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Mechanics Oscillations - 1 David Apsley

MECHANICS, TOPIC F: OSCILLATIONS AUTUMN 2003 1. Introduction 1.1 Overview 1.2 Degrees of freedom 2. Undamped free vibration 2.1 Generalised mass-spring system 2.2 Natural frequency and simple harmonic motion 2.3 Amplitude and phase angle 2.4 Fixing the solution from initial conditions 2.5 Displacement from equilibrium 2.6 Small oscillations 2.7 Derivation of the SHM equation from energy principles 3. Damped free vibration 3.1 The equation of motion 3.2 Levels of damping 4. Forced vibration 4.1 Mathematical expression of the problem 4.2 Static load 4.3 Undamped forced vibration 4.4 Damped forced vibration Examples Answers


1. INTRODUCTION 1.1 Overview Many important dynamical problems arise from the oscillations of systems responding to applied disturbances in the presence of restoring forces. Examples include: • response of a structure to earthquakes; • human-induced structural vibrations (e.g. pop concerts); • flow-induced vibrations (e.g. chimneys, pipelines, power lines); • vibration of unbalanced rotating machinery. We shall see that many systems displaced from a position of equilibrium have a natural frequency of vibration which depends upon its resistance to motion (inertia) and the strength of restoring forces (stiffness). If the system is left to oscillate without further input of energy this is referred to as free vibration. On the other hand, the examples above are mainly concerned with forced vibration; that is, oscillations of a certain frequency are imposed on the system by external forces. If the applied frequency happens to coincide with the natural frequency then there is considerable transfer of energy and resonance occurs with potentially catastrophic consequences. In general, all systems are subject to some degree of (deliberate or natural) frictional damping that removes energy. A freely-vibrating system may be under-damped (oscillates, but with gradually diminishing amplitude) or over-damped (so restricted that it never oscillates). When oscillatory forces act on structures, damping is a crucial factor in reducing the amplitude of vibration. 1.2 Degrees of Freedom Many systems have several modes of vibration. For example, a long power line or suspended bridge deck may “gallop” (bounce up and down) or it may twist and untwist (torsional mode). Here, for simplicity, we shall restrict ourselves to single-degree-of-freedom (SDOF) systems; that is, those whose configuration can be described by a single variable such as a displacement x or an angle 7KH HTXDWLRQ RIPRWLRQ WKHQ GHVFULEHV WKH YDULDWLRQ RI WKDWparameter with time. Examples of dynamical systems and their degree of freedom are: • mass suspended by a spring (vertical displacement); • pivoted body (angular displacement). In both systems, oscillations occur about a position of static equilibrium in which opposing forces balance.


2. UNDAMPED FREE VIBRATION The oscillatory motion of a system displaced from stable equilibrium and allowed to adjust in the absence of externally-imposed forces is termed free vibration. If there are no frictional forces the motion is called undamped free vibration. 2.1 Generalised Mass-Spring System This is a general model for a linear free-vibration problem. It doesn’t physically have to correspond to masses and springs.

mk

x

A generalised mass-spring system is one for which the resultant force following displacement from equilibrium is a function of the displacement and of opposite sign. For ideal springs and, in practice, for small displacements the relationship is linear: kxF −= (1) k is called the stiffness or spring constant or modulus. For a spring, k is measured in Newtons per metre (N m-1). The equation of motion is then

kxt

xm −=

2

2

d

d (2)

2.2 Natural Frequency and Simple Harmonic Motion To solve (2) note that xmktx )/(/dd 22 −= and hence we must look for solutions whose second derivative is proportional to the solution and of opposite sign. The obvious candidates are sine and cosine functions. The general solution (with two arbitrary constants) may be written in either of the forms:

tDtCx

tAx

cossin

)sin(

+=φ+=

(3)

where

inertia

stiffness

m

k == (4)

Any system whose degree of freedom evolves sinusoidally is said to undergo simple harmonic motion (SHM LV FDOOHG WKH natural (circular) frequency and is measured in radians per second (rad s-1). The equation of motion is usually written as

xt

x 22

2

d

d −= or 0d

d 22

2

=+ xt

x (5)

