ESN 459 Applied Mechanics Laboratory...3.Chapters from D. G. Alciatore and M. B. Histand, Introduction to Mechatronics and Measurement Systems, 4th Ed., McGraw Hill, 2012. Requirements

ESN 459 Applied Mechanics Laboratory

Prof. Aleksandar Haber

Department of Engineering Science and Physics

August 30, 2018

Class organization and other information

I Email: [email protected].

I Office hours: On Thursdays from 2-4 PM and by appointment.

I Lectures and experiments.

I 4 credits and 3 laboratory hours.

I Prerequisite or co-requisite: ESN 450. Prerequisite: ENS 249.I Literature:

1. Lecture notes and handouts.2. Chapters from J. P. Holman, Experimental Methods for

Engineers, 8th Ed., McGraw-Hill, 2012.3. Chapters from D. G. Alciatore and M. B. Histand, Introduction

to Mechatronics and Measurement Systems, 4th Ed., McGrawHill, 2012.

Requirements

Class culture:

I Cell phones should be turned off.

I If you are more than 30 minutes late, then it is better to skip theclass.

Requirements:

I If you are absent more than 15 percent of the class hours, you willbe assigned a grade of WU (withdrew unofficially). That is sixhours.

I More than F grade on homework assignments, lab reports, and brieftests.

I You will have to defend your lab reports.

I Late submissions will not be tolerated.

Goals of this class

1. Master the art of measurement and data analysis.

2. Understand the difference between models and real physicalsystems.

3. Identify the sources of model uncertainties.

4. Learn how to clearly and effectively communicate your ideasand learn to the art of writing technical and scientific reports.

5. Understand the basic principles of classical mechanics, fluidmechanics and thermodynamics by performing measurementexperiments.

MATLAB and LaTeX

I Engineers today cannot live without MATLAB!

I Expectation is that all the calculations and graphs are doneusing MATLAB. If you are familiar with C or C++, you canalso use them.

I You are encouraged to write your reports using LaTeX.

Basic MATLAB

I Vectors, matrices, plotting a function, symbolic calculationand graph formating, obtaining help.

I Logarithmic plots and error bar graphs.

I Histograms and box plots.

I Gaussian and radom distribution.

Types of errors

Chapter 3 from the book (from here up to end of the lecture...)

I Fixed errors (systematic or bias errors). Errors that are roughlyindependent from the number of performed measurements.

I Random errors.These kind of errors usually follow a certain statisticaldistribution, but not always!

Which statistical distribution comes to your mindimmediately?

Mean, standard deviation and variance

Say that you measure a temperature and that you obtained nmeasurements, x1, x2, . . . , xn, where xi is the temperaturereading at the ith measurement. Then you might ask the followingquestions:

I What is the ”best” measurement? How to use all themeasurement data?

I What can you say about the accuracy of the measurements?

Mean, standard deviation and variance

I Arithmetic mean (sample mean):

x =1

n

n∑i=1

xi

I Sample standard deviation:

σn =

√√√√ 1

n − 1

n∑i=1

(xi − x)2

What happens when N approaches ∞ ?

I Sample standard variance

σ2n

Probability distributions

-3 -2 -1 0 1 2 3 40

0.5

1

1.5

2

2.5

3

-3 -2 -1 0 1 2 3 40

1

2

3

4

5

6

7

8

9

10

-3 -2 -1 0 1 2 30

5

10

15

20

25

-4 -3 -2 -1 0 1 2 3 40

50

100

150

200

250

300

N=10 N=50

N=100 N=1000

In MATLAB

n = 10

X = randn(n, 1)

X = hist(X )

Gaussian or normal error distribution

f (x) =1√

2σ2πe−

(x−µ)2

2σ2

Hear about the 3-sigma rule?

Why the Gaussian (normal) distribution is important?Because of the central limit theorem!Imagine we perform random measurements of a process which truevalue is µ, and we do not know how the underlying distribution ofthe measurement variable. Imagine you measure the temperatureinside of this room. Say we have n measurements, x1, x2, . . . , xn,and we compute the average:

x =1

n

n∑i=1

xi

I Now, if we repeat n measurements over and over again, everytime x will have a different value! Why?

