Monthly Archives: February 2013

Mass-spring systems should be understood before they are used

Mass-spring systems are often used to model the behavior of an isolated vibrating system.

The mass-spring viewpoint should not be utilized in cases where the system can not be reliably modeled as a mass-spring system. The importance of the static compression of the isolating material, due to gravity, is also overemphasized and often misunderstood.

Some important points:

  • The mass isolates vibrations by resisting change of momentum. Usually: the more mass, the better. The weight (due to gravity) has very little to do with the isolation, it's the mass that matters!
  • The spring's main purpose is to carry static forces (for example the weight). It will also try to resist the displacements of the mass, which means that it will transfer forces to the foundations. From the perspective of isolating vibrations, the spring should be as loose as possible, to allow for the mass to vibrate freely.
  • The mass-spring combination only works on frequencies higher than the resonance frequency.
  • Ideally, there would be no "spring", but it's impossible in practice.

Simplifying rotating parts

The simplification of a rotating machine part

Let's start from the very basic reasoning behind why the mass-spring perspective is so common when designing vibration isolation. Consider case a. We have a machine, with a part rotating around some axis.

Case b shows the varying forces such a movement creates. Perpendicular forces exist relative to the x-axis and y-axis. The varying force in the direction of the x-axis, together with the resisting force at the support, also create a moment of force (kind of like attempting to tip the machine over). The vertical component of the force often creates a moment of force, as well. As does accelerating or decelerating angular motion.

In practical cases, it has been shown that it's often enough to consider the sinusoidal force shown in case c (but it's not always enough!).

Note that this model assumes both the machine and the support to be perfectly rigid. Note also that the force required to keep the rotating mass in bay will rise along with the frequency (\omega^2)! I think I will return to this in another post.

The equilibrium of the system

So far we have simplified the system to a sinusoidally varying vertical force, acting on a mass on a spring.

We will describe the system using these parameters:

  • The mass of the machine
  • A varying force directed on the machine
  • The varying displacement of the machine
  • The spring constant, which we'll assume to be constant (linearly elastic)

We now have five different sources for forces:

  1. F_{m} = ma = m\ddot y, caused by the mass/momentum of the machine which resists acceleration
  2. F_{k} = k y, caused by the compression of the spring
  3. F_{c} = cv = c\dot y, caused by various energy losses which are directly proportional to the speed of the displacement
  4. F_{e}, external forces which we feed into the system.
  5. F_{g}, the force of gravity pulling down the mass

Using the principle of equilibrium (Newton), the sum of these forces must be 0.

F_k + F_c + F_m + F_e + F_g = 0

F_g can be discarded in our calculations. I'll soon explain why.

Calculating the natural frequency of the system

We know that such a system oscillates naturally at a specific frequency. To calculate this frequency, we assume three things:

  1. The motion is sinusoidal
  2. There are no external forces (F_e)
  3. The damping factor (F_c) approaches zero, allowing for the system to oscillate freely

By assuming the motion follows the function y(t) = A\sin(\omega t), we get the following equation:

F_k + F_m + F_g = k(y(t) + y_0) + m\ddot y(t) + mg = k A\sin(\omega t) - m A \omega^2\sin(\omega t) = 0,

where y_0 is the compression caused by the pull of gravity.

k y_0 has to be equal to m g for the system to balance out (Hooke's law), which means that we can throw the effect of gravity out of the equation. Intuitively this can be explained as following: as the effects of gravity and the static compression of the string (k y_0) constantly cancel each other out, only the vibrating part of the displacement (y(t)) makes any difference. The force caused by the vibrating part of the displacement is still directly proportional to the spring constant. Gravity (and as such the static compression of the spring) makes no difference. Gravity only affects the situation if the spring constant changes because of it, not because it compresses the spring or adds a constant downward force to the mass. This holds true for all constant forces, so constant forces should never be included in a vibrating mass-spring system to avoid confusion! They should only be used to calculate the correct spring stiffness, which is almost always very nearly linear when the vibrations are small.

The forces will balance out when \omega = \sqrt{\frac{k}{m}}, or f_0 = \frac{1}{2 \pi}\sqrt{\frac{k}{m}}.

