Question 15.14: A mass m is suspended by a spring with spring constant k in ...

We introduce plane polar coordinates for solving the problem and adopt the radius r and the polar angle \varphi as generalized coordinates.

x = r  sin  \varphi,                         \dot{x} = \dot{r}   sin \varphi −r \dot{\varphi}   cos  \varphi.                              (15.37)

The kinetic energy is given by

T = \frac{1}{2} m(\dot{x}^{2} + \dot{y}^{2}) = \frac{1}{2} m(\dot{r}^{2} +r^{2}  \dot{\varphi}^{2}).                                      (15.38)

The length of the spring in its rest position, i.e., without the displacement caused by the mass m, is denoted by r_{0}. The potential energy then reads

V =−mgy + \frac{k}{2} (r −r_{0})^{2} =−mgr   cos \varphi + \frac{k}{2} (r −r_{0})^{2}.                                        (15.39)

The Lagrangian is then

L = T − V =\frac{1}{2} m(\dot{r}^{2} +r^{2}  \dot{\varphi}^{2})+mgr   cos \varphi + \frac{k}{2} (r −r_{0})^{2}.                                     (15.40)

The equations of motion of the system are obtained immediately via the Lagrange equations:

\frac{d}{dt}(mr^{2} \dot{\varphi}) = −mgr  sin  \varphi.                                          (15.41)

This is just the angular momentum law with reference to the coordinate origin. If we take the time dependence of r into account, then we have

mr \ddot{\varphi} = −mg   sin \varphi −2m\dot{r}  \dot{\varphi}.                                         (15.42)

The last term on the right-hand side is the Coriolis force caused by the time variation of the pendulum length r.

For the coordinate r, one obtains

m\ddot{r} = mr \dot{\varphi}^{2} +mg  cos \varphi −k(r − r_{0}).                                      (15.43)

The first term on the right side represents the radial acceleration, the second term follows from the radial component of the weight force, and the last term represents Hooke’s law. For small amplitudes \varphi the motion appears as a superposition of harmonic vibrations in the r, \varphi-plane.