I So, the empirical mean x is a random variable itself.

I What we can say about the distribution of x?

I Can we say something how accurate is x compared to the”true” value µ?

Central limit theoremSay we have an n random independent and identically distributedrandom variables, x1, x2, . . . , xn, with the same arbitrarydistribution and with the expectation µ and variance σ2. Then asn approaches ∞, the random variable

√n (x − µ)

is approximately normally distributed with a zero mean and avariance of σ2.IMPORTANT:

I Let E [z ], denote the expectation of a random variable z .Then,

E [x ] = µ

I And equally important

σx =σ√n

Important implications

I What happens with the variance of x as n increases?

I What are the practical implications!

I Homework. To be explained at the class.

Least-squares method

Suppose that that x is an independent variable that we can adjust(for example, a position of a valve), and y is a measured quantitythat depends on x . Suppose for a moment that we neglect themeasurement noise in our experimental setup. Furthermore,suppose that x and y are related according to the followingequation:

y = αx2 + βx + c (1)

where α, β and c are the model parameters. The equation (23)can be written in the vector form:

y =[x2 x 1

] αβc

(2)


Suppose that the measurements of y are affected by a noise with aGaussian distribution. Then, our model equation has the followingform:

y =[x2 x 1

] αβc

+ n (3)

where n is a measurement noise. Let us assume that we vary x ,and for each of its values, we obtain a measurement of y . Forexample, say that we have a set of M values of x :x1, x2, . . . , xM. For every value xi , we obtain a value of y ,denoted by yi . That is, we have the following ordered setG = (x1, y1) , (x2, y2) , . . . , (xM , yM).Our goal is estimate the model parameters α, β and c from the setof data G.

Least-squares methodWe can write

y1 = αx21 + βx1 + c + n1

y2 = αx22 + βx2 + c + n2

y3 = αx23 + βx3 + c + n3...

yM = αx2M + βxM + c + nM

Or in the matrix form:y1y2y3...yM

︸︷︷︸

y

=

x21 x1 1x22 x2 1x31 x3 1...

......

x2M xM 1

︸︷︷︸

X

αβc

︸︷︷︸

d

+

n1n2n3...

nM

︸︷︷︸

n


We can write compactly

y︸︷︷︸measurements

= X︸︷︷︸independent

× d︸︷︷︸unknown

+ n︸︷︷︸noise

(4)

Least-squares problem formulation

mind‖y − Xd‖22 (5)

Solution

d =(XTX

)−1XTy (6)







(7)


mind‖y − Xd‖22 (8)

For example,

‖y − Xd‖22 =(y1 − αx21 − βx1 − c − n1

)2+(y2 − αx22 − βx2 − c − n2

)2







(9)


mind‖y − Xd‖22 (10)

Solution

d =(XTX

)−1XTy (11)

Matlab implementation to be shown during the class.

Gravitational accelerationBall of mass m, with an initial condition x(t) = 0 and x(0) = h.When will the ball fall on the ground?

mx = −mg , x = −gx = −gt + C1, x(0) = 0; x = −gt

x = −g t2

2+ C2, x(0) = h, x = −g t

2

2+ h

0 = −g T2

2+ h, T =

√2h

g

Homework-Estimate the constant g using the least-squaresmethod

Use the stopwatch on your cell phone to estimate g . Procedure:for different values of h, measure the time when the ball hits theground (use at least 10 values of h). On the basis of the equation

h = gT 2

2, (12)

Form a least-squares problem, and estimate the constant g . Howaccurate is your estimate? What happens when you increase thenumber of experiments? Do you get a better estimate?Compare the least-squares estimate with the estimate obtained asfollows. First, for every value of h, and a measured value of T ,compute g by solving Eq. (12). Then, find an average of allcomputed g . This average is also an estimate of g . Does thisestimate have a better accuracy than the least-squares estimate?

