Physics - SHM Question with Solution | TestHub

$\mathrm{F}=-20\left(\mathrm{x}-\frac{1}{2}\right)=-20\mathrm{X}\left(\mathrm{X}=\mathrm{x}-\frac{1}{2}\right)$ $\therefore$ Particle will perform SHM about $\mathrm{x}=\frac{1}{2}$ with $\omega =2\mathrm{rad}/\sec \Rightarrow \mathrm{T}=\pi \sec$. $\therefore$ Phase covered in $t=\frac{\pi }{4}$ second $=90^{\circ }$. Given particle is at rest at $\mathrm{x}=1\mathrm{m}\Rightarrow \mathrm{x}=1$ is extreme position. $\therefore$ In $\frac{\pi }{4}$ sec, it will be at equilibrium $\therefore \mathrm{x}=0.5\mathrm{m}$ and momentum $=\mathrm{m}\omega \mathrm{A}=5\times 2\times 0.5=5\mathrm{kg}\mathrm{m}/\mathrm{s}$ Direction will be towards -ve $\mathrm{x}$. Hence option (3)