Equation of Motion for a Single DOF Spring-Mass System

Hello! Welcome to the first lesson in our new module, Introduction to Mechanical Vibrations.

In our previous modules on dynamics, we studied how to predict the motion of bodies under the action of forces. We used Newton's laws, work-energy, and impulse-momentum to analyze translational and rotational motion. Now, we're going to apply those same principles to a special, but extremely important, type of motion: vibration.

Vibration is oscillatory motion about an equilibrium point. In aerospace engineering, understanding vibration is critical. It can be destructive, as in the case of an aircraft wing experiencing flutter, or it can be a nuisance, like the vibration from an engine felt in the cabin. By modeling and analyzing vibrations, engineers can design systems to avoid resonance, isolate sensitive components, and ensure structural integrity.

Today, we will start with the simplest possible model. Your learning outcome is to derive the equation of motion for a single degree-of-freedom spring-mass system. This forms the absolute foundation for everything that follows in the study of vibrations.

1. Modeling a Vibrating System

Real-world structures like an aircraft wing or a helicopter fuselage are complex. To analyze their vibrations, we first simplify them into a conceptual model. The most common approach is the lumped parameter model, where we represent the object's mass as a single point mass ($m$) and its elasticity as a simple spring ($k$).

If the mass can only move in one direction (e.g., up and down), we call it a single degree-of-freedom (SDOF) system. To begin, we will make a few simplifying assumptions: there is no friction or air resistance (no damping), and no external forces are acting on the system.

The following video provides a concise introduction to this modeling concept.

2. Deriving the Equation of Motion

Our goal is to find a mathematical equation that describes the position of the mass, $x$, at any time, $t$. We can achieve this by combining two fundamental principles you've seen before: Newton's Second Law and Hooke's Law.

1. Newton's Second Law: The net force on an object is equal to its mass times its acceleration ($\sum F=ma$). In our SDOF system, acceleration is the second derivative of position with respect to time, so $a=\ddot{x}$. Thus, $\sum F=m\ddot{x}$.
2. Hooke's Law: The force exerted by a spring is proportional to its displacement from its equilibrium position. The formula is $F_s=-kx$. The negative sign is crucial; it signifies that the spring always pulls or pushes the mass back toward the equilibrium point. This is called a restoring force.

By drawing a free-body diagram of the mass and applying Newton's law, we can derive the equation of motion.

As the video shows, by setting the sum of forces equal to mass times acceleration, we get:
$\sum F_x=m\ddot{x}$
$-kx=m\ddot{x}$
Rearranging this gives us the standard form of the equation of motion for an undamped, free SDOF system:
$m\ddot{x}+kx=0$

3. The Solution and Natural Frequency

This equation is a second-order linear homogeneous ordinary differential equation (ODE). Solving it yields the function $x(t)$ that describes the mass's position over time. While we won't go deep into the ODE solution theory, the process reveals the single most important parameter in vibration analysis.

The solution to $m\ddot{x}+kx=0$ describes simple harmonic motion. The general form of the solution is:
$x(t)=C_1\cos ({\omega }_{n}t)+C_2\sin ({\omega }_{n}t)$
The key parameter that emerges from the math is the natural frequency, ${\omega }_n$, defined as:
${\omega }_n=\sqrt{\frac{k}{m}}$
This is the frequency (in radians per second) at which the system will oscillate if displaced from equilibrium and released. It is an intrinsic property of the system, determined solely by its mass and stiffness.

4. The Electrical-Mechanical Analogy

Given your background in electronics engineering, you may have noticed that the equation of motion looks familiar. It is mathematically identical to the equation governing an ideal LC circuit. This is a powerful analogy that can provide great intuition for mechanical systems.