Governors

Figure: Centrifugal governor and a turbine (Credit: Wikipedia).

A governor is a mechanical device used to control the speed of amachine by regulating the fuel or power supply.

Governors

Figure: A more realistic design.

Porter Governor

SM

Figure: Porter governor

Porter Governor

SM


Assumption:I In the revolving plane, the system is in equilibrium.

Sum of moments around the point I should be 0:

F · BD − w · ID − W

2· IC = 0

Porter Governor

SM


F = w · IDBD

+W

2· ICBD

F = w · tan(α) +W

2· (tan(α) + tan(β))

Porter Governor

SM


F = w · IDBD

+W

2· ICBD

F = w · tan(α) +W


Centrifugal force:

F =W

g· ω2 · r

where ω is the angular velocity, g is the gravitational acceleration,and r is the distance between B and the rotational axis.

Porter Governor

F = w · IDBD

+W

2· ICBD

= 0

F = w · tan(α) +W


Centrifugal force:

F =W

g· ω2 · r

where ω is the angular velocity, g is the gravitational acceleration,and r = BS is the distance between B and the rotational axis.After the substitution, we get:

W

g· ω2 · r = w · tan(α) +

W

2· (tan(α) + tan(β)) (13)

Porter Governor

SM


W

g· ω2 · r = w · tan(α) +

W


tan(α) =r

h, tan(β) =

BM

CM(14)

Porter Governor

SM


W

g· ω2 · r = w · tan(α) +

W


tan(α) =r

h, tan(β) =

BM

CM(15)

Impact of a jet

Figure: Water turbine (source Wikipedia)

Impact of a jet

Figure: Water turbine (source Wikipedia)

Impact of a jet

Jockey WeightRetaining Screw

Tally

Flat Plate

Drain PipeSupply Hose

To Weighing Tank

Cover Plate

AdjustingNut

Nozzle

Weigh Beam

Water out

Water in

Figure: Jet impact apparatus

Impact of a jet

Control Volumeof Jet

Flat Plate

uo

u1u1

F

β

Figure: Flat vane

I F , [N] - the force exerted by the jet on the plate in thedirection of the jet.

I u0, [m/s]- velocity of fluid striking the vane.I u1, [m/s]- velocity of fluid leaving the vane.I W , [kg/s]- mass flow rate.I Q, [m3/s]- volume flow rate.

Impact of a jet


Flat Plate

uo

u1u1

F

β

Figure: Flat vane

I W = ρQ, where ρ is a fluid density.

F=rate of change of momentum in the direction of force.

Impact of a jet


Flat Plate

uo

u1u1

F

β

Figure: Flat vane

The rate at which the momentum is entering the control volume inthe vertical direction:

Wu0

The rate at which the momentum is leaving the control volume inthe vertical direction is zero. Hence,

F = Wu0 = ρu0Au0 = ρAu20

Impact of a jet

F

β

x

u1

uo

Figure: Triangular vane

What do you expect, is the magnitude of force higher or lower?

Impact of a jet

F

β

x

u1

uo


What do you expect, is the magnitude of force higher or lower?

Impact of a jet

F

β

x

u1

uo


What is entering the system in the vertical direction?

Wu0

What is exiting the system in the vertical direction?

W

2u1 cos (β) ,

W

2u1 cos (β)

Impact of a jet

F

β

x

u1

uo


So, the total force is

F = Wu0 −W

2u1 cos (β)− W

2u1 cos (β) = Wu0 −Wu1 cos (β)

F = W (u0 − u1 cos (β))

Impact of a jet

F = W (u0 − u1 cos (β))

Now, what is the relation between u0 and u1?

Impact of a jet

F = W (u0 − u1 cos (β))

Now, what is the relation between u0 and u1? Because the massconservation law:

Au0 = A1u1 + A1u1 = 2A1u1

A = 2A1

So we get

u0 = u1

At the end we obtain:

F = ρAu20 (1− cos (β))

Impact of a jet

u1 uo u1

Control Volume of Jet

HemisphericalCup β

F

Figure: Cup-shaped vane

Let us do it together!

