TOPIC E: OSCILLATIONS SPRING 2019

Transcription

1 TOPIC E: OSCILLATIONS SPRING Introduction 1.1 Overview 1.2 Degrees of freedom 1.3 Simple harmonic motion 2. Undamped free oscillation 2.1 Generalised mass-spring system: simple harmonic motion 2.2 Natural frequency and period 2.3 Amplitude and phase 2.4 Velocity and acceleration 2.5 Displacement from equilibrium 2.6 Small-amplitude approximations 2.7 Derivation of the SHM equation from energy principles 3. Damped free oscillation 3.1 The equation of motion 3.2 General solution for different damping levels 4. Forced oscillation 4.1 Mathematical expression of the problem 4.2 Static load 4.3 Undamped forced oscillation 4.4 Damped forced oscillation Mechanics Topic E (Oscillations) - 1 David Apsley

2 1. INTRODUCTION 1.1 Overview Many important dynamical problems arise from the oscillation of systems responding to applied disturbances in the presence of restoring forces. Examples include: response of a structure to earthquakes; human-induced structural oscillations (e.g. grandstands at concerts/sports venues); flow-induced oscillations (e.g. chimneys, pipelines, power lines); vibration of unbalanced rotating machinery. Most systems when displaced from a position of equilibrium have one or more natural frequencies of oscillation which depend upon the strength of restoring forces (stiffness) and resistance to change of motion (inertia). If the system is left to oscillate without further influence from outside this is referred to as free oscillation. On the other hand, the examples above are mainly concerned with forced oscillation; that is, oscillations of a certain frequency are imposed on the system by external forces. If the applied frequency is close to the natural frequency of the system then there is considerable transfer of energy and resonance occurs, with potentially catastrophic consequences. In general, all systems are subject to some degree of frictional damping that removes energy. A freely-oscillating system may be under-damped (oscillates, but with gradually diminishing amplitude) or over-damped (so restricted that it never oscillates). When periodic forces act on structures, damping is a crucial factor in reducing the amplitude of oscillation. 1.2 Degrees of Freedom Many systems have several modes of oscillation. For example, a long power line or suspended bridge deck may gallop (bounce up and down) or it may twist. Often the geometric configuration at any instant can be defined by a small number of parameters: usually displacements x or angles θ. Parameters which are just sufficient to describe the geometric configuration of the system are called degrees of freedom. Here, for simplicity, we shall restrict ourselves to single-degree-of-freedom (SDOF) systems. The equation of motion then describes the variation of that parameter with time. Examples of SDOF 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 applied forces or moments are in balance. 1.3 Simple Harmonic Motion For many systems the forces arising from a small displacement are opposite in direction and proportional in size to the displacement. The equation of motion is particularly simple and solutions take the form of a sinusoidal variation with time. This ubiquitous and important type of oscillation is known as simple harmonic motion (SHM). Mechanics Topic E (Oscillations) - 2 David Apsley

3 2. UNDAMPED FREE OSCILLATION The oscillatory motion of a system displaced from stable equilibrium and then allowed to adjust in the absence of externally-imposed forces is termed free oscillation. If there are no frictional forces the motion is called undamped free oscillation. 2.1 Generalised Mass-Spring System: Simple Harmonic Motion This is a general model for a linear free-oscillation problem. It doesn t physically have to correspond to masses and springs. 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 (i.e. a restoring force). For ideal springs (or, in practice, for small displacements) the relationship is linear: k is called the stiffness or spring constant. It is measured in newtons per metre (N m 1 ). The equation of motion is then kx x (1) (2) To solve (2) note that and that k/m is a positive constant. Thus, it is common to write the equation as Simple Harmonic Motion (SHM) Equation: where or (3) (4) To solve (3) look for solutions x(t) whose second derivative is proportional to the original function, but of opposite sign. The obvious candidates are sine and cosine functions. 1 The general solution (with two arbitrary constants) may be written in either of the forms: Any system whose degree of freedom evolves sinusoidally at a single frequency is said to undergo simple harmonic motion (SHM). ω is called the natural circular frequency and is measured in radians per second (rad s 1 ). 1 In your mathematics classes, when you have covered complex numbers, you will encounter a third form of the general solution involving complex exponentials: x = Ae iωt + Be iωt. (5) (6) Mechanics Topic E (Oscillations) - 3 David Apsley

