Q4: Oscillations of a Particle in a Ring

Physics Problem: Elastic Cords & Ring

Question 4

Three identical elastic cords each of force constant $k$ and almost zero relaxed lengths are connected at their one end to a particle of mass $m$ and the other ends are connected to three equally spaced points A, B, and C on a rigid ring placed on a frictionless tabletop. Mass of the ring is $\eta = 2$ times the mass of the particle. In equilibrium, the particle is at the centre. Now the ring is held and the particle is pulled to point P at distance $r_0$ from the centre O and both are set free.

(a) Find period of the oscillatory motion.
(b) Express position of particle $x_p$ and centre of the ring $x_r$ as function of time $t$.

Solution

Step 1: Effective Spring Constant ($k_{eff}$)

The problem states the cords have zero relaxed lengths. For such a spring, the force is simply $\vec{F} = -k \vec{l}$, where $\vec{l}$ is the length vector.

When the particle is at the center, forces cancel out. When displaced by a position vector $\vec{r}$ from the center, the force exerted by the $i$-th spring connected at position $\vec{R}_i$ on the ring is: $$ \vec{F}_i = -k(\vec{r} – \vec{R}_i) $$ Summing over all three springs (A, B, C): $$ \vec{F}_{net} = \sum \vec{F}_i = -k(3\vec{r}) + k(\vec{R}_A + \vec{R}_B + \vec{R}_C) $$ Since A, B, and C are equally spaced on the ring, their vector sum is zero ($\sum \vec{R}_i = 0$). $$ \vec{F}_{net} = -3k\vec{r} $$ Thus, the system behaves like a single spring with effective constant: $$ k_{eff} = 3k $$

Step 2: Two-Body Problem (Reduced Mass)

The ring (mass $M = \eta m$) and the particle (mass $m$) are free to move. This is a two-body coupled oscillation problem.
The reduced mass $\mu$ of the system is: $$ \mu = \frac{m \cdot M}{m + M} = \frac{m (\eta m)}{m + \eta m} = \frac{\eta m^2}{m(1+\eta)} = \frac{\eta m}{1+\eta} $$

Step 3: Period of Oscillation (Part a)

The angular frequency $\omega$ is given by $\sqrt{k_{eff} / \mu}$: $$ \omega = \sqrt{\frac{3k}{\left( \frac{\eta m}{1+\eta} \right)}} = \sqrt{\frac{3k(1+\eta)}{\eta m}} $$ The time period $T = \frac{2\pi}{\omega}$ is: $$ T = 2\pi \sqrt{\frac{\eta m}{3k(1+\eta)}} $$

Step 4: Position Functions (Part b)

Since the table is frictionless and there are no external horizontal forces, the Center of Mass (COM) of the system remains fixed.

Initial State ($t=0$):
Particle is at $x = r_0$.
Ring center is at $x = 0$ (Origin).

Location of COM ($X_{cm}$): $$ X_{cm} = \frac{m(r_0) + \eta m(0)}{m + \eta m} = \frac{r_0}{1+\eta} $$
Motion relative to COM: Both bodies oscillate about the COM with frequency $\omega$.
For the particle ($x_p$):
Amplitude $A_p = \text{Initial Pos} – X_{cm} = r_0 – \frac{r_0}{1+\eta} = \frac{\eta r_0}{1+\eta}$.
Equation: $x_p(t) = X_{cm} + A_p \cos(\omega t)$ $$ x_p(t) = \frac{r_0}{1+\eta} + \frac{\eta r_0}{1+\eta} \cos\left( \sqrt{\frac{3k(1+\eta)}{\eta m}} t \right) $$
For the ring ($x_r$):
Amplitude $A_r = \text{Initial Pos} – X_{cm} = 0 – \frac{r_0}{1+\eta} = -\frac{r_0}{1+\eta}$.
Equation: $x_r(t) = X_{cm} + A_r \cos(\omega t)$ $$ x_r(t) = \frac{r_0}{1+\eta} – \frac{r_0}{1+\eta} \cos\left( \sqrt{\frac{3k(1+\eta)}{\eta m}} t \right) $$

(a) Period: $$ T = 2\pi \sqrt{\frac{\eta m}{3k(1+\eta)}} $$ (b) Positions: $$ x_p = \frac{r_0}{1+\eta} \left[ 1 + \eta \cos(\omega t) \right] $$ $$ x_r = \frac{r_0}{1+\eta} \left[ 1 – \cos(\omega t) \right] $$ Where $\omega = \sqrt{\frac{3k(1+\eta)}{\eta m}}$