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    06-92-315

    Mechanical Vibrations

    Lecture 2

    Mass Spring System

    and Harmonic Motion

    Dr. Ahmed Deif

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    Spring Mass System

    From strength of

    materials recall:

    A plot of force versus displacement:

    FBD:

    linear

    nonlinear

    experiment fk kx

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    Free-body diagram and

    equation of motion

    Newtons Law:

    Again a 2nd

    order ordinary differential equation

    (1.2)

    m&&x(t) kx(t) m&&x(t) kx(t) 0

    x(0) x0 , &x(0) v0

    )(tXm..

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    Solution to 2nd order DEs

    x(t) Asin(nt)

    Lets assume a solution:

    Differentiating twice gives:

    x(t) nAcos(nt)&&x(t) n

    2Asin(nt) -n

    2x(t)

    Substituting back into the equations of motion gives:

    mn2Asin(nt) kAsin(nt) 0

    mn2 k 0 or n

    k

    mNatural

    frequency

    t

    x(t)

    rad/s

    )(tX..

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    Stiffness and Mass

    Vibration is cause by the interaction of two different forces

    one related to position (stiffness) and one related to

    acceleration (mass).

    m

    k

    x

    Displacement

    Mass Spring

    Stiffness (k)

    Mass (m)

    fk kx(t)

    fm ma(t) m&&x(t)

    Proportional to displacement

    Proportional to acceleration

    statics

    dynamics

    )(tXm..

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    Examples of Single-Degree-of-

    Freedom Systems

    m

    ql =length

    Pendulum

    q(t) g

    lq(t) 0

    Gravity g

    Shaft and Disk

    q

    J&&q(t) kq(t) 0

    TorsionalStiffness

    kMoment

    of inertia

    J

    )(tJ..

    q

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    Initial Conditions

    If a system is vibrating then we must assume that somethingmust have (in the past) transferred energy into to the system

    and caused it to move. For example the mass could have

    been:

    moved a distance x0and then released at t = 0(i.e. given

    Potential energy) or

    given an initial velocity v0(i.e. given Kinetic energy) or

    Some combination of the two above cases

    From our earlier solution we know that:

    x0 x(0) Asin(n 0 ) Asin()

    v0 &x(0) nA cos(n 0 ) nA cos())0(.

    X

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    Initial ConditionsSolving these equation gives:

    v0

    n

    x0

    1

    nn

    2x02 v0

    2

    t

    x(t)

    x0

    Slope

    here is v0

    n

    A 1

    nn

    2x0

    2 v02

    Amplitude

    1 24 4 4 34 4 4

    , tan1nx0

    v0

    Phase

    1 24 4 34 4

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    The total solution for the spring

    mass system is

    (1.10)

    Called the solution to a simple harmonic oscillator

    and describes oscillatory motion, or simple harmonic motion.

    Note (Example 1.1.2)

    x(0) n

    2x0

    2 v02

    n

    nx0

    n2x

    0

    2 v02

    x0

    as it should

    x(t) n

    2x02 v0

    2

    nsin nt tan

    1 nx0

    v0

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    A note on arctangents

    Note that calculating arctangent from acalculator requires some attention. First, allmachines work in radians.

    The argument atan(-/+) is in a different quadrantthen atan(+/-), and usual machine calculationswill return an arctangent in between -p/2 and+p/2, reading only the atan(-) for both of theabove two cases.

    In MATLAB, use the atan2(x,y) function to get

    the correct phase.

    +

    +

    -

    _+

    +

    -

    -

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    Example 1.1.3: wheel, tire suspension m=30

    kg,

    n= 10 hz, what is k?

    k mn2 30 kg 10 cylce

    secg2p rad

    cylce

    1.184 105 N/m

    There are of course more complex models of suspension systems

    and these appear latter in the course as our tools develop

    2

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    Summary of simple harmonic

    motion

    t

    x(t)

    x0 Slopehere is v0

    n

    Period

    T 2p

    n

    Amplitude

    A

    Maximum

    Velocity

    An

    fn n rad/s

    2p rad/cyclen cycles

    2p sn

    2pHz

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    Harmonic Motion

    T 2p rad

    n rad/s

    2p

    ns (1.11)

    The natural frequency in the commonly used units of hertz:

    The period is the time elapsed to complete one complete cylce

    fn n

    2p

    n rad/s

    2p rad/cyclen cycles

    2p sn

    2pHz (1.12)

    For the pendulum:

    n

    g

    l rad/s, T 2pl

    g s

    n k

    Jrad/s, T 2p

    J

    ks

    For the disk and shaft:

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    Relationship between Displacement,

