Question

A mass \(m_{1}\) is suspended from a rigid support on a spring \(S_{1}\) with spring constant \(k_{1}\) and damping constant \(c_{1}\). A second mass \(m_{2}\) is suspended from the first on a spring \(S_{2}\) with spring constant \(k_{2}\) and damping constant \(c_{2}\), and a third mass \(m_{3}\) is suspended from the second on a spring \(S_{3}\) with spring constant \(k_{3}\) and damping constant \(c_{3}\). Let \(y_{1}=y_{1}(t), y_{2}=y_{2}(t),\) and \(y_{3}=y_{3}(t)\) be the displacements of the three masses from their equilibrium positions at time \(t,\) measured positive upward. Derive a system of differential equations for \(y_{1}, y_{2}\) and \(y_{3},\) assuming that the masses of the springs are negligible and that vertical external forces \(F_{1}, F_{2},\) and \(F_{3}\) also act on the masses.

Short answer

Question: Write the system of differential equations for the displacements \(y_{1}(t), y_{2}(t),\) and \(y_{3}(t)\) for a system of three masses connected by springs and dampers. Answer: The system of differential equations for the displacements \(y_{1}(t), y_{2}(t),\) and \(y_{3}(t)\) is: 1. \(m_{1}\frac{d^{2}y_{1}}{dt^{2}} = -k_{1}(y_{1}-y_{2}) - c_{1}\frac{dy_{1}}{dt} + F_{1}\), 2. \(m_{2}\frac{d^{2}y_{2}}{dt^{2}} = k_{1}(y_{1}-y_{2}) - k_{2}(y_{2}-y_{3}) + c_{1}\frac{dy_{1}}{dt} - c_{2}\frac{dy_{2}}{dt} + F_{2}\), 3. \(m_{3}\frac{d^{2}y_{3}}{dt^{2}} = k_{2}(y_{2}-y_{3}) + c_{2}\frac{dy_{2}}{dt} + F_{3}\).

Step-by-step solution

Newton's Second Law for Mass 1

For mass 1, \(m_{1}\), the forces acting on it are the spring force \(F_{s1}\), the damping force \(F_{d1}\), and the external force \(F_{1}\). Apply Newton's second law to mass 1, which states \(F = ma\), where \(F\) is the net force acting on the mass, \(m\) is the mass, and \(a\) is the acceleration. For mass 1, we have: \(m_{1}\frac{d^{2}y_{1}}{dt^{2}} = F_{s1} + F_{d1} + F_{1}\).

Forces acting on Mass 1

Recall the spring force and damping force formulas. The spring force acting on mass 1 is given by \(F_{s1} = -k_{1}(y_{1}-y_{2})\), and the damping force is given by \(F_{d1} = -c_{1}\frac{dy_{1}}{dt}\). Now we can substitute these forces into the equation for mass 1: \(m_{1}\frac{d^{2}y_{1}}{dt^{2}} = -k_{1}(y_{1}-y_{2}) - c_{1}\frac{dy_{1}}{dt} + F_{1}\).

Newton's Second Law for Mass 2

Apply Newton's second law to mass 2, \(m_{2}\). The forces acting on mass 2 are the spring forces \(F_{s1}\) and \(F_{s2}\), the damping forces \(F_{d1}\) and \(F_{d2}\), and the external force \(F_{2}\): \(m_{2}\frac{d^{2}y_{2}}{dt^{2}} = F_{s1} - F_{s2} + F_{d1} - F_{d2} + F_{2}\).

Forces acting on Mass 2

Calculate the spring and damping forces acting on mass 2. We have \(F_{s1} = k_{1}(y_{1}-y_{2})\), \(F_{s2} = -k_{2}(y_{2}-y_{3})\), \(F_{d1} = c_{1}\frac{dy_{1}}{dt}\), and \(F_{d2} = -c_{2}\frac{dy_{2}}{dt}\). Substituting the forces, we get: \(m_{2}\frac{d^{2}y_{2}}{dt^{2}} = k_{1}(y_{1}-y_{2}) - k_{2}(y_{2}-y_{3}) + c_{1}\frac{dy_{1}}{dt} - c_{2}\frac{dy_{2}}{dt} + F_{2}\).

Newton's Second Law for Mass 3

Apply Newton's second law to mass 3, \(m_{3}\). The forces acting on mass 3 are spring force \(F_{s2}\), damping force \(F_{d2}\), and the external force \(F_{3}\): \(m_{3}\frac{d^{2}y_{3}}{dt^{2}} = F_{s2} + F_{d2} + F_{3}\).

