To analyze the scenario you've described, we need to consider the forces acting on the particle and how they influence its motion when displaced. The forces are repulsive and vary with the cube of the distance, which is a key detail in determining the nature of the motion.
Understanding the Forces
Let's denote the two forces acting on the particle as \( F_1 \) and \( F_2 \). The distances from the particle to the centers of these forces are \( a \) and \( b \) respectively. According to the problem, the forces vary as the cube of the distance, which means:
- Force \( F_1 \) can be expressed as \( F_1 = k \cdot \frac{1}{a^3} \)
- Force \( F_2 \) can be expressed as \( F_2 = k' \cdot \frac{1}{b^3} \)
Here, \( k \) and \( k' \) are constants representing the strength of the forces per unit mass. Since the particle is initially in equilibrium, we have:
\( F_1 = F_2 \) when the particle is at rest.
Displacement and Restoring Force
Now, when the particle is slightly displaced towards one of the forces, say towards \( F_1 \), the distance to \( F_1 \) decreases while the distance to \( F_2 \) increases. This change will create a net force that acts to restore the particle back to its equilibrium position.
Let’s denote the small displacement from equilibrium as \( x \). The new distances become \( a - x \) for \( F_1 \) and \( b + x \) for \( F_2 \). The forces now can be expressed as:
- New force \( F_1' = k \cdot \frac{1}{(a - x)^3} \approx k \cdot \frac{1}{a^3} \left(1 + \frac{3x}{a}\right) \text{ (using Taylor expansion)}
- New force \( F_2' = k' \cdot \frac{1}{(b + x)^3} \approx k' \cdot \frac{1}{b^3} \left(1 - \frac{3x}{b}\right) \text{ (using Taylor expansion)}
The net restoring force \( F_{net} \) when displaced is:
\( F_{net} = F_1' - F_2' \)
Establishing Simple Harmonic Motion
For small displacements, we can approximate the net force as:
\( F_{net} \approx -k_{eff} x \)
Where \( k_{eff} \) is an effective spring constant derived from the differences in the forces. The negative sign indicates that the force acts in the opposite direction of the displacement, characteristic of simple harmonic motion (SHM).
Time Period Calculation
The time period \( T \) of simple harmonic motion is given by the formula:
\( T = 2\pi \sqrt{\frac{m}{k_{eff}}} \)
Here, \( m \) is the mass of the particle. To find \( k_{eff} \), we need to differentiate the net force with respect to \( x \) at the equilibrium position. This will yield a value that depends on the constants \( k \) and \( k' \) and the distances \( a \) and \( b \).
After calculating \( k_{eff} \), you can substitute it back into the time period formula to find the exact time period of the oscillation.
In summary, the particle exhibits simple harmonic motion when displaced from equilibrium, and the time period can be derived from the effective spring constant related to the forces acting on it. This approach not only illustrates the principles of equilibrium and motion but also highlights the beauty of physics in explaining natural phenomena.
