About the dynamic performance of 1 mass-spring model

Consider the following undamped 1 mass-spring model with mass \(m\) and spring stiffness \(k\).

Free vibration

The relationship between displacement and force with the point mass in free vibration can be expressed as follows

\(\begin{align}
m \ddot{x} + kx = 0 \tag{1-1} \\
\ddot{x} + \omega_n^2 x = 0 \tag{1-2} \\
\omega_n^2 = \frac{k}{m}　\tag{1-3}
\end{align}\)

\(\omega_n\) is called the natural circular frequency, which has the following relationship with the natural period \(T\) and natural frequency \(f_n\).

\(f_n = \frac{1}{T} = \frac{\omega_n}{2 \pi} = \frac{1}{2 \pi} \sqrt{\frac{k}{m}}
\tag{1-4}\)

Equation (1-4) shows that for constant mass, the natural frequency increases in proportion to the square root of the stiffness, i.e., the higher the stiffness, the higher the natural frequency.

Forced Vibration

Next, consider how the response changes as the natural frequency changes in the following forced vibration problem. For the free vibration problem treated in Equation (1-1), consider the situation where a forced vibration \(F_0\) at an circular frequency of \(\omega\) is given.

The equation of motion of a vibrated point mass from the balance of forces is as follows

\(\ddot{x} + \omega_n^2 x = Fe^{i \omega t} = F_0
\tag{1-5}\)

On the other hand, the vibration is transmitted through this spring, and the reaction force \(F\) generated at the support point side is generated by the term affected by the spring in equation (1-5), which is

\(\omega_n^2 x = F
\tag{1-6}\)

From the above, the transmission coefficient of vibration from the vibration source to the support, \(T_r\), is obtained by taking the ratio of Equations (1-5) and (1-6) as follows

\(\begin{align}
T_r &= \left|\frac{F}{F_0} \right|
= \left|\frac{\omega_n^2 x}{\ddot{x} + \omega_n^2 x} \right|
= \left|\frac{1}{1 + \frac{1}{\omega_n^2} \frac{\ddot{x}}{x}} \right| \\
&= \left|\frac{1}{1 + \frac{-\omega^2}{\omega_n^2}} \right|
= \left|\frac{1}{1 - u^2} \right|
\tag{1-7}
\end{align}\)

\(u\) is the ratio of the excitation frequency to the natural frequency of the 1 mass-spring model (\(u=\frac{\omega}{\omega_n} = \frac{f}{f_n}\)), and \(x\) is solved using the general solution \(x = A e^{i \omega t}\) of the differential equation of the form (1-5).

Equation (1-7) can be graphed as follows

From these we can see the following

• When \(u = 1\), vibrations are transmitted with great amplification (resonance phenomenon)
• Vibrations are always amplified in the range \(u\) less than \(\sqrt{2}\).
• Only when \(u\) exceeds \(\sqrt{2}\), the value of vibration transmission coefficient \(Tr\) falls below 1 and vibration isolation is achieved.

In other words, if you want to isolate vibration, the ratio of frequencies must be greater than \(\sqrt{2}\).

Specific vibration isolation design examples

If we wish to isolate the transmission of vibration generated by a 1 kg device vibrated at 100 Hz so that \(\frac{1}{8}\), the frequency ratio \(u=3\) and the target natural frequency is the following from equation (1-7).

\(3 = \frac{100}{f_n}, \quad 
\tag{1-6}
f_n \fallingdotseq 33 [\rm{Hz}]\)

From equation (1-4), the spring stiffness \(k\) in this case is

\(\frac{100}{3} = \frac{1}{2 \pi} \sqrt{\frac{k}{1}}, \quad
k = \left( \frac{200}{3} \pi \right)^2 \fallingdotseq 43900 [\textrm{N/m}] = 43.9 [\textrm{N/mm}]
\tag{1-7}\)

In actual vibration isolation design, it is rare to find off-the-shelf components that exactly match the performance obtained above, so it is necessary to search for components with similar values in catalogs.
In many cases, vibration isolation component manufacturers provide a selection flow on their websites or in their catalogs, which describes what type of vibration isolation component to select without detailed calculations, or provide tables and graphs to help you find the right component.

Mass vs. frequency

The following table summarizes the results of calculations for vibration isolation for several combinations of mass and excitation frequency.

Thus, we can see that a lower stiffness spring is required to isolate lighter objects even at lower excitation frequencies.
If you are interested in finding out what specific shapes of components are available with this spring stiffness, you can search for components by spring stiffness value at, for example, Monotaro or MISUMI, and so on.。

As an example of a member with the stiffness of 0.439 N/mm required to isolate vibration when a 1 kg mass point in the above table is vibrated at 10 Hz, the following ready-made member of a simple coil spring was found.

Static performance of 1 mass-spring model

In contrast to the problems so far calculated for vibration (dynamic problems), the following is considered for the static state of a member subjected only to gravitational acceleration \(g\).

The deflection under its own weight \(\delta\) is given by the following equation from the balance of forces.

\(\delta = \frac{m g}{k}
\tag{2-1}\)

Equation (2-1) shows that for a constant mass, the greater the stiffness, the smaller the deflection.
Consider the static displacement of the coil spring in the table above, which is determined to be necessary for vibration isolation of a 1 kg spring against 10 Hz excitation.

\(\delta = \frac{1 \times 9.8}{0.44} = 22.2 \textrm{mm}
\tag{2-2}\)

The static deflection is 22.2mm, which means that more than half of the spring's natural length of 40mm has been deformed by the static load.
When the spring is used for vibration isolation, deformation during forced vibration is added from this displacement, so if the spring is greatly deformed in the static state, it may bottom out due to deformation during forced vibration (the spring is deformed 40 mm and the spring makes contact), losing its vibration isolation effect, and a large load from a collision may be applied to the object to be isolated. The object may be subjected to a large load due to a collision.

In the above example, the vibration isolation is 1 kg at 10 Hz, but assuming a lower frequency vibration isolation, the spring stiffness \(k\) is 1/100 of 1 kg at 1 Hz. The amount of deformation in this case is \(22.2 \times 100 = 2220\textrm{[mm]} = 2.22\textrm{[m]}\), an unrealistic value that makes it difficult to achieve vibration isolation with the methods introduced so far.

On the stiffness-mass-frequency tradeoff

Summarizing the dynamic and static characteristics I have shown so far, we can see that there is a tradeoff between

1. Less stiffness
   • Lower first-order natural frequencies allow vibration isolation from lower frequencies
   • Static deflection is increased and excessive deformation may occur and bottom out just by installing the vibration isolating object.
2. Greater stiffness
   • Vibration isolation becomes more difficult due to the higher first-order natural frequencies
   • Static deflection is reduced and even heavy objects can be supported stably.

The following is the result of a previous survey of off-the-shelf anti-vibration components.
Because of this trade-off, we find that there are no off-the-shelf vibration isolation components commercially available in the low-frequency range.
The member at the lower right point of the graph that achieves low-frequency vibration isolation is a member with special spring stiffness, called an air-spring system.

Summary

This article introduced dynamic and static stiffness and the tradeoffs for vibration isolation that can be obtained from them.

In the next article, I will introduce the specifics of one of the structural metamaterials, called QZS, that may be able to break through the tradeoffs introduced in the last section.