Note that this is the combined natural frequency of the spring and the mass! This might seem obvious now, but when we replace the mass and spring constant with something else, it might not be that obvious. The compression of the spring, using Hooke's law, will be mg = k\delta x. Inserting this into the formula will give f_0 = \frac{1}{2 \pi}\sqrt{\frac{g}{\delta x}}. This formula is often used for practical purposes, but effectively hides the real parameters the natural frequency is based on!

Calculating the forced response of an undamped system

We will consider the steady state response, meaning the response of the system after being excited at a specific frequency for a long time (when everything balances out). We will assume the mass will move at the same frequency as the force exciting it.

The equilibrium formula shown earlier will now look like this:

m \ddot y(t) + k y(t) = F(t)

Assuming the machine will be excited according to the function F(t) = F\sin(\omega t) (ignoring everything else for now) and the machine moving according to the function y(t) = A\sin(\omega t) we get the following formula:

-m A \omega^2 sin(\omega t) + k A sin(\omega t) = F_0\sin(\omega t)

Solving for A (the amplitude of the motion caused by the force), we get the following formula:

A = \frac{F_0}{-m \omega^2 + k}

Note that this describes the displacement of the mass. I think the following concept is really important to understand; the force transmitted into the foundations is directly proportional to the displacement. As the spring constant approaches zero (let's imagine the mass floats in space, for example), the system will still vibrate with an amplitude of \frac{F_0}{m \omega^2}. Increasing the mass will decrease the amplitude of the vibration, also when there is no spring.

To get the force transmitted into the foundations of the machine, we need to multiply the displacement with the spring constant k (according to Hooke's law). From this, we get the relation of the transmitted force to the original force \frac{F_{tr}}{F_0}:

\dfrac{F_{tr}}{F_0} = \bigg\vert\dfrac{1}{\frac{-m}{k}\omega^2 + 1}\bigg\vert

The formula nicely shows us a few things:

  • The resonant frequency is at \omega = \sqrt{\frac{k}{m}}, which is the same as earlier
  • The effect of decreasing the stiffness of the spring
  • The effect of increasing the mass of the machine


Please note that this has been a significant simplification of the situation. Ideally, the simplification only works for cases where the vibrating force is purely vertical, and acts in the exact direction of the centre of mass of the object.

The physics of sound in air, from a statistical point of view (part 2)

This is the second part of a study on the statistical nature of sound.

The incident velocity of the collision

Let's consider a particle moving along the x-axis towards the right, in an isolated system with zero average particle velocity. A single particle, moving towards the right, among several particles moving according to the statistical distribution of ideal gases shown in the previous post. Now let's attach the reference point to the particle. This will be equivalent to a situation where the particle stays still, and all the other particles follow a shifted velocity distribution. The probability distribution will be shifted by the velocity of the single particle, but in in the opposite direction.

Note that as we are now investigating things from the perspective of the particle, all collisions will be identical, no matter what the direction of the incoming particle is. Only the velocity of the incoming particle and the angle of incidence matters. This is illustrated in the following image; we can rotate the possible collisions coming from a certain direction any way we like, the situation stays symmetric as the particle doesn't have any velocity from the perspective of the particle.

Collisions only depend on velocity and angle of incidence of other particles if we consider the situation from the perspective of a single particle

Collisions only depend on velocity and angle of incidence of other particles if we consider the situation from the perspective of a single particle

Now we have two things to sort out: how fast will the incoming particle move, and what angle will it arrive from?

Almost all the information we require to solve this is shown in the shifted probability distribution. As an intuitive example: the further we shift the probability distribution to the left, the larger portion of the distribution will be to the left of the particle. In practice this means that when a single particle moves fast enough to the right, almost all particles it encounters will move to the left (relative to the moving particle).