One complete cycle is completed when 2=t . Hence:


period of oscillation: 2=T (6)

natural frequency: 2

1 ==T

f in cycles per second or Hertz (Hz) (7)

Note that it is widespread practice to refer to UDWKHUWKDQf as the natural frequency. Where there is doubt, specify the units as rad s-1 or cycles s-1 (= Hz). 2.3 Amplitude and Phase Angle The general solution (two arbitrary constants) may be written

tDtCx

tAx

cossin

)sin(

+=φ+=

The first of these is called the amplitude/phase-angle form. The two forms are easily interconverted as follows. Expand the amplitude/phase-angle form:

)sincoscos(sin)sin( φ+φ=φ+= ttAtAx &RPSDUHZLWKWKHVHFRQGIRUPDQGHTXDWHFRHIILFLHQWVRIVLQ tDQGFRV t to obtain

φ=φ= sin,cos ADAC

Eliminating φ and A in turn gives:

amplitude 22 DCA +=

phase angle )/(tan 1 CD−=φ

Note that there are two alternative values of φ with the same value of tan φ. These may be distinguished by determining whether C and D are both positive or both negative. Example. Write the following expressions in amplitude/phase-angle form, )sin( φ+tA : (a) tt 3sin53cos12 + (b) tt 2sin32cos4 −

(a) = 3, A = 13, φ = tan-1(12/5) = 1.18 radians or 67.4°. E = 2, A = 5, φ = tan-1(–4/3) = 2.21 radians or 126.9°. (Be careful!)


Example. (Meriam and Kraige). Two fixed counter-rotating pulleys a distance 0.4 m apart are driven at the same angular speed 0. A bar is placed across the pulleys as shown. The FRHIILFLHQW RI NLQHWLF IULFWLRQ EHWZHHQ EDU DQG SXOOH\V LV k = 0.2. Show that, if the bar is initially placed off-centre, then it will undergo SHM and find the period of oscillation.

0.4 m

x

mgω0 ω0

2.01 s


2.4 Fixing the Solution From Initial Conditions The equation of motion gives rise to a second-order ordinary differential equation which, when integrated, yields two arbitrary constants of integration. These may be evaluated if two additional boundary conditions are specified: typically the initial displacement x0 and initial velocity 0)/dd( tx .

Example. For the system shown, what is (a) the equivalent single spring; (b) the natural circular frequency F WKH QDWXUDO IUHTXHQF\ RI YLEUDWLRQ f; (d) the period of oscillation; (e) the maximum speed of the cart if it is displaced 0.1 m from its position of equilibrium and then released.

10 kg

k=100 N/m k=60 N/m

x

(a) k = 160 N m-1; (b) 4 rad s-1; (c) 0.637 Hz; (d) 1.57 s; (e) 0.4 m s-1

2.5 Displacement From Equilibrium For linear systems it is the displacement from a position of static equilibrium that is important, rather than the absolute displacement. Consider a mass suspended by a spring. There is an equilibrium extension ye given by balancing weight mg and restoring force kye:

k

mgykymg ee =⇒=

The general equation of motion, with both elastic and gravitational forces, is

)(

d

d2

2

k

mgyk

mgkyt

ym

−−=

+−=

m

mg

k

m

ky

y


or, writing

uilibriumnt from eqdisplacemek

mgyY =−=

and noting that 2

2

2

2

d

d

d

d

t

y

t

Y = , then

kYt

Ym −=

2

2

d

d

Thus, it is the displacement from static equilibrium that determines the net restoring force. If we work with an equation for Y then the gravitational contribution is eliminated. Example. A block of mass 16 kg is suspended vertically by two light springs of stiffness 200 N m-1. What is (a) the equivalent single spring; (b) the period of oscillation.

