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A spring-mass system is a fundamental model used in physics to explore dynamic behavior. In its simplest form, it consists of one or more masses connected by springs. These masses can oscillate due to external disturbances or initial conditions, and the springs provide a restoring force based on Hooke's Law. In this problem, Fig. 6.29 illustrates a more complex setup with three masses and four springs. Each mass has a specific role:

• Mass 1 is twice the foundational mass, noted as \(m_{1} = 2m\).
• Mass 2 is three times the foundational mass, \(m_{2} = 3m\).
• Mass 3 returns to twice the foundational mass, like Mass 1, \(m_{3} = 2m\).

All springs have a uniform spring constant \(k\), denoted as \(k_{i} = k\). The displacement of each mass is time-dependent, represented as \(x_{1}(t)\), \(x_{2}(t)\), and \(x_{3}(t)\). The study of how these masses move individually or in relation to one another explains the core dynamics of the spring-mass system. Understanding these interactions helps engineers design systems that can withstand or utilize oscillatory motion effectively.

Differential equations are mathematical tools used to model the behavior of dynamic systems. For a spring-mass system like the one in our exercise, we deploy Newton’s second law to derive these equations. Each mass undergoes acceleration, which is a change in velocity over time, caused by the forces exerted by the springs.The equations are crafted as follows for each mass:- **Mass \(m_{1}\)**: The governing differential equation is: \[ 2m \frac{d^2 x_{1}}{dt^2} = -k(x_{1}-x_{2}) \] This shows the acceleration of mass 1 is determined by the difference in displacement between itself and mass 2.- **Mass \(m_{2}\)**: The governing differential equation here is: \[ 3m \frac{d^2 x_{2}}{dt^2} = k(x_{1}-x_{2}) - k(x_{2}-x_{3}) \] Mass 2 is influenced by both Mass 1 and Mass 3, having effects from both sides.- **Mass \(m_{3}\)**: The equation is: \[ 2m \frac{d^2 x_{3}}{dt^2} = -k(x_{2}-x_{3}) \] This represents the influence Mass 2 has on Mass 3.Solving these equations involves calculus and gives us insights into how the masses oscillate over time. They help predict complex motions when linked with initial conditions.

Initial conditions provide the starting values needed to solve differential equations. For systems undergoing motion, these might include initial positions and velocities. In this exercise, the initial conditions include:

• Mass \(x_{1}\) starts at a displaced position \(x_{10}\).
• Masses \(x_{2}\) and \(x_{3}\) start at rest, meaning their initial positions and velocities are zero.
• All initial velocities, such as \(\dot{x}_{1}(0), \dot{x}_{2}(0),\) and \(\dot{x}_{3}(0),\) are zero.

These initial values are crucial because they determine how the system behaves with time. The fact that all initial velocities are zero signifies the system is in free vibration, without external forcing. To find the system's response, these conditions are applied to the general solution: \[ x_{i}(t) = A_{i} \sin (\omega_{i}t) + B_{i} \cos (\omega_{i}t) \]Here, \(A_{i}\) and \(B_{i}\) are constants determined by the initial conditions, and \(\omega_{i}\) is the natural frequency of the system. These determinants make sure the equations accurately model the physical movement of the system from its initial state.