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Physics LibreTexts

23.7: Small Oscillations

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Any object moving subject to a force associated with a potential energy function that is quadratic will undergo simple harmonic motion,

U(x)=U_{0}+\frac{1}{2} k\left(x-x_{e q}\right)^{2} \nonumber

where k is a “spring constant”, x_{e q} is the equilibrium position, and the constant U_{0} just depends on the choice of reference point x_{r e f} for zero potential energy, U\left(x_{r e f}\right)=0,

0=U\left(x_{r e f}\right)=U_{0}+\frac{1}{2} k\left(x_{r e f}-x_{e q}\right)^{2} \nonumber

Therefore the constant is

U_{0}=-\frac{1}{2} k\left(x_{r e f}-x_{e q}\right)^{2} \nonumber

The minimum of the potential x_{0} corresponds to the point where the x -component of the force is zero,

\left.\frac{d U}{d x}\right|_{x=x_{0}}=2 k\left(x_{0}-x_{e q}\right)=0 \Rightarrow x_{0}=x_{e q} \nonumber

corresponding to the equilibrium position. Therefore the constant is U\left(x_{0}\right)=U_{0} and we rewrite our potential function as

U(x)=U\left(x_{0}\right)+\frac{1}{2} k\left(x-x_{0}\right)^{2} \nonumber

Now suppose that a potential energy function is not quadratic but still has a minimum at x_{0}. For example, consider the potential energy function

U(x)=-U_{1}\left(\left(\frac{x}{x_{1}}\right)^{3}-\left(\frac{x}{x_{1}}\right)^{2}\right) \nonumber

(Figure 23.22), which has a stable minimum at x_{0},

clipboard_ed11afbb82ad64aab51fe4416a5a860d7.png
Figure 23.22 Potential energy function with stable minima and unstable maxima

When the energy of the system is very close to the value of the potential energy at the minimum U\left(x_{0}\right), we shall show that the system will undergo small oscillations about the minimum value x_{0}. We shall use the Taylor formula to approximate the potential function as a polynomial. We shall show that near the minimum x_{0} we can approximate the potential function by a quadratic function similar to Equation (23.7.5) and show that the system undergoes simple harmonic motion for small oscillations about the minimum x_{0}.

We begin by expanding the potential energy function about the minimum point using the Taylor formula

U(x)=U\left(x_{0}\right)+\left.\frac{d U}{d x}\right|_{x=x_{0}}\left(x-x_{0}\right)+\left.\frac{1}{2 !} \frac{d^{2} U}{d x^{2}}\right|_{x=x_{0}}\left(x-x_{0}\right)^{2}+\left.\frac{1}{3 !} \frac{d^{3} U}{d x^{3}}\right|_{x=x_{0}}\left(x-x_{0}\right)^{3}+\cdots \nonumber

where \left.\frac{1}{3 !} \frac{d^{3} U}{d x^{3}}\right|_{x=x_{0}}\left(x-x_{0}\right)^{3} is a third order term in that it is proportional to \left(x-x_{0}\right)^{3}, and \left.\frac{d^{3} U}{d x^{3}}\right|_{x=x_{0}},\left.\frac{d^{2} U}{d x^{2}}\right|_{x=x_{0}}, \quad \text { and }\left.\frac{d U}{d x}\right|_{x=x_{0}} are constants. If x_{0} is the minimum of the potential energy, then the linear term is zero, because

\left.\frac{d U}{d x}\right|_{x=x_{0}}=0 \nonumber

and so Equation ((23.7.7)) becomes

U(x) \simeq U\left(x_{0}\right)+\left.\frac{1}{2} \frac{d^{2} U}{d x^{2}}\right|_{x=x_{0}}\left(x-x_{0}\right)^{2}+\left.\frac{1}{3 !} \frac{d^{3} U}{d x^{3}}\right|_{x=x_{0}}\left(x-x_{0}\right)^{3}+\cdots \nonumber

For small displacements from the equilibrium point such that \left|x-x_{0}\right| is sufficiently small, the third order term and higher order terms are very small and can be ignored. Then the potential energy function is approximately a quadratic function,

U(x) \simeq U\left(x_{0}\right)+\left.\frac{1}{2} \frac{d^{2} U}{d x^{2}}\right|_{x=x_{0}}\left(x-x_{0}\right)^{2}=U\left(x_{0}\right)+\frac{1}{2} k_{e f f}\left(x-x_{0}\right)^{2} \nonumber

where we define k_{eff}, the effective spring constant, by

\left.k_{e f f} \equiv \frac{d^{2} U}{d x^{2}}\right|_{x=x_{0}} \nonumber

Because the potential energy function is now approximated by a quadratic function, the system will undergo simple harmonic motion for small displacements from the minimum with a force given by

F_{x}=-\frac{d U}{d x}=-k_{e f f}\left(x-x_{0}\right) \nonumber

At x=x_{0}, the force is zero

F_{x}\left(x_{0}\right)=\frac{d U}{d x}\left(x_{0}\right)=0 \nonumber

We can determine the period of oscillation by substituting Equation (23.7.12) into Newton’s Second Law