4 2.2 Natural Frequency and Period One complete cycle is completed when ωt changes by 2π. Hence: Period of oscillation: (7) Frequency: in cycles per second or hertz (Hz) (8) Important note. It is common in theoretical work to refer to ω rather than f as the natural frequency, because: (a) it avoids factors of 2π in the solution; (b) it is ω rather than f which appears in the governing equation. Thus, one should be quite careful about precisely what is being referred to as frequency ; usually it will be obvious from the notation or units: ω in rad s 1, f in cycles s 1 (Hz). The following example demonstrates that SHM can occur without any elastic forces. Example 1. (Meriam and Kraige) Two fixed counter-rotating pulleys a distance 0.4 m apart are driven at the same angular speed ω 0. A bar is placed across the pulleys as shown. The coefficient of friction between bar and pulleys is μ = 0.2. Show that, provided the angular speed ω 0 is sufficiently large, the bar may undergo SHM and find the period of oscillation. x 0 mg m Mechanics Topic E (Oscillations) - 4 David Apsley

5 2.3 Amplitude and Phase The general solution of the SHM equation (two arbitrary constants) may be written as either: The first is called the amplitude/phase-angle form (A = amplitude, = phase). Whichever form is more convenient may be used. They are easily interconverted as follows. Expand the amplitude/phase-angle form: Compare with the second form and equate coefficients of sin ωt and cos ωt to obtain Eliminating and A in turn gives: amplitude (9) phase angle Note that there are two alternative values of (in opposite quadrants) with the same value of tan. These must be distinguished by the individual signs of C and D. Example 2. Write the following expressions in amplitude/phase-angle form, : (a) (b) (10) In the general solution the two free constants can be determined if initial boundary conditions are given for displacement x 0 and initial velocity. There are two special cases: (1) Start from rest at displacement A: (2) Start from the equilibrium position with initial velocity v 0 : where 1 cos t 0 sin t p/2 p 3p/2 t 2p -1 Mechanics Topic E (Oscillations) - 5 David Apsley

6 Example 3. For the system shown, find: (a) the equivalent single spring; (b) the natural circular frequency ω; (c) the natural frequency of oscillation 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. 100 N/m x 60 N/m 10 kg 2.4 Velocity and Acceleration From the general solution in phase-angle form, Differentiation then gives for the velocity: Squaring and adding, using : Hence, we can find the velocity at any given position in the cycle: (11) Note also the maximum displacement, velocity and acceleration: (12) (13) (14) Example 4. (Exam 2018) (a) A boat is riding the waves in high seas, oscillating with simple harmonic motion in a vertical line. The period of oscillation is 7 seconds, and the height of the boat varies between 2 m and 8 m below a nearby pier. Find: (i) the maximum speed, (ii) the maximum acceleration, of the boat during the oscillations. (b) A box of mass 15 kg sits on the deck of the boat. By considering the forces on the box and the acceleration that it is undergoing, find the normal contact force from the deck on the box: (i) at the top, (ii) at the bottom, (iii) in the middle of the oscillation. Mechanics Topic E (Oscillations) - 6 David Apsley

7 2.5 Displacement From Equilibrium For many oscillating systems it is the displacement from a position of static equilibrium that is important, rather than the absolute displacement. The equilibrium position and the oscillation about it can be obtained simultaneously by: (1) writing down the equation of motion for any convenient degree of freedom; (2) identifying the point of equilibrium as the point where the acceleration is zero; (3) rewriting the equation of motion in terms of the displacement from this point. Consider a mass m suspended by a spring. The equilibrium extension x e could easily be obtained by balancing weight and spring forces: x m k kx m Alternatively, writing down the general equation of motion in terms of the spring extension: mg From this, the position of equilibrium can be identified by setting acceleration d 2 x/dt 2 = 0. Change variables to and note that, since x e is constant,. Then: A constant force (such as gravity) changes the position of equilibrium. A linear-restoring-force system undergoes SHM about the equilibrium position. Example 5. A block of mass 16 kg is suspended vertically by two light springs of stiffness 200 N m 1. By writing down the equation of motion, find: (a) the equivalent single spring; (b) the extension at equilibrium; (c) the period of oscillation about the point of equilibrium. k = 200 N/m k = 200 N/m 16 kg Mechanics Topic E (Oscillations) - 7 David Apsley

8 Example 6. A 4 kg mass is suspended vertically by a string of elastic modulus λ = 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? Example 7. A mass m is hung in the loop of a light smooth cable whose two ends are fixed to a horizontal support by springs of stiffness k and 2k (see figure). Find the period of vertical oscillations in terms of m and k. k 2k m Oscillation about a point of equilibrium can also be observed in multi-spring systems where components are already loaded at equilibrium, as in the following example. Example 8. A particle of mass 0.4 kg is confined to move along a smooth horizontal plane between two points A and B a distance 2 m apart by two light springs, both of natural length 0.8 m. The springs connecting the particle to A and B have stiffnesses k L = 50 N m 1 and k R = 150 N m 1 respectively. (a) Write down the equation of motion of the particle in terms of the distance x from A. (b) (c) Find the position of equilibrium. Show that, if released from rest half way between the walls, the particle undergoes simple harmonic motion and calculate: (i) the period of oscillation; (ii) the maximum speed and maximum acceleration. Mechanics Topic E (Oscillations) - 8 David Apsley

