Mechanical math

Question 1

Consider the mechanical system with three degrees of freedom shown below. The positions of the particles are measured from their equilibrium positions. The system has a normal mode eigenvector (6.57, b, 6.57)T.

 If all particles start from their equilibrium positions and the leftmost and rightmost particles are given a velocity of 34.22 ms-1, the velocity of the middle particle is -47 ms-1, the system will oscillate in a normal mode.
Determine the value of b, giving your answer to 3 decimal places.

Answer:

Question 2

An external sinusoidal force is applied to an oscillating system which can be modelled by a model spring and a damper. The general solution of the equation of the motion is given by,

x =  3 + 3.183cos(2t – ) + 3.18exp(-7.72t) cos(t + )

where , , and  are some constants .
Determine the amplitude of the oscillation when steady-state is reached. Give your answer correct to 3 decimal places.

Answer:

Question 3
Consider two particles of masses m1 and m2 joined to each other and to two fixed walls at both ends by three identical model springs of stiffness k and natural length lo. The matrix equation of motion for the mechanical system is shown below .
For certain choices of m1, m2 and k, a normal mode of the system is given by (x1(t), x2(t))T where a, b, ω and  are some constants. If a = -3.77, b = 15, determine how far (in cm) to the left of its equilibrium position does m1 have be initially displaced if m2  is initially 4.09 cm to the right of its equilibrium position, in order for the system to oscillate as a normal mode. Give your answer correct to 3 decimal places.

Answer:

Question 4

A particle A of mass 2.22 kg collides with another particle B of mass 1.3 kg. Their initial velocities are u1 = 1.55i + 0.88j and u2 = 0.21i + 0.94j just before impact. After collision, they both merged and becomes a composite particle, which travel with a velocity of V = 0.67 i + 0.99j.
Calculate the change in kinetic energy. Give your answer to 3 decimal places.
(Note that if there is a loss in kinetic energy, the answer will be negative.)

Answer:

Question 5

The equation of motion of a forced vibration problem is given by

Given the values m = 7.9, r = 65.49, k = 351.4, P = 8.7 and ω = 5.03.
Determine the steady-state solution, xp(14.99) of the differential equation, giving your answer correct 3 decimal places.

Answer:

Question 6
Two particles A (of mass m) and B (of mass 4m) are connected by a model string which passes over a frictionless pulley as shown. A move downwards while B moves horizontally on a frictionless surface, to the right. Both particles move with a common acceleration.
When m = 9.52, calculate the tension, in N, in the model string, giving your answer to 3 decimal places. Take the magnitude of the acceleration due to gravity to be 9.81 ms-2.

Hint
: Apply Newton’s 2nd Law to each of the two particles A and B.

Answer:

Question 7
A particle A, of mass 1 kg, moves along a frictionless horizontal track. The particle is attached to a fixed point O by a model damper, and to another point B by a model spring as shown in the diagram. The two points O and B are a distance 2 metres apart on the track. The damping constant is 15.73 Ns m-1. The spring has stiffness 11.17 Nm-1 and natural length 0.684 metre. The displacement of the particle, x, is measured from O.
Calculate the equilibrium position of the particle, measured from O, giving your answer to 3 decimal places.

Answer:

Question 8
A particle A of mass 1.4 kg collides with another particle B of mass 2.18 kg. Their initial velocities are u1 = 1.65i + 0.71j and u2 = 0.54i + 1.33j just before impact. After collision, they both merged and become a composite particle, which travel with a velocity of V.
Calculate the speed |V|, a scalar, in ms-1. Give your answer to 3 decimal places.
Answer:

Question 9

A mass m of 44.59 kg is travelling towards the left, on a straight frictionless, horizontal track onto buffers at a constant speed of 2.23 ms-1 when at t = 0, it hits the buffer. The buffers are to be modelled by a model spring, with stiffness 135.97 Nm-1, together with a model damper with damping constant 207.73 Nsm-1. The x-axis is chosen directed away from the buffers down the track (in the direction opposite to the incoming mass), with origin at the fixed end of the model spring. The natural length of the model spring is 0.53 m.
The equation of motion is given by, where a, b and c are some constants.

where a, b and c are some constants.
When m = 44.59, determine is the value of c? Give your answer to 3 decimal place.
The system of model damper and model spring is shown below.

