\documentclass[a4paper,12pt]{article}
\newcommand{\ds}{\displaystyle}
\newcommand{\pl}{\partial}
\begin{document}
\parindent=0pt


{\bf Question}

A spring pendulum consists of a mass $m$ on the end of a light
spring of stiffness $k$,the other end of which is fixed to a
stationary support.  The natural length of the spring is $l$. Find
the Langrangain of the system in terms of $r$ and $\theta$, and
hence derive the equations of motion. Putting $r = l +
\frac{mg}{k} + \epsilon$ and neglecting all terms of second order
of smallness in $\epsilon$ and $\theta$ show that the equations of
motion reduce to $$m \ddot \epsilon + k \epsilon = 0,
\hspace{.2in} \left(l + \frac{mg}{k}\right) \ddot \theta + g
\theta = 0.$$  Deduce that $\omega_\epsilon$, the frequency of
radial oscillations, must be greater than $\omega_\theta$, the
frequency of angular oscillations and that $\omega_\epsilon = 2
\omega_\theta$ when $k = \frac{3mg}{l}$.

\vspace{.25in}

{\bf Answer}

${} $
\begin{center}
\setlength{\unitlength}{.25in}
\begin{picture}(5,8)

\put(0,8){\line(1,0){6}}

\put(1,8){\line(1,-2){1}}

\put(4,2){\line(-1,2){1.1}}

\put(1,8){\line(0,-1){.5}}

\put(1,7){\line(0,-1){.5}}

\put(1,6){\line(0,-1){.5}}

\put(1,5){\line(0,-1){.5}}

\put(1,4){\line(0,-1){.5}}

\put(1.3,6.9){\makebox(0,0){$\theta$}}

\put(3.9,1.9){$\bullet$}

\put(4,2){\vector(0,-1){2}}

\put(5,1){\makebox(0,0){$mg$}}

\put(2,6){\line(1,0){.5}}

\put(2.5,6){\line(-1,-1){.6}}

\put(1.9,5.4){\line(1,0){.8}}

\put(2.7,5.4){\line(-1,-1){.6}}

\put(2.1,4.8){\line(1,0){.8}}

\put(2.9,4.8){\line(-1,-1){.6}}

\put(2.3,4.2){\line(1,0){.6}}

\put(2.5,8){\vector(1,-2){3}}

\put(2.5,8){\vector(-1,2){0}}

\put(4.5,5){\makebox(0,0){$r$}}
\end{picture}
\end{center}
\begin{eqnarray*} K.E. & = & \frac{1}{2}m(\dot r^2 + r^2 \dot
\theta^2) \\ \\ P.E. & = & -mgr\cos \theta + \frac{1}{2}k(r-l)^2
\\
\\ L & = & \frac{1}{2}m((\dot r^2 + r^2 \dot \theta^2) + mgr\cos \theta
- \frac{1}{2}k(r-l)^2 \\ \\ \frac{\pl L}{\pl \theta} & = &
-mgr\sin \theta \hspace{.9in} \frac{\pl L}{\pl \dot \theta}  =
mr^2 \dot \theta \\ \frac{\pl L}{\pl r} & = & mr\dot \theta^2 +
mg\cos\theta \hspace{.5in} \frac{\pl L}{\pl \dot r} = m\dot r
\end{eqnarray*}

\newpage
Euler-Lagrange equations:

$\ds \theta: \hspace{.2in} \frac{d}{dt} \left(\frac{\pl L}{\pl
\dot \theta}\right) - \frac{\pl L}{\pl \theta} = 0 \Rightarrow r^2
\ddot \theta + 2r \dot r \dot \theta + gr \sin \theta = 0$


$\ds r:\hspace{.2in} \frac{d}{dt} \left(\frac{\pl L}{\pl \dot
r}\right) - \frac{\pl L}{\pl r} = 0 \Rightarrow \ddot r - r \dot
\theta^2 - g \cos \theta + \frac{k}{m}(r - l) = 0$

Put $\ds r=l + \frac{mg}{k} + \epsilon \Rightarrow \dot r = \dot
\epsilon, \ddot r = \ddot \epsilon $ in the Euler-Lagrange
equations and neglect quadratic terms in $\theta, \dot \theta,
\ddot \theta, \epsilon, \dot \epsilon, \ddot \epsilon,$ giving

$\ds \left(l + \frac{mg}{k}\right) \ddot \theta + 0 + g\theta = 0
\Rightarrow \left(l+\frac{mg}{k}\right) \ddot \theta + g\theta =
0$

$\ds \ddot \epsilon - 0 - g + \frac{k}{m} \epsilon = 0 \Rightarrow
\ddot \epsilon + \frac{k}{m} \epsilon = 0$

Whence both $\theta$ and $\epsilon$ undergo simple harmonic motion
with frequencies

$\ds \omega_\theta^2 = \frac{g}{(l+\frac{mg}{k})},\ \
\omega_\epsilon^2 = \frac{k}{m}$

Therefore $\ds \frac{\omega_\epsilon^2}{ \omega_\theta^2 } = 1 +
\frac{kl}{mg}>1 \Rightarrow \omega_\epsilon>\omega_\theta$

Put $\ds k = \frac{3mg}{l} \Rightarrow  \frac{\omega_\epsilon^2}{
\omega_\theta^2 } = 4 \Rightarrow  \omega_\epsilon =
2\omega_\theta$



\end{document}