Impact of a jet

Figure: Cup-shaped vane

Pelton turbine experiment

Figure: Pelton turbine

Modified Porter governor-experiments

Figure: Modified governor. We have modified the governor by pituing aplate of mass m3.


M

G


2T3 cos (β)− (m2 + m3)g = 0

T3 cos (β) =1

2(m2 + m3)g


M

G

Equilibrium: Sum of moments around the point B should be 0.

T3AB cos(β) + m1gMB − FCM = 0

1

2(m2 + m3)gAB + m1gMB − FCM = 0

F =1

2(m2 + m3)g

AB

CM+ m1g

MB

CM


F =1

2(m2 + m3)g (tan(α) + tan(β)) + m1g tan(α)

Centrifugal force:

F = m1ω2EC , EC =

√l22 − x2

F = m1ω2√

l22 − x2

On the other hand,

tan(α) =

√l22 − x2

x,

tan(β) =

√l22 − x2 − l3√

l21 − (l22 − x2) + 2√

l22 − x2l3 − l23


F =1


G (x) =1


m1ω2√

l22 − x2 = G (x)

ω =

√√√√ G (x)

m1

√l22 − x2

Angular velocity measurements-Photgate device

The photogate works by projecting an infrared beam from one armof the sensor to the other arm. When the beam is blocked thesensor stops sending a signal, which illuminates an LED on the topof the gate as well as triggering Logger Pro to display a blockedmessage in the data collection area.

Motion detector

This Motion Detector emits short bursts of ultrasonic sound wavesfrom the gold foil of the transducer. These waves fill a cone-shapedarea about 15 to 20 off the axis of the centerline of the beam. TheMotion Detector then listens for the echo of these ultrasonic wavesreturning to it. The equipment measures how long it takes for theultrasonic waves to make the trip from the Motion Detector to anobject and back. Using this time and the speed of sound in air, thedistance to the nearest object is determined.

Fluid dynamics

So, what is a fluid?

I A fluid is a substance that deforms continuously under theapplication of a shear (tangential) stress no matter how smallthe shear stress may be.

I In other words, fluid is any substance that cannot sustain ashear stress when at rest.

Newtonian fluidWe have two plates. Bottom one is fixed and the top one ismoving.

I When the fluid is sheared, it begins to move at a strain rateinversely proportional to a property called its coefficient ofviscosity µ.

I u is a fluid velocity, τ is a shear stress, and θ is a shear angle.

Newtonian fluid is characterized by a linear relation

τ = µdu

dy

Inviscid flow, streamlines, steady flow

I We say that a flow is inviscid if µ = 0.

I A streamline is a line everywhere tangent to the velocityvector at a given instant.

I A steady flow is one in which the conditions (velocity, pressureand cross-section) may differ from point to point but do notchange with time.

Bernoulli equation

Assumptions

I Steady flow.

I Incompressible flow.

I Frictionless flow.

I Flow along a streamline.

Static, Stagnation, and Dynamic PressuresBernoulli equation

p

ρ+

V 2

2+ gz = const

I pressure p is the thermodynamic pressure-it is commonlycalled the static pressure.

I The stagnation pressure is obtained when a flowing fluid isdecelerated to zero speed by a frictionless process.

I

Static, Stagnation, and Dynamic Pressures

From the Bernoulli equation

p0ρ

+V0

2=

p

ρ+

V 2

2

Velocity is zero, it implies

p0 = p +1

2ρV 2

I p0 is the total pressure.

I The term 12ρV

2 is the dynamic pressure.

We can express the velocity as follows

V =

√2(p0 − p)

ρ(16)

The conclusion is thatBy measuring the static and total pressure, we can obtain a fluidvelocity.

V =

√2(p0 − p)

ρ(17)

Measure p0 and p, and directly calculate V .

Standing in a steady wind holding up your hand. What do you feel?Fact: There is no pressure variation normal to straight streamlines.

Bernoulli obstruction theory

Define the parameter β = d/D.

Continuity equation Q = πD2V14 = π

4D22V2.

