m\omega}^{2} \mathrm{a}=\mathrm{m} \times(2 \pi \mathrm{f})^{2} \mathrm{a}=60 \times\left(2 \pi \times \frac{2}{\pi}\right)^{2} \times 0.1 \mathrm{N}$
$=60 \times 16 \times 0.1=96 \mathrm{N}=\frac{96}{9.8} \approx 10 \mathrm{kgf}$ and this
force is towards mean position.
The reaction of the force on the platform away from the mean position. It readuces the weight of man on upper extreme, i.e., net weight $=(60-10) \mathrm{kgf}$
This force adds to the weight at lower extreme position i.e., net weight becomes $=(60+10) \mathrm{kgf}$ Therefore, the reading the weight recorded by spring balance fluctuates between $50$ $kgf$ and $70$ $\mathrm{kgf}$
Statement $I :$ A second's pendulum has a time period of $1$ second.
Statement $II :$ It takes precisely one second to move between the two extreme positions.
In the light of the above statements, choose the correct answer from the options given below:

$y = \frac{1}{{\sqrt a }}\,\sin \,\omega t \pm \frac{1}{{\sqrt b }}\,\cos \,\omega t$ will be
