A particle is moving on x-axis has potential energy U=2−20x+5x2J along x-axis. The particle is released at x = - 3. The maximum value of 'x' is _________.[x is in meters and U is in joule]

Given that U=2−20x+5x2Interaction force on particle is given asF=−dUdx=20−10xAs F is linearly varying with -x, particle is executing SHMAt equilibrium position F = 0⇒ 20−10x=0⇒ x=2Since particle is released at x = - 3, therefore amplitude of particle is 5.It will oscillate about x = 2 with an amplitude of 5.⇒ Maximum value of x will be 7.