Bernoulli: p0 = p1 + 12ρV

21 = p2 + 1

2ρV22 .

At the end we get

Q

A2= V2 ≈

√[2(p1 − p2)

ρ(1− D42/D

4)

]

Flow measurementsAn effective way to measure the flowrate through a pipe is to placesome type of restriction within the pipe and to measure thepressure difference between the low-velocity, high-pressureupstream section, and the high-velocity, low-pressure downstreamsection.

Flow measurementsAn effective way to measure the flowrate through a pipe is to placesome type of restriction within the pipe and to measure thepressure difference between the low-velocity, high-pressureupstream section, and the high-velocity, low-pressure downstreamsection.

A more realistic version of the Bernoulli equation

It should be kept in mind that the Bernoulli equation is valid underthe following two assumptions:

I Incompressible fluid.

I Inviscid flow (no friction).

However, in reality

I Fluids are compressible. Compress a fluid particle => pressureincreases => temperature increases.

I Friction converts mechanical energy to thermal energy.

Since we will mostly deal with water, we will neglect thecompressibility effects and focus only on friction.


What creates friction?

Newtonian fluid is characterized by a linear relation

τ = µdu

dy

Work is done by the shear force, this energy is converted to heat.


Friction is created by

I Fluid rubbing against the walls of the pipe.

I Fluid layers rubbing against each other.

I Turbulence.

How do we observe friction:

I Sound

I Heat


Assuming a flow throughout a pipe, the amount of frictional loss isaffected by the following parameters

I The length of the pipe. How does it depend?

I The roughness of the pipe walls.

I The diameter of the pipe. How does it depend?

I The velocity of the fluid. How does it depend?

I The type of flow of the fluid. Laminar? Turbulent?

I Change in the shape or section of the pipe.


p1ρ

+V1

2+ gz1 =

p2ρ

+V2

ρ+ gz2 + ∆h

What we have modified?

I We have taken into account all the friction losses by the term∆h = ∆h1 + ∆h2.

I ∆h1-major loss-This is essentially an internal energy increase,that is, mechanical energy is transferred to the thermal energy.

I ∆h2-minor losses resulting from entrances, fittings, areachanges, and etc.

Reynolds number

Re = ρVD

µ(18)

where Re is the Reynolds number, ρ is the fluid density, D- is thepipe diameter, V is the average velocity, and µ- is the dynamicviscosity.

Losses due to friction

I For a pipe with the constant diameter, we have

Q =π∆pD4

128µL(19)

where Q is the flow, ∆p is the pressure drop, D is thediameter, L is the length.

I Major loss for a laminar flow.

∆h =

(64

Re

)L

D

V 2

2(20)

Minor loss

The flow in a piping system may be required to pass through avariety of fittings, bends, or abrupt changes area. Additional lossesare encountered, primarily as a result of flow separation.

I Flow separation-The fluid flow becomes detached from thesurface of the object, and instead takes the forms of eddiesand vortices.

Minor loss

∆h2 = KV 2

2(21)

Minor loss

Fourier’s law and Heat equationHow is the heat transferred from one end to another end of anobject? What is the mechanism of transfer?

I Internal energy diffuses as rapidly moving or vibrating atomsand molecules interact with neighboring particles, transferringsome of their microscopic kinetic and potential energies.

I Heat is transferred by conduction when adjacent atoms ormolecules collide, or as several electrons move backwards andforwards from atom to atom in a disorganized way so as notto form a macroscopic electric current, or as phonons collideand scatter.

I Conduction is greater in solids because the network ofrelatively close fixed spatial relationships between atoms helpsto transfer energy between them by vibration.

I The inter-molecular transfer of energy could be primarily byelastic impact, as in fluids, or by free electron diffusion, as inmetals, or phonon vibration, as in insulators. In insulators, theheat flux is carried almost entirely by phonon vibrations.

Fourier’s law

The law of heat conduction, also known as Fourier’s law, statesthat the time rate of heat transfer through a material isproportional to the negative gradient in the temperature and to thearea, at right angles to that gradient, through which the heat flows.

q = −k∇T (22)

where

I q is the local heat flux density Wm−2.

