A particle is moving on x-axis has potential energy $\mathrm{U}=2-20\mathrm{x}+5{\mathrm{x}}^2$ Joules along x-axis. The particle is released at x = 3. The maximum value of ' x.' will be: [x is in meters and U is in joules]

$\mathrm{U}=2-20\mathrm{x}+5{\mathrm{x}}^2$

$\mathrm{F}=-\frac{\mathrm{dU}}{\mathrm{dx}}=20-10\mathrm{x}$

At equilibrium position; F = 0

20 - 10x = 0

$\Rightarrow$x =2

It will oscillate about x = 2 m.

Now m $$\begin{gathered} \stackrel{..}{mx = F =-10 (X-2)} \\ \Rightarrow \stackrel{..}{x}+10(x-2) =0 . \\ The above equation represents an SHM whose equation is given by \\ x-2 = A \cos (\omega t+\phi ) \\ At t=0 , x=3m and at t=0 , v= \frac{dx}{dt}=0 \\ Solving A= 1m and \phi =0 . \\ Therefore x-2=1 \cos \omega t . Since maximum value of \cos \omega t is +1, \\ we get x_{max}=(2+1 ) m=3m. \end{gathered}$$

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