91Ó°ÊÓ

Newton's second law of motion is the foundation for understanding the movement of objects. It states that the force acting on an object is equal to the mass of the object times its acceleration: \[ F = ma \].
In our problem, the point mass P is subject to two different forces in parts (α) and (β).
For part (α), the force is purely due to the proportional restoring force, given by \( F = -8x \).
Using Newton's second law, we equate the restoring force to the mass times acceleration, resulting in the differential equation \[ m \frac{d^2 x}{dt^2} = -8x \]. Since \( m = 2 \), we simplify to \[ \frac{d^2 x}{dt^2} + 4x = 0 \].
For part (β), an additional damping force is added. This force depends on the velocity \(v\) of the mass and is given by \( F_d = -8v \).
Again, using Newton's second law, we get \[ m \frac{d^2 x}{dt^2} + 8 \frac{dx}{dt} + 8x = 0 \], which simplifies to \[ \frac{d^2 x}{dt^2} + 4 \frac{dx}{dt} + 4x = 0 \].
Understanding Newton's second law helps us form these critical equations for solving the problem.

A damping force opposes the motion of an object. It is usually proportional to the object's velocity.
In our scenario, the damping force is given in part (β) as \( F_d = -8v \).
This means that for every unit increase in velocity, the damping force increases with a factor of 8. The damped motion is modeled with Newton's second law:
\[ m \frac{d^2 x}{dt^2} + 8 \frac{dx}{dt} + 8x = 0 \].
The presence of the damping force makes the system lose energy over time, usually leading to an exponential decay in the amplitude of oscillations. This effect is evident in the solution: \( x(t) = 10 e^{-2t}(1 + 2t) \).
The term \( 10 e^{-2t} \) indicates an exponential decay of the amplitude due to the damping force.

A harmonic oscillator describes a system that experiences a restoring force proportional to its displacement.
For part (α), the force is entirely due to the restoring force \( F = -8x \).
This results in the differential equation \[ \frac{d^2 x}{dt^2} + 4x = 0 \], a classic simple harmonic oscillator equation.
Such systems oscillate at a specific natural frequency without any external damping forces. The solution to this equation is \( x(t) = 10 \cos(2t) \), where the cosine function represents the periodic motion typical of undamped oscillators.
The term \( 2t \) represents the angular frequency, showing the solution oscillates with frequency \( 2 \).

The characteristic equation is derived from the homogeneous form of a linear differential equation.
It plays a crucial role in determining the behavior of the solution. For instance, in part (β), with the differential equation \[ \frac{d^2 x}{dt^2} + 4 \frac{dx}{dt} + 4x = 0 \], we form the characteristic equation by substituting \( x = e^{rt} \). This yields:
\[ r^2 + 4r + 4 = 0 \].
Solving it, we find the roots \( r = -2 \) and \( r = -2 \), indicating a repeated root.
The solution to this type of equation is \[ x(t) = (C_1 + C_2 t) e^{-2t} \].
The characteristic equation's roots thus inform us about the nature of the damped motion, suggesting an exponentially decaying solution characteristic of an over-damped system.

Initial conditions are essential for finding specific solutions to differential equations.
They provide the necessary information to solve for constants in the general solution. In our problem, we are given:
- At \( t = 0 \), the position \( x(0) = 10 \)
- The mass is at rest, so the initial velocity \( v(0) = 0 \).
For part (α), these initial conditions are used in the general solution \( x(t) = A \cos(2t) + B \sin(2t) \).
Applying \( x(0) = 10 \), we find \( A = 10 \), and \( v(0) = 0 \) ensures \( B = 0 \), giving us \( x(t) = 10 \cos(2t) \).
For part (β), in the form \( x(t) = (C_1 + C_2 t) e^{-2t} \), applying initial conditions yields:
\( x(0) = 10 = C_1 \), and \( v(t) = -2C_1 + C_2 = 0 \), solving to \( C_2 = 20 \). Finally, our specific solution is \( x(t) = 10 e^{-2t}(1 + 2t) \).
Initial conditions tailor the general solutions to fit the specific scenario in the problem.