I k is the metal conductivity Wm−1K−1

I ∇T is the temperature gradient.

Heat equation-one dimensional

∂T

∂t= α

∂2T

∂x2

where α is the thermal diffusivity constant.Our goal is to numerically solve this equation and to compare itssolution to the experimental measurements. You will start toappreciate the difference between the mathematical model and areal life experiment.

Finite-difference approximation

Discretize the space and time:

Tmi ≈ T (m∆t, i∆x)

where ∆t is the time discretization step

Partial derivatives approximation:

∂T

∂t≈

Tmi − Tm−1

i

∆t,

∂2T

∂x2≈

Tmi+1 − 2Tm

i + Tmi−1

(∆x)2


Discretize the space and time:

Tmi ≈ T (m∆t, i∆x)

where ∆t is the time discretization step


∂T

∂t≈

Tmi − Tm−1

i

∆t,

∂2T

∂x2≈

Tmi+1 − 2Tm

i + Tmi−1

(∆x)2


∂T

∂t= α

∂2T

∂x2


∂T

∂t≈

Tmi − Tm−1

i

∆t,

∂2T

∂x2≈

Tmi+1 − 2Tm

i + Tmi−1

(∆x)2

Tmi − Tm−1

i

∆t= α

Tmi+1 − 2Tm

i + Tmi−1

(∆x)2

After some transformation:

−k1Tmi−1 + (1 + 2k1)Tm

i − k1Tmi−1 = Tm−1

i

where

k1 =α∆t

(∆x)2


After some transformation:

−k1Tmi−1 + (1 + 2k1)Tm

i − k1Tmi−1 = Tm−1

i

Write down equations:

(1 + 2k1)Tm1 − k1T

m2 = Tm−1

1

− k1Tm1 + (1 + 2k1)Tm

2 − k1Tm3 = Tm−1

2

− k1Tm2 + (1 + 2k1)Tm

3 − k1Tm4 = Tm−1

3


Write down equations:

(1 + 2k1)Tm1 − k1T

m2 = Tm−1

1

− k1Tm1 + (1 + 2k1)Tm

2 − k1Tm3 = Tm−1

2

− k1Tm2 + (1 + 2k1)Tm

3 − k1Tm4 = Tm−1

3

Matrix form:

(1 + 2k1) −k1 0 0 . . .−k1 (1 + 2k1) −k1 0 0 . . .

0 −k1 (1 + 2k1) −k1 0 . . .. . . . . . . . . . . . . . .. . . 0 0 −k1 (1 + 2k1)

Tm1

Tm2

Tm3

Tm4...

TmN

=

Tm−11

Tm−12

Tm−13

Tm−14...

Tm−1N


(1 + 2k1) −k1 0 0 . . .−k1 (1 + 2k1) −k1 0 0 . . .

0 −k1 (1 + 2k1) −k1 0 . . .. . . . . . . . . . . . . . .. . . 0 0 −k1 (1 + 2k1)

Tm1

Tm2

Tm3

Tm4...

TmN

︸︷︷︸

xm

=

Tm−11

Tm−12

Tm−13

Tm−14...

Tm−1N

︸︷︷︸

xm−1

Axm = xm−1

xm = Bxm−1, B = A−1


At the end of the rod, the temperature is constant. Incorporatingthe boundary conditions:

Tm1 = T 0

1 , TmN = T 0

N , ∀m ∈ N0

We modify the second and (N − 1)th equation:

−k1T 01 + (1 + 2k1)Tm

2 − k1Tm3 = Tm−1

2

−k1T 0N + (1 + 2k1)Tm

N−1 − k1TmN−2 = Tm−1

N−1


Now the vector dimension is x ∈ RN−2

A1xm = xm−1 +

k1T

01

0...0

k1T0N

︸︷︷︸

c1

xm = B1xm−1 + B1c1, B1 = A−11

Frequency response of mechanical systems



mx = −kxmx + kx = 0

where m is mass and k is a spring constant.


