A particle is moving in a potential $V(x)=V_0\vert x \vert$. I need to get the angular frequency and the period of the movement of the particle.

This is what i have done.

The equation of motion is $$ \DeclareMathOperator{\sgn}{sgn}\begin{align} m\ddot x &= -\dfrac{\partial V}{\partial x} \\ &= -V_0 \sgn (x) \end{align}$$


My problem is: How to compare the equation of motion of this system with the equation of motion of a harmonic oscillator in order to get the angular frequency $\omega$?


closed as off-topic by John Rennie, stafusa, Kyle Kanos, Jon Custer, sammy gerbil Oct 24 '17 at 14:20

This question appears to be off-topic. The users who voted to close gave this specific reason:

  • "Homework-like questions should ask about a specific physics concept and show some effort to work through the problem. We want our questions to be useful to the broader community, and to future users. See our meta site for more guidance on how to edit your question to make it better" – John Rennie, stafusa, Kyle Kanos, Jon Custer, sammy gerbil
If this question can be reworded to fit the rules in the help center, please edit the question.

  • $\begingroup$ I don't believe you need to solve the equation to get $\omega$. Simple harmonic motion / oscillator has the general 2nd-order ODE: $m \ddot{x} = -kx$,where $\omega = \sqrt{k/m}$. That being said, by your first equation above, $\omega = \sqrt{V_o /m}$. The period is just then $T = 2\pi / \omega$. $\endgroup$ – Dr. Ikjyot Singh Kohli Oct 23 '17 at 3:17
  • $\begingroup$ But the period have to depend on the amplitude, because it is a inclined plane. $\endgroup$ – Gabriel Sandoval Oct 24 '17 at 0:21
  • $\begingroup$ Possible duplicate of physics.stackexchange.com/q/60202 $\endgroup$ – sammy gerbil Oct 24 '17 at 14:23
  • $\begingroup$ It is similar but is not the same. $\endgroup$ – Gabriel Sandoval Oct 24 '17 at 17:49

The general problem $V\left(x\right) \propto \left|x\right|^n$ is discussed here.

For your problem $\left(n=1\right)$, if the particle is released from rest at $x=A$ at $t=0$, where $A$ is the amplitude, then the particle will cross $x=0$ at $T/4$, where $T$ is the period.

As you found, from $x=A$ to $x=0$, the force is $-V_0$, and the acceleration is $-V_0/m$, so

$$ \begin{eqnarray} x\left(t\right) &=& x\left(0\right) + v\left(0\right) t + \frac{1}{2} a t^2 \\ &=& A - \frac{V_0}{2m} t^2 \end{eqnarray} $$

I'll leave the rest for you to work out.

If you want the period in terms of the energy $E$ instead of the amplitude $A$, note that since there is no kinetic energy at $x=A$, $V_0 A = E$.

Finally, just use $\omega = 2 \pi / T$ for the angular frequency.

  • $\begingroup$ Why $x=0 \rightarrow t=T/4$ I was thinking that the period is twice the fall time $x=0=A-\dfrac{V_0}{2m}t'^2 \Rightarrow t'=2\sqrt{\dfrac{A2m}{V_0}} $ $\endgroup$ – Gabriel Sandoval Oct 23 '17 at 22:12
  • $\begingroup$ @GabrielSandoval After a time $2\sqrt{2mA/V_0}$ the particle will be at $x=-A$, not $x=A$. The period is the time it takes for the particle to return to the same position. $\endgroup$ – Eric Angle Oct 24 '17 at 1:45
  • $\begingroup$ Wow, i didn't figured out. Thanks a lot! $\endgroup$ – Gabriel Sandoval Oct 24 '17 at 2:06

This problem is a bit subtle, so let's start from your equation of motion: $$m\ddot{x} = -\frac{\partial V}{\partial x} \; .$$ Before working with the explicit form of the potential, write $\ddot{x} = d\dot{x}/dt$ and multiply by $\dot{x}$ to find: $$m \dot{x} \ddot{x} = - \dot{x} \frac{\partial V}{\partial x} \; .$$ This can be expressed as a total time derivative $$\frac{d}{dt} \left[ \frac{1}{2} m \dot{x}^2 + V(x) \right] = 0 \; ,$$ or, equivalently, $$\frac{1}{2} m \dot{x}^2 + V(x) = \frac{1}{2} m v_0^2 \; ,$$ where $v_0$ is an integration constant, equal to the velocity at $x=0$. Now, solving for $\dot{x}$ gives $$\dot{x} = \pm \sqrt{v_0^2-2V(x)/m} \; .$$ So, in terms of differentials $$dt = \pm\frac{dx}{\sqrt{v_0^2-2V(x)/m}}$$ In $1/4$ period, the oscillator will travel from $x=0$ to the maximum displacement $x_m$. Note that for the given potential, the maximum displacement is $x_m = m v_0^2/(2V_0)$, which is the point at which the velocity vanishes. Thus, the period $\tau$ is $$\tau = \frac{4}{v_0} \int_0^{x_m} \frac{dx}{\sqrt{1-x/x_m}}\; .$$ Note that $|x|=x$ above since we are computing the integral over a region where $x \ge 0$. The integral is elementary, and the final result is $$\tau = \frac{4}{v_0} (2 \, x_m) = \frac{4m v_0}{V_0},$$ such that the angular frequency is then $\omega = 2\pi/\tau$.

  • $\begingroup$ Why one quarter of the period is the time to go from $x=0$ to $x=x_{max}$??? I don't get it $\endgroup$ – Gabriel Sandoval Oct 24 '17 at 0:42
  • $\begingroup$ You're welcome. I know you understand now, but perhaps to clarify for others, in each quarter period you move from: (0 to $x_m$) ($x_m$ to 0) (0 to -$x_m$)(-$x_m$ to 0). Note also that this potential is a special case, and there is a very simple approach as shown by Eric Angle. However, if you plan on taking more physics courses, I highly recommend practicing the "general" method above. $\endgroup$ – jcandy Oct 24 '17 at 2:54

Not the answer you're looking for? Browse other questions tagged or ask your own question.