9 2.6 Small-Amplitude Approximations Many oscillatory systems do not undergo exact simple harmonic motion (where the restoring force is proportional to a displacement), but their motion is approximately SHM provided that the amplitude of oscillation is small. Such small-amplitude approximations are particularly common for rotational motion about a fixed point (where the restoring torque is often provided by gravity or elasticity) and rely on the approximations (15) (16) when θ is measured in radians. These may be derived formally by power-series expansions, but their essential validity is easily seen geometrically (see right). 1 sin The following table shows that the approximation for sin θ is accurate to about 1% or better for angles as large as 15. If a structural member were actually displaced by this much then oscillation would be the least of your worries! θ (degrees) θ (radians) sin θ Important warning: duplication of notation When considering rotational oscillations we will run into problems with the same symbols being used for different quantities. T is used for both torque and period of oscillation (and sometimes tension); in this section we will use it to mean torque and write period out in full. ω is used for both natural circular frequency and angular velocity; while dealing with oscillations we shall reserve it to mean the natural circular frequency and write or dθ/dt for the angular velocity Rotational Oscillations Driven by Gravity Compound Pendulum In a simple pendulum all the mass is concentrated at one point. This can be treated using either the momentum or angular-momentum equations. In a compound pendulum the mass is distributed; the system must be analysed by rotational dynamics. In the absence of friction, 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. A L G Let the distance from axis to centre of gravity be L and let the angular displacement of line AG from the vertical be θ. Mg The line of action of the weight Mg lies at a distance L sin θ from the axis of rotation, and hence imparts a torque (moment of force) of magnitude and in the opposite Mechanics Topic E (Oscillations) - 9 David Apsley

10 sense to θ. The rotational equation of motion is torque = moment of inertia angular acceleration Hence, For small oscillations,, so that This is SHM with natural circular frequency Example 9. (Exam 2017) A uniform circular disk of mass 3 kg and radius 0.4 m is suspended from a horizontal axis passing through a point on its circumference and perpendicular to the plane of the disk. A small particle of mass 2 kg is attached to the other side of the disk at the opposite end of a diameter. (a) (b) Find the moment of inertia of the combination about the given axis. Find the period of small oscillations about the axis. Axis 3 kg 2 kg Mechanics Topic E (Oscillations) - 10 David Apsley

11 Example 10. 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. 0.5 m The Dog and Duck 0.5 m 0.5 m Rotational Oscillations Driven By Elastic Forces If the rotational displacement θ is small then the additional extension of a spring at distance r from an axis is essentially the length of circular arc: This gives a force of magnitude opposing the displacement and hence a torque of magnitude Fr acting in the opposite direction to the rotational displacement: axis r r Example 11. 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 natural circular frequency for small oscillations. L k k Mechanics Topic E (Oscillations) - 11 David Apsley

12 2.7 Derivation of the SHM equation from Energy Principles For a body of mass m, subject only to elastic forces 2 with stiffness k, total energy is constant: Differentiating with respect to time gives Apply the chain rule to each term: Replacing v by dx/dt: Dividing by dx/dt: Hence, Example 12. 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 a vertical plane without rotating, the rods making an angle θ with the vertical. (a) Write exact expressions for the potential energy and kinetic energy of the sign in terms of M, L, g, the displacement angle θ and its time derivative. (b) If the sign is displaced an angle θ = p/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,. (d) Show that, for small-amplitude oscillations, the assumption of constant total energy leads to simple harmonic motion, and find its period. L L M 2 Actually, far more general. For small displacements about a stable equilibrium point in an arbitrary potential: the last because, for stable equilibrium, (dv/dx) 0 = 0, whilst (d 2 V/dx 2 ) 0 is a positive constant. Mechanics Topic E (Oscillations) - 12 David Apsley