16 kg

k=200 N/mk=200 N/m

(a) k = 400 N m-1; (b) T = 1.26 s Example. A 4 kg mass is suspended vertically by a string of elastic modulus E = 480 N and unstretched length 2 m. What is its extension in the equilibrium position? If it is pulled down from its equilibrium position by a distance 0.2 m, will it undergo SHM?

0.1635 m; no (string becomes unstretched).


2.6 Small Oscillations “Small-oscillations” problems typically occur when there is restricted motion about a fixed point – i.e rotation – and rely on the approximations

)-1 sometimes, or,(1cos

sin2

21≈

≈ (8)

when LV PHDVXUHG LQ UDGLDQV 7KHVH PD\ EHderived formally by power-series expansion, but their essential validity is easily seen geometrically (see right). The following table shows that the approximation for sine is accurate to about 1% or better for angles as large as 15Û ,IDVWUXFWXUDOPHPEHUZHUHDFWXDOO\GLVSODFHGE\WKLVPXFKWKHQoscillation would be the least of your worries!

GHJUHHV 5 10 15 20 UDGLDQV 0.0873 0.175 0.262 0.349 VLQ 0.0872 0.174 0.259 0.342

2.6.1 Oscillations Driven by Gravity The simplest small-oscillation problems are pendulums – objects swinging freely under gravity. In a simple pendulum all the mass is concentrated at one point. This can be treated using the force-momentum principle. In a compound pendulum the mass is distributed; the system must be analysed by rotational dynamics. Many bodies are effectively pivoted about one point and, therefore, constitute a compound pendulum. Simple Pendulum Two forces act on the mass: its weight mg and the tension in the string. The latter can be eliminated if we resolve in the tangential direction:

R( ) 2

2

d

dsin

tmLmg =−

sind

d2

2

L

g

t−=⇒

For small oscillations, sin ≈ , so that

d

d2

2

L

g

t−=

This is SHM with natural frequency L

g= .

mg

θ

L

Lθ..T

Lθ. 2

θsin θ

1

θ


Compound Pendulum Since this is an extended body and not a point mass it must be treated by rotational mechanics. Two forces act on the body – the weight of the body (which acts through the centre of gravity) and the reaction at the axis. Only the former has any moment about the axis. Let the distance from axis to centre of gravity be L and let the angular displacement of line AG from the vertical be The line of action of the weight Mg lies at a distance L sin IURPWKHaxis of rotation, and hence imparts a torque (or moment) sinMgL in WKHRSSRVLWHVHQVHWR

2

2

d

dsin

tIMgL

momentumangularofchangeofratetorque

=−⇒

=

Making the small-oscillation approximation sin ≈ , one has

0d

d2

2

=+I

MgL

t

This is SHM with natural circular frequency I

MgL= .

Example. A uniform circular disc of radius 0.5 m is suspended from a horizontal axis passing through a point halfway between the centre and the circumference. Find the period of small oscillations.

1.74 s

L

Mg

θ

A

G


Example. A pub sign consists of a square plate of mass 20 kg and sides 0.5 m, suspended from a horizontal bar by two rods, each of mass 5 kg and length 0.5 m. The sign is rigidly attached to the rods and swings freely about the bar. Find the period of small oscillations.

1.70 s 2.6.2 Oscillations Driven By Elastic Forces Example. A uniform bar of mass M and length L is allowed to pivot about a horizontal axis though its centre. It is attached to a level plane by two equal springs of stiffness k at its ends as shown. Find the frequency of small oscillations.

L

k k

M

k6=

The Dogand Duck

0.5 m

0.5 m

0.5 m


2.7 Derivation of the SHM equation from Energy Principles For a body, of mass m, subject only to elastic forces with stiffness k, the total (i.e. kinetic + potential) energy is constant:

constantkxmv =+ 2212

21

and, hence, differentiating with respect to time gives

0)(d

d 2212

21 =+ kxmv

t

or, by the chain rule for each term:

0d

d)(

d

d

d

d)(

d

d 2212

21 =+

t

xkx

xt

vmv

v

0d

d

d

d =+⇒t

xkx

t

vmv

Replacing v by dx/dt:

0d

d

d

d

d

d2

2

=+t

xkx

t

x

t

xm

Finally, dividing by dx/dt:

0d

d2

2

=+ kxt

xm

Example. (Examination, January 2003). A sign of mass M hangs from a fixed support by two rigid rods of negligible mass and length L (see below). The rods are freely pivoted at the points shown, so that the sign may swing in DYHUWLFDOSODQHZLWKRXWURWDWLQJWKHURGVPDNLQJDQDQJOH ZLWKWKHYHUWLFDO (a) Write down exact expressions for the potential energy and kinetic energy of the sign

in terms of M, L, g WKHDFHOHUDWLRQGXHWRJUDYLW\WKHGLVSODFHPHQWDQJOH DQGLWVtime derivative & .

(b) ,IWKHVLJQLVGLVSODFHGDQDQJOH = π/3 radians and then released, find an expression for its maximum speed.

(c) Find an expression for the total (i.e. kinetic + potential) energy using the small-angle

approximations sin ≈ , 2211cos −≈ .

(d) Show that, for small-amplitude oscillations, the assumption of constant total energy leads to simple harmonic motion, and find its period.

θ

M

LL


3. DAMPED FREE VIBRATION All real dynamical systems are subject to friction, which opposes relative motion and consumes mechanical energy. For a system undergoing free vibration, we shall show that: • moderate damping ⇒ decaying amplitude and reduced frequency; • excessive damping prevents oscillation altogether. The level at which oscillation is just suppressed is called critical damping. 3.1 The Equation of Motion

The direction of friction is such as to oppose relative motion. For a frictional force that depends on velocity, the damping force is often modelled as

t

xcFd d

d−= (9)

c is the viscous damping coefficient. If x is a displacement then c has units of N s m-1. In practice, c may vary with velocity, but a useful analysis may be conducted by assuming it is a constant, in which case the damping is termed linear.

m

k

xc

The equation of motion for a damped mass-spring system is

t

xckx

t

xm

d

d

d

d2

2

−−=

or

0d

d

d

d2

2

=++ kxt

xc

t

xm (10)

Dividing by m, this can be written

0d

d)(

d

d 22

2

=++ xt

x

m

c

t

x (11)

where mk/= is the natural frequency of the undamped system. The undamped system (c = 0) has solutions of the form )sin( φ+tA . In the presence of damping we expect the solution to decay in magnitude and have a slightly different frequency. Hence we try solutions of the form )sin( φ+= − tAex d

t (12)

where A and φ DUH DUELWUDU\ FRQVWDQWV DQG DQG d are to be found. Successive differentiations give, after some algebra,

)]cos(2)sin()[(d

d

)]cos()sin([d

d

22

2

2

φ+−φ+−=

φ++φ+−=

−

−

ttAet

x

ttAet

x

ddddt

dddt


Substituting these into (11) and collecting coefficients of )cos( φ+td gives:

02 =+− dd m

c

whence

0)2( =−m

cd

+HQFHHLWKHU d = QRRVFLOODWLRQRUWKHH[SRQHQWLDOGHFD\FRQVWDQW LVJLYHQE\

m

c

2= (13)

Similarly, collecting coefficients of )sin( φ+td gives:

0222 =+−−m

cd (14)

$VVXPLQJLQLWLDOO\WKDWRVFLOODWLRQLVSRVVLEOHZHPD\VXEVWLWXWHIRU DQGUHDUUDQJHIRU 2d :

])2

(1[ 222

m

cd −=

The RHS is positive - and hence oscillation is possible – if and only if the damping ratio

2m

c= (15)

is less than 1. ,I > 1 then an oscillatory solution of the form (12) is not possible: the general solution is a combination of two decaying exponentials

tt BeAex 21 −− += ZKHUH 1DQG 2 are the roots of (14ZKHQ d = 0. ,I = 1 the above two roots are equal: the appropriate solution is then of the form

teBtAx )( −+= This is the point where the system just fails to oscillate. We say that it is critically damped. 3.2 Levels of Damping Summarising the analysis of Section 3.1, the damped mass-spring equation