    Velocity and Acceleration

    0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1

    -1

    0

    1

    x

    0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1-20

    0

    20

    v

    0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1-200

    0

    200

    Time (sec)

    a

    A=1, n=12

    x(t) Asin(nt)

    x(t) nAcos(nt)

    x(t) n2Asin(nt)

    Note how the relative magnitude of each increases forn>1

    Displacement

    Velocity

    Acceleration

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    Example 1.2.1Hardware store spring, bolt: m=49.2x10-3 kg, k=857.8 N/m and x0 =10 mm. Compute n

    and the max amplitude of vibration.

    n k

    m

    857.8 N/m

    49.210-3 kg132 rad/s

    fn

    n

    2p 21 Hz

    T2p

    n

    1

    fn

    1

    21cyles

    sec

    0.0476 s

    x(t)max A 1

    nn

    2x0

    2 v02 x0 10 mm

    0

    Note: common

    Units are Hertz

    To avoid Costly errors use fnwhen working in Hertz and nwhen in rad/s

    Units depend on system

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    Compute the solution and max velocity and

    acceleration

    v(t)max nA 1320 mm/s = 1.32 m/s

    a(t)max n2A 174.24 103 mm/s2

    =174.24 m/s2 17.8g!

    tan1nx0

    0

    p

    2

    rad

    x(t) 10sin(132tp/ 2) 10cos(132t) mm

    g= 9.8 m/s2

    2.92 mph

    90

    ~0.4 in max

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    Does gravity matter in spring

    problems?

    Let be the deflection caused byhanging a mass on a spring

    ( = x1-x0 in the figure)Then from static equilibrium: mg kNext sum the forces in the vertical for some point x> x1 measured

    from m&&x k x mg kx mg k

    01 24 34 m&&x(t) kx(t) 0

    So no, gravity does not have an effect on the vibration

    (note that this is not the case if the spring is nonlinear)

    X..

    X..

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    Example Pendulums & measuring g

    A 2 m pendulum

    swings with a period

    of 2.893 s

    What is the

    acceleration due to

    gravity at that

    location?

    g4p

    2

    T

    2l

    4p2

    2.893

    2

    s

    22 m

    g= 9.434 m/s2

    This is gin Denver, CO USA, at 1638m

    and a latitude of 40

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    Review of Complex Numbers and

    Complex Exponential

    b

    a

    A

    c a jb Aejq

    A complex number can be written with a real and imaginary

    part or as a complex exponential

    Where

    a Acosq,b Asinq

    Multiplying two complex numbers:

    c1c2 A1A2ej(q1 q2 )

    Dividing two complex numbers:

    c1

    c2

    A1

    A2ej(q1 q2 )

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    Equivalent Solutions to 2nd

    order Differential Equations

    x(t) Asin(nt)x(t) A1 sinnt A2 cosnt

    x(t) a1ejn t a2e

    jn t

    All of the following solutions are equivalent:

    The relationships between A and , A 1and A 2, and a1and a2can be found in Window 1.4 of the course text, page 17.

    Sometimes called the Cartesian form

    Called the magnitude and phase form

    Called the polar form

    Each is useful in different situations

    Each represents the same information

    Each solves the equation of motion

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    Derivation of the solution

    This approach will be used again for more complicated problems

    (1.18)

    Substitute x(t) aet

    into m&&x kx 0 m

    2ae

    t kaet 0

    m2 k 0

    k

    m

    k

    mj n j

    x(t) a1en jt and x(t) a2e

    n jt

    x(t) a1en jt a2e

    n jt

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    Is frequency always positive?

    From the preceding analysis, = + n thenx(t) a1e

    n jt a2en jt

    Using the Euler relations for trigonometric functions, the

    above solution can be written as

    x(t) Asin nt (1.19)It is in this form that we identify as the natural frequency nand this is positive, the + sign being used up in the transformation

    from exponentials to the sine function.

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    Calculating RMS

    A peak value

    x limT 1Tx(t)dt = average value

    0

    T

    x 2 limT

    1

    Tx2 (t)dt

    0

    T

    = mean-square value

    xrms x2 = root mean square value

    Proportionalto energy

    Not very useful since for

    a sine function the

    average value is zero

    May need to be limited dueto physical constraints

    Also useful when the vibration is random

    (1.21)

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    The Decibel or dB scale

    It is often useful to use a logarithmic scale to plot vibrationlevels (or noise levels). One such scale is called the decibelor

    dB scale. The dB scale is always relative to some reference

    value x0. It is define as:

    For example: if an acceleration value was 19.6m/s2 then relative

    to 1g(or 9.8m/s2) the level would be 6dB,

    10log1019.6

    9.8

    2

    20log10 2 6dB

    Or for Example 1.2.1: The Acceleration Magnitude

    is 20log10(17.8)=25dBrelative to 1g.

    dB 10log10 xx0

    2

    20log10 xx0

    (1.22)