Chapter 4, Problem 4.1P

To determine

The potential corresponding to the motion of particle.

Answer to Problem 4.1P

The potential corresponding to the motion of particle is$U\left(x\right)\cong \left[\frac{kd}{l}\right]x^2+\left[\frac{k\left(l-d\right)}{4l^3}\right]x^4$.

Explanation of Solution

Consider the figure below.

Unstretched length is$l_0$. The mass m is stretched by a distance d. When the system moves a horizontal distance x, net force on the mass is,

    $F=-2k\left(s-l_0\right)\sin \theta$

Here, F is the force on the mass.

Potential is given by,

    $U\left(x\right)=-{\displaystyle \int Fdx}$

Here, U is the potential.

From the diagram,

    $\begin{array}{c}\sin \theta =\frac{x}{s}\\ =\frac{x}{\sqrt{x^2+l^2}}\end{array}$

The unstretched length is given by,

    $l_0=l-d$

Replace$\sin \theta$by$\frac{x}{\sqrt{x^2+l^2}}$and$l_0$by$l-d$in the expression for F.

    $\begin{array}{c}F=-2k\left(s-l+d\right)\left(\frac{x}{\sqrt{x^2+l^2}}\right)\\ =-2k\left(\sqrt{x^2+l^2}-l+d\right)\left(\frac{x}{\sqrt{x^2+l^2}}\right)\\ =-\frac{2kx}{\sqrt{x^2+l^2}}\left(\sqrt{x^2+l^2}-l+d\right)\\ =-2kx\left[1-\left(\frac{l-d}{l}\right){\left[1+\frac{x^2}{l^2}\right]}^{-1/2}\right]\end{array}$

Using Binomial theorem and expressing up to first two terms,

    $\begin{array}{c}F=-2kx\left[1-\left(\frac{l-d}{l}\right)\left[1-\frac{x^2}{2l^2}\right]\right]\\ =-\frac{2kd}{l}x-\frac{k\left(l-d\right)}{l^3}{x}^3\end{array}$

Conclusion:

Substitute for F in the expression for U and integrating,

    $\begin{array}{c}U=-{\displaystyle \int \left[-\frac{2kd}{l}x-\frac{k\left(l-d\right)}{l^3}{x}^3\right]}dx\\ =\frac{2kd}{l}{\displaystyle \int x}dx+\frac{k\left(l-d\right)}{4l^3}{\displaystyle \int x^3}dx\\ =\left[\frac{kd}{l}\right]x^2+\left[\frac{k\left(l-d\right)}{4l^3}\right]x^4\end{array}$

Therefore, potential corresponding to the motion of particle is$U\left(x\right)\cong \left[\frac{kd}{l}\right]x^2+\left[\frac{k\left(l-d\right)}{4l^3}\right]x^4$.