A particle is moving on x-axis has potential energy U=2−20x+5x2 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]

U=2−20x+5x2F=−dUdx=20-10xAt equilibrium position; F = 020 - 10x = 0⇒x =2It will oscillate about x = 2 m.Now m mx = F =-10 (X-2).. ⇒x..+10(x-2) =0 . The above equation represents an SHM whose equation is given by               x-2 = A cos (ωt+φ) At t=0 , x=3m and at t=0 , v= dxdt=0 Solving A= 1m and φ=0 . Therefore x-2=1 cos ωt .  Since maximum value of cos ωt is +1,  we get xmax=(2+1 ) m=3m...