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

$$x=x_0+v_0t-\dfrac{V_0}{m}\sgn(x)\dfrac{t^2}{2}$$

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 $\omega$?

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  • $\begingroup$ 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 $T = 2\pi / \omega$. $\endgroup$ Commented Oct 23, 2017 at 3:17
  • $\begingroup$ But the period have to depend on the amplitude, because it is a inclined plane. $\endgroup$ Commented Oct 24, 2017 at 0:21
  • $\begingroup$ Possible duplicate of physics.stackexchange.com/q/60202 $\endgroup$ Commented Oct 24, 2017 at 14:23
  • $\begingroup$ It is similar but is not the same. $\endgroup$ Commented Oct 24, 2017 at 17:49

2 Answers 2

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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.

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  • $\begingroup$ 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}} $ $\endgroup$ Commented Oct 23, 2017 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
    Commented Oct 24, 2017 at 1:45
  • $\begingroup$ Wow, i didn't figured out. Thanks a lot! $\endgroup$ Commented Oct 24, 2017 at 2:06
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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 $\omega = 2\pi/\tau$.

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  • $\begingroup$ Why one quarter of the period is the time to go from $x=0$ to $x=x_{max}$??? I don't get it $\endgroup$ Commented Oct 24, 2017 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
    Commented Oct 24, 2017 at 2:54

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