0d

d

d

d2

2

=++ kxt

xc

t

xm

has solutions that depend on

undamped natural frequency: m

k=

damping ratio: 2m

c=


Case (1): = 0. Undamped system (c = 0) The general solution may be written

)sin( φ+= tAx (16) Case (2): 0 < < 1. Under-damped system (0 < c < 2m

The general solution may be written

)sin(2 φ+=−

tAex dm

ct

(17)

where

21−=d (18)

Note that there is oscillation with:

reduced amplitude, decaying exponentially as te−

reduced frequency, 21−=d

The amplitude reduction factor over one cycle ( 2=td or dt /2= ) is

1

2exp

2−− (19)

and this may be used to determine the damping ratio experimentally. &DVH = 1. Critically-damped system (c = 2m

The general solution has the form

teBtAx )( −+= (20) There is no oscillation and the amplitude decays to zero.

&DVH > 1. Over-damped system (c > 2m The general solution is of the form

tt BeAex )1()1( 22 −−−−+− += (21) There is no oscillation and the amplitude decays to zero – more slowly with greater damping.

Key Points

(1) The effect of damping depends on the damping ratio 2m

c= .

(2) ,I 1 the system is under-damped, with:

decaying-amplitude te−∝

reduced frequency 21−=d

(3) ,I > 1 the system is over-damped and no oscillation occurs. (4) The fastest return to equilibrium occurs when the system is critically damped = 1).


Example m = 1 kg, k = 64 N m-1

⇒ = 8 rad s-1 x0 = 0.05 m, 0)/dd( 0 =tx .

Cases: c = 1.6 N s m-1 = 0.1) c = 16 N s m-1 = 1) c = 64 N s m-1 = 4) Exercise. Use Microsoft Excel or any other computer package to compute and plot the solution for various combinations of mass, stiffness, damping = m, k, c. Example. Analyse the motion of the system shown. What is the damping ratio? Does it oscillate? If so, what is the period? What value of c would be required if the system is to be critically damped?

40 kg

c=60 N s/m

k=700 N/m

It oscillates with = 0.179, T = 1.53 s; for critical damping, c = 335 N s m-1 Example. Find the solution of the equation

010d

d2

d

d2

2

=++ xt

x

t

x

such that x = 3 and 0/dd =tx at t = 0. By what factor is the amplitude reduced over each cycle?

)3cos33(sin ttex t += − . The amplitude is reduced by a factor 3/2−e at each oscillation.


4. FORCED VIBRATION Systems that oscillate about a position of equilibrium under restoring forces at their own “preferred” or natural frequency are said to undergo free vibration. Systems that are perturbed by some externally-imposed oscillatory forcing are said to undergo forced vibration. Examples which may be encountered in civil engineering are: • concert halls and stadiums; • bridges – both traffic- and wind-induced oscillations; • earthquakes – the lateral oscillations of the foundations are equivalent to oscillatory

forcing in a reference frame moving with the surface. The natural frequency and damping ratio are important parameters for systems responding to externally-imposed oscillations. Large-amplitude vibrations occur when the imposed frequency is close to the natural frequency (the phenomenon of resonance). In general, the system’s free-vibration properties will affect both amplitude and phase of response to external forcing. 4.1 Mathematical Expression of the Problem

m

k

xc

F sin t0 Ω

The general form of the equation of motion with harmonic forcing is

tFkxt

xc

t

xm sin

d

d

d

d02

2

=++ (22)

or

tm

Fx

t

x

m

c

t

xsin

d

d)(

d

d 022

2

=++ (23)

where the undamped natural frequency is given by

m

k=2

Second-order differential equations of the form (23) will be dealt with in more detail in your mathematics course next semester. The general solution is the sum of a complementary function (containing two arbitrary constants and obtained by setting the RHS = 0) and a particular integral (which is any particular solution of the equation and is obtained by a trial function based on the nature of the RHS). The complementary function is a free-vibration solution and, if there is any damping at all, will decay exponentially with time. The large-time behaviour of the system, therefore, is determined by the particular integral which, from the form of the RHS, should be a combination of sin t and cos t.


