# Mechanical Vibration

## Web Book

## Introduction

**Learning objectives**

- Recognize phenomena and applications of vibration in practice.
- Model a vibrating system with mass, spring and damper elements.
- Understand the concept of amplitude, frequency and phase.

**Sections**

**1.4 Amplitude, frequency and phase **

When we discuss about vibration, we are talking about a motion which fluctuates around an equilibrium point. Observe the vibration of a vertical motor pump in Figure 1.4.1 below.

**Figure 1.4.1 **A vibrating, vertical motor pump.

Someone may ask, ‘What is the level of vibration?’

You reply, ’8 mm/s’.

‘At what frequency?’

’At 13 Hz’

‘How does the machine vibrate?’

You reply again, ‘Mostly in horizontal direction (x-axis), rocking motion’.

The conversation reveals three important variables in vibration:

Amplitude: represents the level of vibration

(How severe?)

Frequency: represents the speed of vibration

(How fast?)

Phase: represents the behaviour of vibration

(How does it vibrate?)

1.4.1 Amplitude

Let us see an example of motion of a mass on a spring, where its centre moves with a distance from the equilibrium line (position when the mass is not in motion). Suppose there is a pen fixed to the mass and is placed on top of a paper which runs from left to right (with the same speed as that of the mass), we can see in Animation 1.4.1 that the line drawn on that paper looks like a **sinusoidal function**.

So from our observation, the instantaneous displacement y(t) of the moving mass can be expressed as

y(t)=A\displaystyle\cos\left(\frac{2\pi t}{T}\right)=A\cos\left(\omega t\right)\phantom{xxxxxxx}{\color{red}(1.4.1)}

where T is the period (i.e. the time taken from a peak to the next peak, in second), t is the instantaneous time and \omega=2\pi f is the angular frequency (in rad/s), with f the frequency (in Hertz).

**Animation 1.4.1 **Motion of a mass-spring system showing a sinusoidal pattern.

Since maximum of \cos(\omega t) across the time is 1, thus the maximum of the displacement of the mass from Eq. (1.4.1) is y_{max}= A . This is called the **peak amplitude** of vibration.

In Eq. (1.4.1), the position of the mass is changing with time. And therefore instead of stating the amplitude of vibration with displacement, we can also state it with **velocity**.

From Eq. (1.4.1). the velocity is given by

v(t)=\displaystyle\frac{dy(t)}{dt}=-A\omega\sin(\omega t)\phantom{xxxxxxxxx}{\color{red}(1.4.2)}

The maximum amplitude of the velocity is |v_{max}|=\omega A

We do not state the vibration amplitude in ‘negative’ value. This is a dynamic problem where mass is oscillating around the equilibrium position. We are only interested in its magnitude, i.e. how much the mass goes beyond the equilibrium line, either in positive or negative directions.

In terms of acceleration, the amplitude is given by

a(t)=\displaystyle\frac{d^2y(t)}{dt^2}=A\omega^2\cos(\omega t)\phantom{xxxxxxxxxx}{\color{red}(1.4.3)}

where the peak acceleration is |a_{max}|=\omega^2 A .

## Example Problem 1.4.1

A system vibrating harmonically at 2 Hz has a maximum displacement of 2 mm.

What is the maximum velocity?

Solution

For a harmonic motion, the displacement can be expressed as y(t)=A\cos\left(\omega t\right) as shown in Eq. (1.1).

The **peak displacement** is thus A = 2 mm.

The velocity is

v(t)=\displaystyle \frac{dy(t)}{dt}=-A\omega \sin(\omega t)

From this, the **peak velocity** is therefore

v_{max}=|A\omega|=A(2\pi f)=2(2\pi 2)=8\pi mm/s.

## Example Problem 1.4.2

What is the peak displacement of a system known to have peak velocity of 10 mm/s at frequency 4 Hz? The system is known to vibrate harmonically.

Solution

From Example Problem 1.4.1, we know that the peak velocity is

v_{max}=\omega A

The peak displacement is thus

A=y_{max}=\displaystyle\frac{v_{max}}{\omega}=\frac{10}{2\pi(4)}=0.4 mm

## Example Problem 1.4.3

What is the peak displacement of a system with harmonic vibration with amplitude of acceleration of 50 mm/s^2 at 10 Hz?

Solution

From Eq. (1.3), we know that the peak acceleration is

a_{max}=\omega^2 A

The peak displacement is therefore

A=y_{max}=\displaystyle\frac{a_{max}}{\omega^2}=\frac{50}{(2\pi(10))^2}=0.013 mm

Root-mean-square (r.m.s) amplitude

The type of motion which is governed by sine or cosine functions is called **simple harmonic motion**. We can determine the peak amplitude easily. In practice, the vibration in many cases is not always like a sine function, but rather has a stationary random amplitude. Since this type of vibration has a rapid change of amplitude across the time, it is meaningless to define its maximum amplitude at an instantaneous time, t .