Can you solve this equation?

mx + kx = 0

As an engineer, can you guess the mathematical form of thesolution?


mx + kx = 0 (23)

Assume x = ert , and substitute in (23), to obtain(mr2 + k

)ert = 0 (24)

And we have

mr2 + k = 0, r = ±ωj , ω =

√k

m(25)


x(t) = K1eωtj + K2e

−ωtj

x(t) = C1 cos(ωt) + C2 sin(ωt)

x(t) = A cos(ωt + δ)

where

A =√C 21 + C2, cos(δ) =

c1A, sin(δ) = −c2

A(26)


Forced oscillations

mx + kx = F0 cos (Ωt) (27)

The solution should look like

x(t) = A cos(ωt + δ) + xp(t)

We have two cases, depending on Ω,do you know them?


First case, Ω 6= ω, the particular solution should have the followingform

xp(t) = A1 cos(Ωt) + B1 sin(Ωt)

Substitute in the differential equation, to obtain

xp =F0

m(ω2 − Ω2)cos(Ωt)

The complete solution has the following form

x(t) = A cos(ωt + δ) +F0

m(ω2 − Ω2)cos(Ωt)


Second case, Ω = ω, the particular solution should have thefollowing form

xp(t) = A1t cos(Ωt) + B1t sin(Ωt)

And the solution has the following form

x(t) = A cos(ωt + δ) +F0t

2mωcos(ωt)

The addition of the t in the particular solution will mean that weare going to see an oscillation that grows in amplitude as tincreases. This case is called resonance and we would generally liketo avoid this at all costs.



mx = k (y − x)− bx

mx + bx + kx = ky


General form

mx + cx + kx = 0

x +c

mx +

k

m= 0

Introduce:

ω =

√k

m, ζ =

c

2√km

At the end we get

x + 2ζωx + ω2x = 0

How would you find the solution of this equation?


x +c

mx +

k

m= 0

Solution

x(t) = C1es1t + C2e

s2t + tC3es1 + tC4e

s2︸︷︷︸additional elements if necessary

where s1 and s2 are the roots of the characteristic polynomial

s2 + 2ζωs + ω2s = 0

Roots

s1,2 = ω(−ζ ±

√ζ2 − 1

)


x(t) = C1es1t + C2e

s2t + tC3es1 + tC4e

s2︸︷︷︸additional elements if necessary

s1,2 = ω(−ζ ±

√ζ2 − 1

)Three cases

I ζ > 1 Over-damped system.

I ζ = 1 Critically damped system.

I ζ < 1 Under-damped system.


Over-damped system ζ > 1

x(t) = C1e−ωt

(ζ+√ζ2−1

)+ C2e

−ωt(ζ−√ζ2−1

)

Critically damped system ζ = 1

x(t) = C1e−ωtζ + C2te

−ωtζ

Under damped system ζ < 1

x(t) = e−ωtζ(C1e√

1−ζ2j + C2e−√

1−ζ2j)

RLC circuits

VR + VL + VC = V (t) (28)

RLC circuits

VR + VL + VC = V (t) (29)

RI + LdI

dt+

1

C

∫ t

−∞I (τ)dτ = V (t) (30)

Let Vc be a voltage at the capacitor, then the capacitor current isI = C dVc

dt . The equations have the following form

RCdVc

dt+ LC

d2Vc

dt2+ Vc = V (t)

d2Vc

dt2+

R

L

dVc

dt+

1

LCVc =

1

LCV (t) (31)

RLC circuits

d2Vc

dt2+

R

L

dVc

dt+

1

LCVc =

1

LCV (t)

d2Vc

dt2+ 2ζω

dVc

dt+ ω2Vc = ω2V (t)

where ω = 1√LC

and ζ = R2

√CL .

Documents

ESN 459 Applied Mechanics Laboratory...3.Chapters from D. G. Alciatore and M. B. Histand, Introduction to Mechatronics and Measurement Systems, 4th Ed., McGraw Hill, 2012. Requirements