(a) Find numerical values of the constants and computer plot together on the same axes graphs of (5.30),(5.31) and (5.32) in order to compare overdamped, critically damped, and oscillatory motion. Suggested numbers: Let \(\omega=1,\) and \(b=13 / 5,1,5 / 13\) for the three kinds of motion. Let \(y(0)=1\) and \(y^{\prime}(0)=0\). (b) Repeat the problem with the same set of \(\omega\) and \(b\) values and with \(y(0)=1\) but with \(y^{\prime}(0)=1\). (c) Again repeat, with \(y^{\prime}(0)=-1\).

Differential equations describe how a quantity changes over time. In this exercise, we deal with a second-order linear differential equation, specifically for damped harmonic motion.
This can be written as: \(\ddot{y} + 2b\dot{y} + \omega^2 y = 0\), where y represents the displacement from an equilibrium position.
Here, \(\ddot{y}\) is the acceleration, \(\dot{y}\) is the velocity, and y is the displacement. The term 2b\dot{y} represents damping (resistance to motion). Depending on the value of b, the system can be overdamped, critically damped, or underdamped.

Initial conditions specify the state of the system at the start of observation. They are crucial for solving differential equations because they allow us to find a specific solution that fits a given context. In our problem, we have sets of initial conditions:
\(\begin\{cases\}y(0)=1\ y\prime(0)=0\, y(0)=1\ y\prime(0)=1\, y(0)=1\ y\prime(0)=-1\end\{cases}\).
These initial conditions mean we start with the displacement of 1 and various initial velocities (0, 1, and -1). Solving the differential equation with these will show how the system behaves over time for each scenario.

Mechanical vibrations occur when a system oscillates about an equilibrium position. This is often due to forces that try to restore the system to its rest state.
In a damped harmonic oscillator, these vibrations are also influenced by damping, which is a resistance force proportional to the velocity. The damping force reduces the amplitude (height) of the vibrations over time.
Mathematically, mechanical vibrations are described by our differential equation: \(\ddot{y} + 2b\dot{y} + \omega^2 y = 0\). This equation results from Newton’s second law, considering a restoring force proportional to displacement, a damping force proportional to velocity, and an inertial force proportional to acceleration.

Overdamping occurs when the damping constant b is large enough to prevent oscillations. In this case, the system returns to equilibrium without oscillating.
For instance, if \(\b=\frac{13}{5}\), our differential equation becomes \(\ddot{y} + \frac{26}{5}\dot{y} + y = 0\). Solving this will show a smooth, non-oscillatory return to equilibrium.
Overdamped systems take longer to return to equilibrium compared to critically damped systems, as the high damping force initially opposes the motion before gradually letting the displacement reduce.

Critical damping is the exact amount of damping that allows the system to return to equilibrium as quickly as possible without oscillating. This happens when the damping constant b equals a specific critical value.
In our problem, with \(\b=1\), we get the equation \(\ddot{y} + 2\dot{y} + y = 0\) for critical damping. Solving it shows the fastest return to equilibrium without overshooting.
Critical damping is ideal for systems wanting a quick return to rest, like car shock absorbers, where too much or too little damping would prolong the settling time or cause oscillations.

Underdamping happens when the damping constant b is small and the system oscillates while gradually returning to equilibrium. The amplitude of oscillations decreases over time due to the damping force.
For \(\b=\frac{5}{13}\)\ff, our differential equation becomes \(\ddot{y} + \frac{10}{13}\dot{y} + y = 0\). Solving this shows oscillatory motion with gradually reducing peaks.
Underdamped systems are common in many mechanical and electrical systems where some oscillation is permissible or even desired, such as in clocks or musical instruments.