We are rather interested in its ‘statistical’ value across a range of time T_r . We can define an amplitude called root-mean-square (r.m.s) value.

If y_r(t) is a stationary random amplitude, the r.m.s value is defined by

y_{\text{rms}}=\sqrt{\displaystyle\frac{1}{T_r}\int_0^{T_r} y_r^2(t)dt}\phantom{xxxxxxxxx}{\color{red}(1.4.4)}

The illustration of the calculation process is shown in Animation 1.4.2.

**Animation 1.4.2 **The step-by-step calculation of the root-mean-square (r.m.s) value.

Note that in principle, the r.m.s amplitude can be either for displacement (in mm), velocity (in mm/s) or acceleration (in mm/s ^2 or in g).

This r.m.s amplitude in practice is used as the **alarm limit** to indicate the severity of vibration of a rotating machine. For example in ISO 10816-3, a newly commissioned, rotating machine with power in the range of 15 -300 kW, installed on a rigid foundation, must ideally have r.m.s amplitude of no more than 0.7 mm/s.

If the r.m.s amplitude is between 2.8-3.5 mm/s, then it must be restricted in operation and this should trigger the engineer to inspect the machine and to perform necessary maintenance before the machine falls into the zone of ‘damage occurs’.

Figure 1.4.2 shows the chart from ISO 10816-3: Mechanical vibration – Measurement and evaluation of machine vibration.

Note: ISO 10816-3 has been replaced with the updated ISO 20816-3:2022.

**Figure 1.4.2 **The vibration severity chart from ISO 10816-3.

1.4.2 Frequency

Back to Animation 1.4.1, the time taken for the center of the mass to return to the same location is t=T (from peak to peak, or dip to dip). This is called **period**. The inverse of the period is **frequency**, which means how many times the centre of mass returns the same location in one second, which is expressed as

f=\displaystyle\frac{1}{T} \qquad (\text{in Hertz})\phantom{xxxxxxxxxxxxxxx}{\color{red}(1.4.5)}

In a rotating machinery, frequency can be associated with the turning speed of the shaft and is often denoted as RPM (rotation per minute) or CPM (cycle per minute) where

1~\text{RPM} =\displaystyle \frac{1~\text{rotation}}{1~\text{minute}}\phantom{xxxxxxxxxxxx}{\color{red}(1.4.6)}

Thus 600 RPM means 600 rotation/60 seconds = 10 Hz.

For an object rotating around a centre of rotation, the frequency is called **angular frequency** or **angular speed** which defines the angular displacement per second written as

\omega=2\pi f \qquad (\text{in radian/s})\phantom{xxxxxxxxx}{\color{red}(1.4.7)}

For the gear in Animation 1.4.3 rotating with the angular speed, \omega the angular displacement of a point on the gear from the starting position to a position at t second, is given by

\theta=\omega t \qquad (\text{in radian})\phantom{xxxxxxxxx}{\color{red}(1.4.8)}

**Animation 1.4.3 **Angular speed of a gear.

For one complete cycle when t=T , the angular displacement from Eq. (1.4.8) is therefore \theta=\omega T=2\pi fT=2\pi T/T=2\pi radians.

Rotational to translational motion

If it is an angular frequency (which should correspond to angular motion), why can it also be used to represent a **translational motion** as shown in Animation 1.4.1?

How does an angular motion relate to a translational motion?

Animation 1.4.4 shows the concept of the Bourke engine which utilises the skotch-yoke mechanism where the motion of the piston (translational motion) is converted into rotational motion. The speed of the piston relates to the angular frequency, \omega of the rotating crank, while the translational amplitude of the piston relates to the length of the crank, A .

**Animation 1.4.4 **The scotch yoke mechanism which converts the linear motion into rotational motion (press the red STOP button to see the math).

Does vibration always have a single frequency, as shown in Animation 1.4.4? The answer is absolutely ‘No’. Vibration can be generated from multiple frequency components. The random vibration as in Animation 1.4.2 actually consists of multiple number of frequencies.

Say we have two machines with two different operating speeds, \omega_A and \omega_B (see Animation 1.4.5). If only one machine operates at one time, the vibration amplitude measured at the corresponding running machine is given by

y_A=A\cos(\omega_At)\phantom{xxxxxxxxxxxxxx}{\color{red}(1.4.9)}

y_B=B\cos(\omega_Bt)\phantom{xxxxxxxxxxxxxx}{\color{red}(1.4.10)}

If both machines are now operated at the same time and by assuming that the vibration from each machine is transferred completely to one another, the total vibration measured at each machine is mathematically given by

y_{AB}(t)=y_A(t)+y_B(t)=A\cos(\omega_At)+B\cos(\omega_Bt)\phantom{xxx}{\color{red}(1.4.11)}

Animation 1.4.5 shows how the vibration amplitude changes with the total summation of two vibration with different frequencies.