4.2 Static Load A useful quantity for comparison is the displacement under a steady load F0, rather than a variable load tF cos0 . The steady-state displacement xs is given by

0Fkxs =

i.e.

2

00

m

F

k

Fxs == (24)

This is simply the position of static equilibrium. 4.3 Undamped Forced Vibration In the idealised case of no frictional damping, the equation of motion is

tm

Fx

t

xsin

d

d 022

2

=+

The form of the forcing function suggests a particular integral of the form tCx sin= . By substitution, one finds that this satisfies the equation of motion if

2222

0

/1)( −=

−= sx

m

FC (25)

where, as above, 20 /mFxs = is the displacement under a static load of the same amplitude.

Hence the magnitude of forced oscillations is Mxs, where the amplitude ratio or magnification factor M is given by

22 /1

1

−=M (26)

Key Points (1) The response of the system to external forcing depends on the ratio of the forcing

frequency WRWKHQDWXUDOIUHTXHQF\ (2) There is resonance (M → ∞) if the forcing frequency approaches the natural

IUHTXHQF\ → (3) ,I < WKH RVFLOODWLRQV DUH in phase with the forcing (because the system can

respond fast enough). (4) ,I > WKHRVcillations are 180° out of phase with the forcing (because the imposed

oscillations are too fast for the system to respond).


4.4 Damped Forced Vibration The ultimate, most complete analysis of vibrations for a SDOF oscillating system is for forced motion with damping. The equation of motion is

tm

Fx

t

x

m

c

t

xsin

d

d)(

d

d 022

2

=++

Because of the dx/dt term, a trial solution for the PI must contain both tsin and tcos . Substituting a trial solution of the form

tDtCx cossin += produces a solution (optional exercise) of the form

ss xDxC22222222

22

)/2()/1(

/2,

)/2()/1(

/1

+−−=

+−−= (27)

where the damping ratio is

2m

c=

and the displacement under static load is 2

0 /mFxs =

This is most conveniently written in the amplitude/phase-angle form )sin( φ−= tMxx s (28)

where the amplitude ratio M is given by

2222

22

)2()/1(

1

/DCM

+−=+= (29)

and the phase lag φ by

22 /1

/2tan

−=−=φ

C

D (30)

Key Points (1) The response of the system to forcing

depends on both the ratio of forcing to natural IUHTXHQFLHV DQGWKHGDPSLQJUDWLR

(2) Damping prevents the blow-up (M → ∞) as

→ (3) The imposed frequency at which the

maximum amplitude oscillations occur is reduced below the undamped natural IUHTXHQF\ ,QIDFW

2max 21−=

(4) 7KHSKDVHODJYDULHVIURPDV → WR DV → ∞). However, the phase lag is

DOZD\V ZKHQ = LUUHVSHFWLYHRIWKHlevel of damping.


Example. Write down general solutions for the following differential equations:

(i) 124d

d2

2

=+ yt

y

(ii) txt

x

t

x2sin130155

d

d4

d

d2

2

+=++

(i) 32cos2sin ++= tDtCy (or, equivalently, 3)2sin( ++= tAy ).

(ii) tttAex t 2cos162sin23)sin(2 −++φ+= − (or equivalent)

Example. (Meriam and Kraige) The seismometer shown is attached to a structure which has a horizontal harmonic vibration at 3 Hz. The instrument has a mass m = 0.5 kg, a spring stiffness k = 20 N m-1 and a viscous damping coefficient c = 3 N s m-1. If the maximum recorded value of x in its steady-state motion is 2 mm, determine the amplitude of the horizontal movement xB of the structure.

x (t)B

mc

x

k

1.89 mm


Examples Mathematics of Oscillating Systems Q1. Find general solutions for the following differential equations:

(a) 025d

d2

2

=+ xt

x

(b) 09d

d2

d

d2

2

=++tt

Undamped Free Vibration Q2. A 4 kg mass is suspended vertically by a spring of stiffness k = 900 N m-1. Find: (a) the static displacement at equilibrium; (b) the natural circular frequency of vibration.

