Question 5.2: A bead of mass m slides along a smooth horizontal cylindrica......

Taking the horizontal displacement x as a generalised coordinate, the Lagrangian is

L=\frac{m}{2}\dot{x}^{2} -\frac{k}{2}\left(\sqrt{a^{2}+x^{2}}-l \right)^{2} .  (5.14)

The equilibrium positions satisfy

\frac{dV}{dx}=k \left(\sqrt{a^{2}+x^{2}}-l \right)\frac{x}{\sqrt{a^{2}+x^{2}}}=0,  (5.15)

whose solutions are x = 0 and x = ±\sqrt{l^{2}-a^{2}} (if l > a). In order to investigate stability we must determine the sign of

\frac{d^{2}V}{dx^{2}}=k \frac{x^{2}}{a^{2}+x^{2}}+k \left(\sqrt{a^{2}+x^{2}}-l \right)\frac{a^{2}}{\left(a^{2}+x^{2}\right)^{{3}/{2}}}  (5.16)

at the equilibrium points. (1) Case a > l. Only x = 0 is an equilibrium point, which is stable since

\left(\frac{d^{2}V}{dx^{2}}\right)_{x=0}=k \left(1- \frac{l}{a}\right) > 0 .   (5.17)

The frequency of small oscillations about x = 0 is

ω = \sqrt{\frac{k}{m}\left(1- \frac{l}{a}\right)}.  (5.18)

(2) Case a<l. There are three equilibrium positions: x = 0 or x = ±\sqrt{l^{2}-a^{2}}. The equilibrium at x = 0 is now unstable because V^{\prime \prime}(0) < 0. On the other hand, the equilibrium positions x = ±\sqrt{l^{2}-a^{2}} are stable because

\left(\frac{d^{2}V}{dx^{2}}\right)_{x=±\sqrt{l^{2}-a^{2}}}=k \left(1- \frac{a^{2}}{l^{2}}\right) > 0 ,   (5.19)

with

ω = \sqrt{\frac{k}{m}\left(1- \frac{a^{2}}{l^{2}}\right)}  (5.20)

being the frequency of small oscillations about these positions. Note that if a = l we are confronted with the previously mentioned anomalous situation: the only equilibrium position is x = 0, which is stable, but V^{\prime \prime}(0) = 0 and the fourth-order term dominates the expansion of the potential about x = 0.