-k_{eff}\left(x-x_{0}\right)=m_{eff} \frac{d^{2} x}{d t^{2}} \nonumber

where m_{eff} is the effective mass. For a two-particle system, the effective mass is the reduced mass of the system.

m_{eff}=\frac{m_{1} m_{2}}{m_{1}+m_{2}} \equiv \mu_{red} \nonumber

Equation (23.7.14) has the same form as the spring-object ideal oscillator. Therefore the angular frequency of small oscillations is given by

\omega_{0}=\sqrt{\frac{k_{e f f}}{m_{e f f}}}=\sqrt{(\left.\frac{d^{2} U}{d x^{2}}\right|_{x=x_{0}})/{m_{e f f}}} \nonumber

Example 23.6: Quartic Potential

A system with effective mass m has a potential energy given by

U(x)=U_{0}\left(-2\left(\frac{x}{x_{0}}\right)^{2}+\left(\frac{x}{x_{0}}\right)^{4}\right) \nonumber

where U_{0} and x_{0} are positive constants and U(0)=0 (a) Find the points where the force on the particle is zero. Classify these points as stable or unstable. Calculate the value of U(x) / U_{0} at these equilibrium points. (b) If the particle is given a small displacement from an equilibrium point, find the angular frequency of small oscillation.

Solution: (a) A plot of U(x) / U_{0} as a function of x / x_{0} is shown in Figure 23.23.

clipboard_e7ee2c68c7c6f419360ad7f9bb7a60b39.png
Figure 22.23 Plot of U(x) / U_{0} as a function of x / x_{0}

The force on the particle is zero at the minimum of the potential energy,

\begin{array}{l} 0=\frac{d U}{d x}=U_{0}\left(-4\left(\frac{1}{x_{0}}\right)^{2} x+4\left(\frac{1}{x_{0}}\right)^{4} x^{3}\right) \\ =-4 U_{0} x\left(\frac{1}{x_{0}}\right)^{2}\left(1-\left(\frac{x}{x_{0}}\right)^{2}\right) \Rightarrow x^{2}=x_{0}^{2} \text { and } x=0 \end{array} \nonumber

The equilibrium points are at x=\pm x_{0} which are stable and x = 0 which is unstable. The second derivative of the potential energy is given by

\frac{d^{2} U}{d x^{2}}=U_{0}\left(-4\left(\frac{1}{x_{0}}\right)^{2}+12\left(\frac{1}{x_{0}}\right)^{4} x^{2}\right) \nonumber

If the particle is given a small displacement from x=x_{0} then

\left.\frac{d^{2} U}{d x^{2}}\right|_{x=x_{0}}=U_{0}\left(-4\left(\frac{1}{x_{0}}\right)^{2}+12\left(\frac{1}{x_{0}}\right)^{4} x_{0}^{2}\right)=U_{0} \frac{8}{x_{0}^{2}} \nonumber

(b) The angular frequency of small oscillations is given by

\omega_{0}=\sqrt{\left.\frac{d^{2} U}{d x^{2}}\right|_{x=x_{0}} / m}=\sqrt{\frac{8 U_{0}}{m x_{0}^{2}}} \nonumber

Example 23.7: Lennard-Jones 6-12 Potential

A commonly used potential energy function to describe the interaction between two atoms is the Lennard-Jones 6-12 potential

U(r)=U_{0}\left[\left(r_{0} / r\right)^{12}-2\left(r_{0} / r\right)^{6}\right] ; r>0 \nonumber

where r is the distance between the atoms. Find the angular frequency of small oscillations about the stable equilibrium position for two identical atoms bound to each other by the LennardJones interaction. Let m denote the effective mass of the system of two atoms.

Solution: The equilibrium points are found by setting the first derivative of the potential energy equal to zero,

0=\frac{d U}{d r}=U_{0}\left[-12 r_{0}^{12} r^{-13}+12 r_{0}^{6} r^{-7}\right]=U_{0} 12 r_{0}^{6} r^{-7}\left[-\left(\frac{r_{0}}{r}\right)^{6}+1\right] \nonumber

The equilibrium point occurs when r=r_{0} The second derivative of the potential energy function is

\frac{d^{2} U}{d r^{2}}=U_{0}\left[+(12)(13) r_{0}^{12} r^{-14}-(12)(7) r_{0}^{6} r^{-8}\right] \nonumber

Evaluating this at r=r_{0} yields

\left.\frac{d^{2} U}{d r^{2}}\right|_{r=r_{0}}=72 U_{0} r_{0}^{-2} \nonumber

The angular frequency of small oscillation is therefore

\omega_{0}=\sqrt{\left.\frac{d^{2} U}{d r^{2}}\right|_{r=r_{0}} / m} \nonumber

=\sqrt{72 U_{0} / m r_{0}^{2}} \nonumber


This page titled 23.7: Small Oscillations is shared under a not declared license and was authored, remixed, and/or curated by Peter Dourmashkin (MIT OpenCourseWare) via source content that was edited to the style and standards of the LibreTexts platform.

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