Mechanical Vibration

Web Book

Chapter 1

Introduction

Learning objectives

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

Sections

advanced divider

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 1.15 below. 

Figure 1.15 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 1.19  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}(1.1)}

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 1.9 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. (1.1) is y_{max}= A .  This is called the peak amplitude of vibration. 

In Eq. (1.1), 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. (1.1). the velocity is given by

v(t)=\displaystyle\frac{dy(t)}{dt}=-A\omega\sin(\omega t)\phantom{xxxxxxxxx}{\color{red}(1.2)}

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}(1.3)}

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)}  

The illustration of the calculation process is shown in Animation 1.10.

Animation 1.10 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 1.16 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 1.16 The vibration severity chart from ISO 10816-3.

1.4.2  Frequency

Back to Animation 1.9, 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}(1.5)}  

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

  1~\text{rpm/cpm} =\displaystyle \frac{1~\text{rotation/cycle}}{1~\text{minute}}\phantom{xxxxxxxxxxxx}{\color{red}(1.6)}  

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}(1.7)}