I am currently working on a problem where a particle of mass $m$ can move along the x-axis with no friction. The particle is moving in a gravitational field and the potential energy is given by:


I want to find the equation of motion for this particle by using the law of conservation of energy.


$$\iff -2Cm\frac{1}{\sqrt{x^2+(ab)^2}}+\frac{1}{2}m(\dot{x})^2=constant$$

I didn't know how to deal with the constant term so I just used the fact that the derivative of any constant is $0$ and differentiated both sides.

$$\implies \frac{\partial}{\partial x}(-2Cm\frac{1}{\sqrt{x^2+(ab)^2}})+\frac{\partial}{\partial x}(\frac{1}{2}m(\dot{x})^2)=0$$

$$\iff Cm\frac{2x\color{red}{\dot{x}}}{(\sqrt{x^2+(ab)^2})^3}+m\dot{x}\color{red}{\ddot{x}}=0$$

Question 1: Am I differentiating correctly here? I am not sure if I have to implicitly differentiate $x$ and $\dot{x}$

Question 2: Does my approach even make sense or am I just wasting my time here?


$$m\dot{x}\space (\ddot{x}+\frac{C}{(x^2+(ab)^2)^\frac{3}{2}}x)=0$$

$$\iff \ddot{x}+\frac{C}{(x^2+(ab)^2)^\frac{3}{2}}x=0$$

  • $\begingroup$ How do you get $\dot{x}$ in the first term if you are taking the derivatives of $x$? $\endgroup$ – Kyle Kanos Apr 21 '15 at 17:11
  • $\begingroup$ @KyleKanos I differentiated $x^2$ under the square root. This is why I was asking if I need to implicitly differentiate because then I would get $2x\cdot \dot{x}$ $\endgroup$ – qmd Apr 21 '15 at 17:14
  • $\begingroup$ That doesn't make sense. What is $\dot{x}$? $\endgroup$ – Kyle Kanos Apr 21 '15 at 17:17
  • $\begingroup$ I would say that is the velocity. $\endgroup$ – qmd Apr 21 '15 at 17:18
  • $\begingroup$ So how do you get a velocity ($dx/dt$) when taking the derivative with respect to space? $\endgroup$ – Kyle Kanos Apr 21 '15 at 17:18

You actually don't need to take any time derivatives here. Since the energy, $E$, is a constant, you only need to know it at one moment in time (say at t=0 as an initial condition), and you know it at all other times. Thus, just solve for $\dot{x}(t)$ in terms of the other quantities ($x,E,...)$ in the second equation that you have to get the equation of motion. Like all differential equations, you have to supply initial conditions to get the particular solution or trajectory.

  • $\begingroup$ Thanks. So I don't need the second derivative in my equation at all? I always thought they have to be in the form $\ddot{x}+Cx=0$ $\endgroup$ – qmd Apr 21 '15 at 18:43
  • $\begingroup$ An equation of motion is really just any equation involving the trajectory x(t) and its derivatives. $\endgroup$ – mr blick Apr 21 '15 at 18:46
  • $\begingroup$ But when you say "its derivatives", does it have to be all of them ($\ddot{x}(t), \dot{x}(t)$) or can it be just the first derivative $\dot{x}(t)$ $\endgroup$ – qmd Apr 21 '15 at 19:00
  • $\begingroup$ It can just be the first one. Really any combination is an equation of motion. $\endgroup$ – mr blick Apr 21 '15 at 19:05

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