Animated Solution for Physics - Oscillations: Comprehension Passage

Imagine two identical balls, $A$ and $B$, constrained to slide inside a smooth circular pipe of radius $R=0.06\text{ m}$. They are connected by two identical springs, each of natural length $l_0=0.06\pi \text{ m}$.

Let us first check the geometry of this system. The total circumference of the circular pipe is given by:

Since each spring has a natural length of $0.06\pi \text{ m}$, which is exactly half of the total circumference, the springs in their unstretched state span exactly a semicircle. This means that in the equilibrium position, the two balls must lie at diametrically opposite points, $P$ and $Q$.

When both balls are displaced by an angle $\theta$ in the same direction (say, upwards towards the top), we disturb this equilibrium. Let us analyze the resulting forces and oscillations.

When both balls are displaced upwards by an angle $\theta$, each ball travels an arc distance of:

Because both balls move towards each other at the top, the top spring is compressed from both ends. The total compression of the top spring is:

Similarly, because the balls move away from each other at the bottom, the bottom spring is stretched from both ends by the same amount:

Now, let us look at the forces acting on ball $A$. The compressed top spring pushes ball $A$ downwards along the tangent with a force:

At the same time, the stretched bottom spring pulls ball $A$ downwards along the tangent with a force:

Since both forces act in the same direction to restore ball $A$ to its equilibrium position, the net restoring force along the tangent is:

To find the angular acceleration $\alpha$, we write the torque equation about the center $O$:

Using Newton's second law for rotation, $\tau =I\alpha$, where the moment of inertia of a single ball of mass $m$ at distance $R$ is $I=m{R}^2$:

Simplifying this, we get:

This is the classic equation of angular simple harmonic motion, $\alpha =-{\omega }^2\theta$, where the angular frequency $\omega$ is:

The linear frequency of oscillation $f$ is given by:

Substituting the given values, $k=0.1\text{ N/m}$ and $m=0.1\text{ kg}$:

This is our final answer for part (a).

To find the speed of the balls when they pass through their equilibrium positions (the ends of diameter $PQ$), we can use the Conservation of Mechanical Energy.

In the initial displaced position, both balls are at rest, and the energy is purely potential. The total potential energy of the system is the sum of the potential energies of both springs:

When the balls return to their mean positions, the springs are unstretched ($PE=0$), and the energy is purely kinetic. The total kinetic energy of both balls is:

By conservation of energy:

Substituting the given values, $R=0.06\text{ m}$, $\theta =\pi /6\text{ rad}$, $k=0.1\text{ N/m}$, and $m=0.1\text{ kg}$:

This is our final answer for part (b).

The total mechanical energy $E$ of the system is constant and equal to the maximum kinetic energy at the mean position:

Substituting $m=0.1\text{ kg}$ and $v=0.02\pi \text{ m/s}$:

Using ${\pi }^2\approx 9.87$:

This is our final answer for part (c).