Question

A massless spring (force constant \(k_{1}\) ) is suspended from the ceiling, with a mass \(m_{1}\) hanging from its lower end. A second spring (force constant \(k_{2}\) ) is suspended from \(m_{1}\), and a second mass \(m_{2}\) is suspended from the second spring's lower end. Assuming that the masses move only in a vertical direction and using coordinates \(y_{1}\) and \(y_{2}\) measured from the masses' equilibrium positions, show that the equations of motion can be written in the matrix form \(\mathbf{M y}=-\mathbf{K y},\) where \(\mathbf{y}\) is the \(2 \times 1\) column made up of \(y_{1}\) and \(y_{2} .\) Find the \(2 \times 2\) matrices \(\mathbf{M}\) and \(\mathbf{K}\).

Short answer

Matrices \(\mathbf{M} = \begin{bmatrix} m_1 & 0 \\ 0 & m_2 \end{bmatrix}\) and \(\mathbf{K} = \begin{bmatrix} k_1 + k_2 & -k_2 \\ -k_2 & k_2 \end{bmatrix}\).

Step-by-step solution

Understand the System Configuration

The system consists of two masses connected by two springs. The first spring with force constant \(k_1\) is attached to the ceiling and has mass \(m_1\) hanging from it. The second spring with force constant \(k_2\) is attached to the mass \(m_1\) and has a second mass \(m_2\) hanging from its free end.

Define Coordinates and Equilibrium

Set coordinates \(y_1\) and \(y_2\) to measure the displacement of masses \(m_1\) and \(m_2\) from their equilibrium positions. Here, the equilibrium position of each mass is where they are when the system is at rest.

Write the Force on Each Mass

For mass \(m_1\), the force due to spring \(k_1\) is \(-k_1 y_1\) and due to spring \(k_2\) is \(k_2(y_2 - y_1)\). Combined, this gives a net force of \(-k_1 y_1 + k_2(y_2 - y_1)\).For mass \(m_2\), the force is \(-k_2(y_2 - y_1)\).

Apply Newton's Second Law

According to Newton's second law, the sum of forces equals the mass times acceleration. Therefore, we have:\( m_1 \ddot{y}_1 = -k_1 y_1 + k_2(y_2 - y_1) \)\( m_2 \ddot{y}_2 = -k_2(y_2 - y_1) \)

Express Equations in Matrix Form

The equations can be written in matrix form as \(\mathbf{M} \ddot{\mathbf{y}} = - \mathbf{K} \mathbf{y}\). Define \(\mathbf{y} = \begin{bmatrix} y_1 \ y_2 \end{bmatrix}\). The matrices \(\mathbf{M}\) and \(\mathbf{K}\) are given by:\[\mathbf{M} = \begin{bmatrix} m_1 & 0 \ 0 & m_2 \end{bmatrix}\]\[\mathbf{K} = \begin{bmatrix} k_1 + k_2 & -k_2 \ -k_2 & k_2 \end{bmatrix}\]

Finalize Matrix Equation Form

With the matrices \(\mathbf{M}\) and \(\mathbf{K}\) defined, the equations of motion are expressed as \(\mathbf{M} \ddot{\mathbf{y}} = - \mathbf{K} \mathbf{y}\), where \(\mathbf{y}\) is the vector of displacements of the masses.

Key concepts

Equations of Motion

In Lagrangian mechanics, the equations of motion are fundamental for describing how physical systems evolve over time. These equations help us understand how forces affect the motion of objects. For the spring-mass system in this problem, each mass is subject to forces exerted by the springs, which can be described using Newton's second law.

• For the first mass (\(m_1\)), it is connected to both the ceiling and the second spring. The forces act in such a way that the resulting motion can be expressed as \(m_1 \ddot{y}_1 = -k_1 y_1 + k_2(y_2 - y_1)\).
• For the second mass (\(m_2\)), it is influenced by the second spring alone, described by \(m_2 \ddot{y}_2 = -k_2(y_2 - y_1)\).

These equations are derived assuming no damping or external forces are present, focusing purely on the spring forces and mass inertia. This symmetric approach simplifies the analysis but effectively captures the system's dynamics.

Spring-Mass System

The spring-mass system in this exercise consists of two springs and two masses arranged vertically. This setup provides a classic and insightful example of mechanical oscillations and vibrations in physics. The properties of the springs and masses determine how the system behaves.

• The first spring with the constant \(k_1\) is connected to a fixed point (the ceiling), supporting the first mass (\(m_1\)).
• The second mass (\(m_2\)) is attached via a second spring with constant \(k_2\), which extends downward from \(m_1\).

This system demonstrates how varying spring constants and mass distributions influence the overall dynamics, leading to different equilibrium positions and oscillation patterns. The interaction between \(k_1\) and \(k_2\) makes the analysis particularly interesting because each spring affects both masses differently, making this a coupled oscillation system.

Matrix Formulation

Matrix formulation offers a powerful and elegant way to express equations of motion for systems with multiple degrees of freedom. For this spring-mass setup, the motion equations can be encapsulated using matrix algebra, which greatly simplifies solving the system numerically or analytically.
The displacement vector \(\mathbf{y} = \begin{bmatrix} y_1 \ y_2 \end{bmatrix}\) represents the displacement of each mass from its equilibrium. The equation \(\mathbf{M} \ddot{\mathbf{y}} = - \mathbf{K} \mathbf{y}\) unites the system's dynamics into a compact form:

• \(\mathbf{M} = \begin{bmatrix} m_1 & 0 \ 0 & m_2 \end{bmatrix}\) is the mass matrix, incorporating each mass.
• \(\mathbf{K} = \begin{bmatrix} k_1 + k_2 & -k_2 \ -k_2 & k_2 \end{bmatrix}\) is the stiffness matrix, defining the strength of the force interactions between the masses and the springs.

This approach enables efficient computation and clear visualization of how masses and springs influence each other, setting a foundation for more complex mechanical systems analysis.