Answer:

Question 10
Two particles X and Y, joined by a rigid rod of negligible mass, move along a smooth horizontal plane. Particle X has mass 6.31 kg and is pushed by an external force Fx of magnitude 42.73 N. Particle Y has mass 7.35 kg and is pulled along the plane by an external force Fy of magnitude 43.95 N. It is found that the two particles accelerate along the plane a fixed distance apart.
Calculate the common acceleration of the particles. Give your answer to 3 decimal places.

Answer:

Incture 5

Cove{fleg fof Lpctur,e 5

1. Unit 6 * Normal modes.
o Modelling free undamped vibrations with

two degrees of freedom
o Finding equilibrium positions
. Setting up two equations of motion

2. Normal mode natural frequencies
(eigenvalues).

3. Normal mode displacement ratios
(eigenvectors).
r in-phase and phase opposed motion
r the general displacement equations and

initial conditions

Page I

it Lecturr 5

Unit6 – NormalModes

This Unit is about vibrations and it continues from where
MTH2L3 Units I wtd Mfi{215 Unit 5 left off. Its aim is to
extend your ability to model systems in which vibration play
an important role. It will use some of the mathematical
techniques you have learnt, such as systems of differential
equations, matrices and eigenvalues.

In our previous study of vibration, only one particle of mass
m is involved, and only one displacement x is needed to
completely describe its motion. However, in real mechanical
systems, more than one particle with different masses are
moving and thus more than one displacement needed to
define the positions of the particles.

The number of degrees offreedom of a system is the
smallest number of coordinates which are required to
speciff the configuration of the system at any instant. The
importance of the number of degrees of freedom is that it is
equal to the number af equations af motion. (Refer to

For a one degree offreedora system without damplng and
forcing, the homogeneous equation of motion has the form,

mti+kx=0
The solution to this equation of motion is,

x(t) : Acos(ott + 0)
which is always sinusoidal, cortmonly called simple
harmonie motion (or SHM in short).

We2

I.€gfinE 5

For a mechanical system with two degree offreedom, the
motion of the two particles is different from the much easier
simple harmonic motion described previously. In general,
the motion of & two (or greater) degree offreedora system is
not sinusoidal.

However, there are situations (depending on the inrtial
positions andvelocities of the individual particles) where
each part of the system does oscillate sinusoidally with the
same frequency. The particles complete each cycle together
with the same period. These particular solutions to the
equation of motion are called normal modes.

Definition : A normsl mode is a type of motion of a system
of particles in which the displacemenrs of the
particles all vary sinusoidally (with SHM) at the
s ame angular frequency.

The number of normal modes correspond to the number of
degrees offreedom.

The angular frequency of the sinusoidal motion is called
normal mode angular frequency.

The two particles in the system can move in the same
direction and such motion is said to be in-phose. When the
two particles move in the opposite direction, then the motion
is said to be phase-opposed.

Exercise 1.2 in page 9 illushates some of the above
definitions.

Pag€ 3

This is a model with two degrees of freedom having two
masses and two springs.

In the configuration shown aboveo the particles arc at their
natural lengths.

The spring forces acting on the two particles are,

Hr: – k7.xe,i
Hz: &n(x, -x7^)i
H3:-H2–ku(xn-xii

Apply Newton’s 2’d Law to,

particle A, ffi.a x.n= H1 + H2
: – k6xai + frs (rs – x7*) i.

particle B, ms )c.a = H3
: -lcy(xr-xa)

I

Resolving in the I – direction and grouping the terms,

ffi.a fr.a= – (ko+ ks) xe + ks xs
ke xA, ks xs

I
txDH

I

i6l Particle A Particlc 8

ffinxg=

Pege 4

Ircture 5

The two equations of motion can be written in matrix form.

ru)r::
x =[;;] =

Hr l-, H2 H3 1″2 H.l*, **,
ons of motion in matrix form

l- 2k k I
=[i;] =l t !”1[;l

L *, mr)

(kn + kr)
mA

b ]:Ax
-ffin

The matrix A is called the dynamic matrix of the system.

