# acceleration and potential energy due to conservative force

how can we find acceleration due to a conservative force as a function of time when potential energy due to the conservative force is given as function of position(e.g. U(x)=x^2)

• I think you need a time dependent potential to do the trick. (not exactly sure what you mean though) Oct 28, 2016 at 5:33

The force $F$ is minus the potential energy gradient $\frac{dU}{dx}$ for a conservative field.
$F =-\dfrac {dU}{dx}$
For a conservative field, the force $F$ can be calculated from the potential energy gradient $dU/dx$ $$F = -\frac{dU}{dx} = -2x$$ where $U(x) = x^2$ in your example. If the mass, $m$, is known, Newton's second law gives $$\ddot{x} = \frac{F}{m} = -\frac{2}{m} x$$ which is the differential equation for an undamped oscillator. This linear differential equation can easily be solved for $x(t)$: $$x(t) = a\cos(\omega t-\phi)$$ where \begin{eqnarray*} a &=& \sqrt{x_0^2+(\dot{x}_0/\omega)^2} \\ \omega &=& \sqrt{2/m} \\ \phi &=& \tan^{-1}\left(\frac{\dot{x}_0}{\omega x_0}\right) \end{eqnarray*} $x_0$ and $\dot{x}_0$ are the initial conditions, $x_0=x(t_0)$ and $\dot{x}_0=\dot{x}(t_0)$.