Physics - W.E.P. Question with Solution | TestHub

$U=\frac{x^4}{4}-\frac{x^2}{2}$(given)

$For\; maxima\; or\; minima\; of\; PE\; =\frac{dU}{dx}=(x^3-x)$

$\therefore x\left(x^2-1\right)=0\Rightarrow x=0$or$\pm 1$

$\frac{d^{2}U}{d{x}^2}=3x^2-1$

at$x=\pm 1$

$\frac{d^{2}U}{dx}=+vei.e.,atx=\pm 1,$

P.E. is minimum

$U=-\frac{1}{4}$

$E=K_{\mathrm{m}\mathrm{a}\mathrm{x}}+V_{\mathrm{m}\mathrm{i}\mathrm{n}}\Rightarrow 2=K_{\mathrm{m}\mathrm{a}\mathrm{x}}-\frac{1}{4}$

$\therefore \frac{1}{2}m{v}_{\mathrm{m}\mathrm{a}\mathrm{x}}^2=\frac{9}{4}$

$\because m=1 \mathrm{kg}$(given)$\Rightarrow v_{\mathrm{m}\mathrm{a}\mathrm{x}}=\frac{3}{\sqrt{2}}$$\mathrm{m} {\mathrm{s}}^{-1}$$$