The same procedure can be applied to two masses and three
springs as in Example I.I on pg 10.

la ,o lo
rl

becomes,

ll
FI

or\

The equati

.r

Page 5

_ Ax

Lecture 5

Remember, in Unit 7,for SHM,

i : -crzx
Notice that if we can express the matrix equation in the
form,

e: Ax : l’x
we can then write down the characteristic equation of the
dynamic matrix A and find the two distinct real eigenvalues
l,r and l,z such that,

0t=Ja and @z=m
Note that the eigenvalues of the dynamic matrix of an
oscillating mechanical system (without damping and
forcing) are always real and Ag@.

We also learnt n Unit 15, that the general solution to this
type of equations of motion can be written in the form,

x(/): v 1 (D 1 cos(co 1 r)+D2s in(rrl 1 /))+ v2(D3 cos (rD 2 r)+D4sin(ot 2/))

where, v1 and v2 0,ro the eigenvectors colresponding to the
eigenvalues l,r and 1,2 of the dynamic matrix A

and Dr, De, D3 and Da are the arbitrary constants.

The general solution can also be rewritten as,

x(0 : Cr v1 cos(o)r, + 0r) + Czvzcos(co2l + $z)

where, Cu Cz,$r and 0, are the arbitary constants which
can be found from known initial conditions.

Page 6

I.€c.ture 5

As can be seen, the general solution x(0 is a linear
combination of linearly independent normal modes of the
system. (Theorem 7.7, pg 12)

Go through Procedure 7.1, pg 14 ano’Analysing oscillating
mechanical systems”.

Coqrditiou$Jor qor{nal moSes to occur
i.e. both particles must oscillate sinusoidally with same (0.

. For systems starting from rest i.e. ;L(0) = 0 (pS 19)
the two particles must be given displacements equal to
the normal mode eigenvector.

. For systems sturting with particles at origin (pS 23)
i.e. x1(0) : xz(0) : 0.
the two particles must be given initial velocities equal to
the normal mode eigenvector.

If the above conditions are not satisfied, then the two
particles will exhibit motions that are not in a sinusoidal
manner and with different angular frequencies as shown
in the two displacement responses, xa and xs below.

An efample.of nor-slnfso#a/ responpe from the 2 padicles

PweT

L€ctur€5

The motion ofparticles inthefirst normal mode is given
by,

x(r) : Cr v1 cos(or/ + 0r)

or,
[;8] =c,[;]cos(a’;,r

+d,)

Similarly, the motion of particles in the second normal
mode is given by,

x(/) : Cz v2 cos($2t + Q2)

[,’lill – cz[- lcos(a,, t +oz)or’
Lxz\t ) J Lvz J

Non$al modes disp,lacement ratio. R tpy 21)

Consider say, the first normal mode. Then the displacements
of the two particles are given by,

xr(/) = Cr v1 cos({or, + 0r)
xzf): Cr v2 cos(olrf + 0r)v

The normal mode displaeement ratio,is defined as,

.’4″ – b – ,R giving xz4)- ^R rr(r)xr(t) vt Y

Hence, the eigenvector v, can be expressed in terms of the
normal mode displacement ratio. i.e.

v :
L;]

= v,[;] since vz:R vr

Pag€ 8

L€ctur€ 5

Relative motion of two particles (Theorem 1.2, pg 2l)

In-phase – if the normal mode displacement ratio is positive
Phase-opposed * if the normal mode displacement ratio is

negative

1

O’, -trr

Hr H,

m1

H, lIu

Particles displaged from equilibrium (pS 25)

F*-)
ot xt

m,
+-+)

Here, we consider changes in spring forces AII, due to
displacements of particles from their equilibrium positions.

The changes in spring forces for the above configuration ffo,
AH1 : -ktr1i
AH2 : – k2 (x, – xr) (-l) : kz (xz – x) i
AH3: -AH2: -kz(xz-x)i
AII4 : – fr: (0 – xz) (-D : – kt xzi

Applying Newton’s 2″d Law, the equations of motion ffio,
ffir*ri = AHr * AHz – – kr h L+ kz @z- x) i
nz *zl= AIIr + AII4 – – kz (xz- x) i – kt xzi

Resolving in the j -direction gives,
ffir*r=-(h+k) rr * kzxz
ffiz iz* – k2 x1* (kz + kixz

PagP 9

7

lxcture 5

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o._i

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AET

*2k
B&

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air

rasista nce ls ne/e,ctenf’
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