Q3. A block of mass 16 kg is suspended vertically by two light springs of stiffness 100 N m-1 and 300 N m-1 hung end-to-end as shown. Find: (a) the stiffness of an equivalent single spring; (b) the period of oscillation.

Q4. (a) Assuming that it does not slip, what mass m must be placed on the top of the 20 kg cart in order that the period of oscillation be 2 s? (b) What is the minimum coefficient of static frictiRQ s such that the mass will not slip if the cart is pulled 0.2 m away from the equilibrium position and then released?

4 kg

k=900 N/m

16 kg

k=100 N/m

k=300 N/m

20 kg

mk=300 N/m


Oscillating Systems Other than Mass-Spring Q5. (a) A floating buoy consists of a cylinder of mass 3 kg and radius 0.1 m. If it is pushed down (but not submerged) and then released, show that it undergoes SHM and find the natural circular frequency of RVFLOODWLRQ7KHGHQVLW\RIVHDZDWHU w, is approximately 1050 kg m-3; neglect its motion and assume that the buoy is stable). (b) If the buoy is replaced by an inverted cone of mass m and semi-YHUWH[ DQJOH ZULWH GRZQ DQ HTXDWLRQ RI PRWLRQ LQ WHUPV RIsubmerged depth and show that any oscillatory motion is not SHM.

Small Oscillations Q6. (Examination, January 2002) A light rigid rod of length L = 0.8 m is able to swing in a vertical plane about a pivot P at a distance L/4 from one end (see Figure). At the lower end is a concentrated mass M = 5 kg. The top of the rod is joined by two springs of stiffness k = 200 N m-1 to fixed supports. (a) By considering the combined torque of all elastic and

gravitational forces (or otherwise), calculate the period of small oscillations.

(b) If the rod is turned an angle 0.2 radians from the vertical and then released, calculate the maximum speed of the mass M.

Q7. A uniform rod of mass 5 kg and length 0.9 m supports a concentrated mass of 10 kg at one end and is pivoted at the other. It is supported by two springs, each of stiffness k = 2000 N m-1, positioned as shown. Its equilibrium position is horizontal. Calculate the period of small oscillations.

0.1m

y

P

M

k k

L

L

14

34

0.9 m

0.3 m0.3 m

10 kg

5 kg


Damped Free Vibrations Q8. (Meriam and Kraige) The cannon fires a 4.5 kg cannonball with velocity of 250 m s-1 at 20° to the horizontal. The combined mass of the cannon and its cart are 750 kg. If the recoil mechanism consists of a spring of constant k = 27 kN m-1 and a damper with viscous coefficient c = 9000 N s m-1, determine the maximum recoil deflection xmax of the cannon unit. Forced Vibrations

Q9. (Meriam and Kraige) The 30 kg cart is acted upon by a harmonic force tF cos25= Newtons as shown. Determine the ranges(s) of the driving frequenF\ IRU ZKLFK WKH DPSOLWXGH RI WKHsteady-state response is less than 75 mm if c takes the values (a) 0 and (b) 36 N s m-1.

20o

k

cm

250 m/s

k=1080 N/m

c 30 kgF=25 cos tΩ


Answers A1. (a) )5sin( BtAx += or tDtCx 5cos5sin +=

(b) )22sin( BtAe t += − or )22cos22(sin tDtCe t += − A2. (a) 0.0436 m; (b) 15 rad s-1 A3. (a) k = 75 N m-1 (b) T = 2.90 s A4. (a) 10.4 kg; (b) 0.201 A5. (a) 10.4 rad s-1

(b) 3231

2

2

)tan(d

dyg

t

ym w−=

A6. (a) 1.25 s (b) 0.603 m s-1 A7. 0.644 s A8. 86.4 mm A9. D rad s-1 DQG > 6.86 rad s-1 E < 5.18 rad s-1DQG > 6.62 rad s-1

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