**Animation 1.4.5 ** Two machines operating with very close operating speeds (frequencies) [top] and summation of vibration amplitudes of two different frequencies [bottom].

Using trigonometry relation where \cos X+\cos Y=2\cos[(X+Y)/2]\cos[(X-Y)/2] , Eq. (1.4.11) can be written as

y_{AB}(t)=2AB\cos[(\omega_A+\omega_B)t/2]\cos[(\omega_A-\omega_B)t/2]\phantom{xx}{\color{red}(1.4.12)}

The total amplitude is now a product of two cosine functions, each with different frequency, namely the large frequency \omega_A+\omega_B and the small frequency \omega_A-\omega_B=\epsilon .

As seen in Figure 1.4.3, the second term in Eq. (1.4.12), which is \cos(\epsilon t/2) determines the period of the interchange between the ‘soft’ and ‘loud’ noise (as simulated in Animation 1.4.5).

**Figure 1.4.3 **A typical feature of a beating vibration amplitude.

1.4.3 Phase

Let us look at the mathematic first before we proceed to the meaning of the variable ‘phase’. The amplitude of a vibrating system can come with an **absolute phase** information \phi written as

y(t)=A\cos(\omega t+\phi)\phantom{xxxxxxxxxxx}{\color{red}(1.4.13)}

where \phi is in radian.

The absolute phase in Eq. (1.4.13) only shows how the system starts to vibrate at t=0 , which in vibration analysis does not have any meaning. We are interested in the condition of vibration at all times.

Now let us see two different systems P and Q, running with the same frequency, but have different absolute phases.

y_P(t)=A\cos(\omega t+\phi_P)\phantom{xxxxxxxxxxx}{\color{red}(1.4.14)}

and

y_Q(t)=A\cos(\omega t+\phi_Q)\phantom{xxxxxxxxxxx}{\color{red}(1.4.15)}

The phase difference between them is called **relative phase**, where

\Delta \phi=\phi_P-\phi_Q\phantom{xxxxxxxxxxxxx}{\color{red}(1.4.16)}

**Animation 1.4.6 ** Two systems vibrating with different phase [tap on each mass to change its level of amplitude].

Phase and mode shape

With the same concept in mind, the phase information (as well as the amplitude) obtained from measurement can also be used to construct the motion of a vibrating continuous system, for example a vibrating beam as shown in Animation 1.4.7. This shows the vibration behaviour (called **mode shape**) of a cantilever beam at its first natural frequency (1^{st} mode) and second natural frequency (2^{nd} mode).

For the 1^{st} mode, point 1 and 2 are in-phase, but different displacement amplitude. For the 2nd mode, point 1 and 2 have almost the same displacement, but are out-of-phase.

**Animation 1.4.7 **Role of phase in determining the mode shape of a continuous structure (example: cantilevered beam).

In order to construct the vibration mode shape in more details, the more points (with phase and amplitude information) are therefore required at a particular frequency. In engineering practice, the techniques used to visualise the mode shape are namely **modal testing** or **operating deflection shape** (ODS). The concept is similar to the application of numerical calculation (simulation technique) using the Finite Element Method as shown in Figure 1.3.4.

Phase and vibration control

With similar concept as in Animation 1.4.5, the vibration of two systems with different phase (but of same amplitude) can be superimposed together and the total amplitude can turn into zero (canceled out) for the relative phase of \Delta \phi=\pi radians. Play the graph in Animation 1.4.8 by changing the phase of y_Q.

We can see that the total amplitude y_P+y_Q=0 is when \phi_Q=-1.57 radians, or -\pi/2 radians.

Thus the relative phase between system *P* and *Q* is

\Delta\phi=\phi_P-\phi_Q=\pi/2-(-\pi/2)=\pi radians.

This is the fundamental concept applied in **active vibration control**, where to cancel out vibration amplitude of, say *A* mm/s, we have to inject an external vibration amplitude of also *A* mm/s, but with relative phase of 180 degrees (out-of-phase).

**Animation 1.4.8 **The effect of ‘relative’ phase on the total vibration amplitude.

SUMMARY

The amplitude, frequency and phase are three variables in vibration which provide us the understanding on how we approach the vibration problems and how to solve them. These variables, in practice, are measured using vibration instruments, and after signal processing, they will be visualized in a graph or even in a single data number.

Animation 1.4.9 is the summary on how to visualize what amplitude, frequency and phase are in a vibrating system.

**Animation 1.4.9 **Visualization of amplitude, frequency and phase in a vibrating system.