Forces acting on Mass 3

Calculate the spring and damping forces acting on mass 3. We have \(F_{s2} = k_{2}(y_{2}-y_{3})\) and \(F_{d2} = c_{2}\frac{dy_{2}}{dt}\). Substituting the forces, we get: \(m_{3}\frac{d^{2}y_{3}}{dt^{2}} = k_{2}(y_{2}-y_{3}) + c_{2}\frac{dy_{2}}{dt} + F_{3}\). Now we have the system of differential equations for the displacements \(y_{1}(t), y_{2}(t),\) and \(y_{3}(t)\): 1. \(m_{1}\frac{d^{2}y_{1}}{dt^{2}} = -k_{1}(y_{1}-y_{2}) - c_{1}\frac{dy_{1}}{dt} + F_{1}\), 2. \(m_{2}\frac{d^{2}y_{2}}{dt^{2}} = k_{1}(y_{1}-y_{2}) - k_{2}(y_{2}-y_{3}) + c_{1}\frac{dy_{1}}{dt} - c_{2}\frac{dy_{2}}{dt} + F_{2}\), 3. \(m_{3}\frac{d^{2}y_{3}}{dt^{2}} = k_{2}(y_{2}-y_{3}) + c_{2}\frac{dy_{2}}{dt} + F_{3}\).

Key concepts

Newton's Second Law

One of the cornerstones in understanding motion is Newton's Second Law. This law connects the dots between force, mass, and acceleration. It states that the force acting on an object is equal to the mass of that object multiplied by its acceleration: \( F = ma \). This simple yet powerful formula helps us solve problems where we know some forces acting on a system and need to find the resulting motion.
Let's break it down in the context of mechanical vibrations:

• The mass ( m) is a measure of how much matter is in the object. It's typically measured in kilograms.
• Acceleration ( a) is how fast the object's velocity is changing with time.
• Force ( F) is what pushes or pulls on the mass, and can be a result of gravity, tension, or any applied pressure.

By rearranging the equation \( a = \frac{F}{m} \), we can understand how an object will accelerate given a certain force. In the study of vibrations, mass-spring-damper systems often utilize this fundamental equation to derive more complex outcomes.

Damping Forces

In the real world, not all mechanical systems vibrate indefinitely. This is where damping forces come into play. Damping forces are those that dissipate energy, usually in the form of heat, and work to reduce the amplitude of vibrations over time.
Damping can originate from several sources:

• Friction between moving parts.
• Viscous damping, like in shock absorbers, where a liquid or gas resists motion.
• Internal resistance of materials to deformation.

Mathematically, damping forces are often represented by a term proportional to the velocity of the moving part, usually understandable as \( F_d = -c\frac{dy}{dt} \), where \( c \) is the damping coefficient. Here, \( \frac{dy}{dt} \) represents the velocity of the component being damped.
The negative sign indicates that the damping force always acts in the opposite direction of the motion. It is crucial in stabilizing systems and ensuring they return to equilibrium swiftly, and is fundamental in many systems to keep oscillations at bay.

Spring Constants

In any spring system, the spring constant, often represented by \( k \), plays a critical role. It measures the stiffness of a spring; the larger the spring constant, the stiffer the spring and the more force required to stretch or compress it over a distance.
According to Hooke's Law, the force exerted by a spring is directly proportional to its displacement: \( F_s = -k \times x \). Here,

• \( F_s \) is the spring force,
• \( k \) is the spring constant,
• \( x \) is the displacement from the spring's equilibrium position.

The negative sign shows that the spring force always acts in the opposite direction of the displacement. In practical scenarios, understanding \( k \) helps in predicting how a spring system will behave under loads.
In mechanical vibration setups, multiple springs might be used, each with its unique constant, influencing the overall dynamics of the system. Hence, a precise calculation of \( k \) assures safety and performance in engineering designs.

Mechanical Vibrations

Mechanical vibrations refer to the oscillatory movements of bodies resulting from energy exchanges within or between systems. They can be beneficial, like in musical instruments or clocks, but also need to be controlled, as they can cause damage when excessive, like in buildings during an earthquake.
Here's a quick look at their defining features:

• Amplitude: The peak value of displacement from the equilibrium position.
• Frequency: How many cycles occur per unit of time, measured in Hertz (Hz).
• Period: The time it takes to complete one full cycle of vibration.

In systems involving springs and damping, vibrations arise naturally from energy displacement and restoring forces. The interconnectedness of masses and springs can lead to complex vibration patterns often described using differential equations.
Understanding these properties allows engineers to predict behavior under various conditions and design solutions accordingly. Like reducing unwanted vibrations through damping mechanisms, ensuring that systems operate smoothly and safely.