The faster some other particle moves, the more probable it is that it will hit the static particle we're observing within a given time. This relationship is very nearly linear, which can be based on the following: the area a single particle, somewhere in the box, will cover in a given time is almost directly proportional to its speed. Thus the chance that the area will overlap the area required for a collision is also directly proportional to its speed (except at very small velocities, which we can safely ignore here, as the average distance between collisions is significantly larger than the area of one particle, these small velocity collisions happen very rarely and almost no kinetic energy is transferred when they happen). This probabilistic relationship is shown in the figure below. The figure shows a random particle and the probability that it will hit another particle, purely on the basis of its velocity. Note that this figure will approach infinity if there are an infinite number of velocities (the probability will continue rising forever with increasing velocity), so the statistical distribution will always need to be confined into some specific range of velocities.

The probability of another randomly distributed particle hitting the particle in the center, as a function of velocity

The relative probability of another randomly distributed particle hitting our particle, as a function of velocity

Note also that the statistical distribution shown to the left is purely based on velocity. It doesn't take into account the distribution of velocities (the chance that a random particle has a certain velocity, based on the Maxwell speed distribution). The distribution of the velocities is taken into account in the shifted probability distribution.

To get the probability distribution for how likely it is that a random particle with some velocity will hit our particle, we do the following: we multiply the shifted particle velocity distribution with the probability distribution based purely on velocity (and normalize this distribution so that the statistics are valid, but never mind this now). This can be seen, in the following figure, to show what intuitively can be expected; if the particle is moving to the right, it is much more likely that a collision will happen with a particle which is moving to the left relative to the particle. Note that this statistical distribution will look different depending on the velocity of the particle we are investigating!

Probability of a particle colliding with a static particle, given the velocity

Probability of a  random particle colliding with a static particle if a certain average particle velocity exists, given the velocity of the random particle

To be continued (some day)!


The physics of sound in air, from a statistical point of view (part 1)

This is the first part of a study on the statistical nature of sound.

Considering the phenomena of sound propagation at the very basic level, I think, is the best way for me to understand the very nature of sound. I'm planning on investigating the nature of sound in gases at standard temperature and pressure, yet resorting to as little math as possible (to keep things as clear as I can). I think surprisingly many attributes of sound, which otherwise might be difficult to grasp, can be explained from a statistical point of view (this area of physics is called statistical mechanics). And then there's always the saying "If you can't explain it simply, you don't understand it well enough". So I shall give it a try! Simplicity can also be relative, something to keep in mind 🙂

An ideal gas

I'll start from the assumption that sound propagates in an ideal gas. I've found that the concept of an ideal gas is best understood through the kinetic theory of gases.

The movement of molecules in an ideal gas, according to the kinetic theory of gases (source: Wikipedia)

By viewing the great picture provided by Wikipedia, one can see that an ideal gas consists of particles (molecules) which are constantly in motion. By isolating the system (like in the picture) and assuming that no energy is lost in any of the collisions, the particles will continue colliding forever. This seems like a starting point which is relatively easy to understand.

A small portion of air, like in the box, contains a certain amount of kinetic energy, distributed throughout the particles in the form of different velocities. The velocities of the particles in an isolated system will follow a statistical distribution called the Maxwell speed distribution. I will not investigate the derivation of this distribution any further, and just assume it holds true.

Probability distribution of the velocity of a single particle

Probability distribution of the velocity of a single particle

We can illustrate the statistical distribution further by doing a plot of the most probable velocity a random particle will have in the system. We place the particle in the middle, and consider the color surrounding it as describing the most probable place to draw a velocity vector, with red corresponding to the most probable choice and purple the least probable choice.

There are a few familiar ways to measure the amount of kinetic energy stored in the isolated particle system:

  • pressure, which describes the average force per unit area, caused by the particles colliding with the walls of the container.
  • temperature, which describes the average kinetic energy of the particles.
Probability distribution of the velocity of a single particle with an average particle velocity to the left

Probability distribution of the velocity of a single particle with an average particle velocity to the left

There's also another really important concept  here which relates to acoustics; the average velocity of the particles. In the case of the isolated system above, the average particle velocity will be 0, as the velocities are equally distributed in all directions. But if the box was moving relative to some fixed point, the average particle velocity would differ from 0, and instead be equal to the velocity of the box. This is illustrated in the figure to the right, where the added average velocity simply shifts the probability distribution to some direction by the amount of the average particle velocity. The average velocity is relative; it's fully dependent on the point of reference.