Consider the equation for an LC circuit, derived from Kirchhoff's voltage law:
$L\frac{di}{dt}+\frac{1}{C}q=0$
Since current $i$ is the rate of change of charge $q$ (i.e., $i=\dot{q}$ and $\frac{di}{dt}=\ddot{q}$), we can write this as:
$L\ddot{q}+\frac{1}{C}q=0$

Now, compare the two equations:

• Mechanical System: $m\ddot{x}+kx=0$
• Electrical System: $L\ddot{q}+\frac{1}{C}q=0$

The analogy is clear:

Mechanical Quantity	Symbol	Electrical Analogy	Symbol
Mass (Inertia)	$m$	Inductance	$L$
Spring Stiffness	$k$	Inverse Capacitance	$1/C$
Displacement	$x$	Charge	$q$
Velocity	$\dot{x}$	Current	$i$

Just as an inductor resists changes in current, a mass resists changes in velocity (acceleration). And just as a capacitor stores energy in an electric field, a spring stores potential energy by being stretched. We'll see later that mechanical damping is analogous to electrical resistance.

5. What About Gravity? The Vertical System

So far, we've used a horizontal system to avoid dealing with gravity. What happens if the mass is hanging vertically from the spring?

Let's derive the equation of motion for this vertical system.
First, at static equilibrium (position b), the spring stretches by an amount ${\delta }_{st}$ to support the weight of the mass. The forces are balanced:
$F_{spring}=W⟹k{\delta }_{st}=mg$
Now, let's displace the mass by a distance $y$ downwards from this equilibrium position (to position c) and apply Newton's second law. The total stretch in the spring is now $({\delta }_{st}+y)$. Taking downwards as positive:
$\sum F_y=m\ddot{y}$
$mg-k({\delta }_{st}+y)=m\ddot{y}$
Distribute the spring term:
$mg-k{\delta }_{st}-ky=m\ddot{y}$
From our equilibrium analysis, we know that $mg=k{\delta }_{st}$. Substituting this in, these two terms cancel out:
$(k{\delta }_{st})-k{\delta }_{st}-ky=m\ddot{y}$
$-ky=m\ddot{y}$
Rearranging this gives us:
$m\ddot{y}+ky=0$
This is the exact same equation of motion as the horizontal case! This is a critical result: for a vertical spring-mass system, if you measure displacement from the static equilibrium position, the effect of gravity is cancelled out, and the analysis is identical to the horizontal system.

Test your understanding!

An avionics package with a mass of $m=5$ kg is mounted in a transport aircraft on a suspension system that can be modeled as a single spring with stiffness $k=4500$ N/m.

1. Write the equation of motion for the system, assuming it vibrates only in the vertical direction.
2. Based on this equation, what is the system's natural frequency of vibration, ${\omega }_n$?

Show answer

1. Equation of Motion:
   As we've shown, for a vertical system where displacement ($y$) is measured from the static equilibrium position, the equation of motion has the standard form:
   $m\ddot{y}+ky=0$
   Substituting the given values:
   $5\ddot{y}+4500y=0$

2. Natural Frequency:
   The formula for natural frequency is ${\omega }_n=\sqrt{k/m}$.
   ${\omega }_n=\sqrt{\frac{4500\text{ N/m}}{5\text{ kg}}}=\sqrt{900}=30\text{ rad/s}$
   So, if disturbed, the avionics package will oscillate at a natural frequency of 30 radians per second.

Conclusion

In this lesson, you've taken the first and most important step in understanding mechanical vibrations. We've seen how to simplify a physical object into a basic mathematical model and derive the governing equation that describes its motion.

Key Takeaways:

• Vibrating systems are often modeled as lumped parameter systems (mass $m$ and spring $k$).
• The equation of motion for an undamped, single degree-of-freedom system is derived using Newton's Second Law and Hooke's Law, resulting in $m\ddot{x}+kx=0$.
• The solution to this equation describes simple harmonic motion at the system's natural frequency, ${\omega }_n=\sqrt{k/m}$.
• For vertical systems, gravity's effect is nullified if displacement is measured from the static equilibrium position.
• There is a direct and useful analogy between mechanical (mass-spring) and electrical (inductor-capacitor) systems.

Next Lesson Preview:
Now that we have derived the fundamental equation and the formula for natural frequency, the next lesson will focus on applying these concepts. You will learn to calculate the natural frequency and period for undamped free vibrations in a variety of engineering scenarios, including systems with springs in series and parallel.