13 3. DAMPED FREE OSCILLATION All real dynamical systems are subject to friction, which opposes relative motion and consumes mechanical energy. For a system undergoing free oscillation, we shall show that: moderate damping decaying amplitude and reduced frequency; large damping oscillation prevented. The level at which oscillation is just suppressed is called critical damping. 3.1 The Equation of Motion Friction always acts in a direction so as to oppose relative motion. For a frictional force that depends on velocity, the damping force is often modelled as 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. (17) The equation of motion ( F = ma ) for a damped mass-spring system is or c k x m (18) Dividing by m, this can be written (19) where is the natural frequency of the undamped system. Example 13. (Exam, May 2016) A carriage of mass 20 kg is attached to a wall by a spring of stiffness 180 N m 1. When required, a hydraulic damper can be attached to provide a resistive force with magnitude proportional to velocity; the constant of proportionality c = 40 N/(m s 1 ). (a) If the carriage is displaced, write down its equation of motion, in terms of the displacement x, in the case when the hydraulic damper is in place. (b) (c) If there is no damping, find the period of oscillation. If the hydraulic damper is attached, find the period of oscillation and the fraction by which the amplitude is reduced on each cycle. c = 40 N/(m/s) x 20 kg k = 180 N/m Do this question from first principles, rather than the damping formulae that follow. Mechanics Topic E (Oscillations) - 13 David Apsley

14 3.2 General Solution For Different Damping Levels The undamped system (c = 0) has solutions of the form. In the presence of damping we expect the solution to decay in magnitude and, possibly, have a slightly different frequency. Hence we might anticipate solutions of the form where A and are arbitrary constants and λ and ω d are to be found. If you don t like the analysis which follows, you can try simply substituting this into equation (19) and (after quite a lot of algebra) deriving the same results as below. Equation (19) is a homogeneous, linear, second-order differential equation with constant coefficients. Seek solutions of the form: (20) where p is a constant to be evaluated. Substituting in (19) we obtain the auxiliary equation with roots (21) (For convenience) define (damping ratio) (22) Then (23) As you know from your maths course, there are 3 possibilities, depending on the sign of the quantity under the square root (here, ) in equation (23): 1. two complex conjugate roots if < 1; 2. two distinct negative real roots if > 1; 3. two equal (negative, real) roots if = 1. For complex conjugate roots ( )the general solution can be written where A and B (or C and D) are arbitrary constants. The last bracket also has an equivalent amplitude/phase-angle form. Mechanics Topic E (Oscillations) - 14 David Apsley

15 Case 1: < 1 ( c < 2mω): under-damped system The general solution may be written where (24) (25) There is oscillation with reduced amplitude (decaying exponentially as frequency, ω d. ) and reduced The amplitude reduction factor over one cycle ( or ) is This may be used to determine the damping ratio experimentally. Case 1 also includes the special case of no damping ( = 0), in which case λ = 0 and ω d = ω. (26) Case 2: > 1 (c > 2mω): over-damped system The general solution is of the form where (27) λ ω (28) There is no oscillation and, as both of the roots p are negative (i.e. λ positive), the displacement decays to zero, more slowly as the damping ratio increases. Case 3: = 1 (c = 2mω): critically-damped system The general solution has the form: There is no oscillation and the amplitude decays rapidly to zero. (29) Mechanics Topic E (Oscillations) - 15 David Apsley

16 displacement (m) Summary The damped mass-spring equation has solutions that depend on: undamped natural frequency: damping ratio: If < 1 the system is under-damped, oscillating with exponentially-decaying amplitude and reduced frequency. If > 1 the system is over-damped and no oscillation occurs. The fastest return to equilibrium occurs when the system is critically damped ( = 1). Example. m = 1 kg, k = 64 N m 1 ω = 8 rad s 1 x 0 = 0.05 m,. Cases in the graph: c = 1.6 N s m 1 ( = 0.1) c = 16 N s m 1 ( = 1) c = 64 N s m 1 ( = 4) over-damped critically damped time (s) under-damped Exercise. Use any other computer program or tool to compute and plot the solution for various combinations of m, k, c. Example 14. 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 for the system to be critically damped? c = 60 N s/m k = 700 N/m 40 kg Mechanics Topic E (Oscillations) - 16 David Apsley

17 4. FORCED OSCILLATION Systems that oscillate about a position of equilibrium under restoring forces at their own preferred or natural frequency are said to undergo free oscillation. Systems that are perturbed by some externally-imposed periodic forcing are said to undergo forced oscillation. Examples which may be encountered in civil engineering are: concert halls and stadiums; bridges (e.g. traffic- or wind-induced oscillations); earthquakes. The natural frequency and damping ratio are important parameters for systems responding to externally-imposed forces. Large-amplitude oscillations occur when the imposed frequency is close to the natural frequency (the phenomenon of resonance). In general, the system s freeoscillation properties affects both amplitude and phase of response to external forcing. 4.1 Mathematical Expression of the Problem c x The general form of the equation of motion with harmonic forcing is k m F sin t 0 (30) or (31) where the undamped natural frequency is given by Second-order ODEs of the form (31) are covered in your mathematics courses. The general solution is the sum of a complementary function (containing two arbitrary constants and obtained by setting the RHS of (31) to 0) and a particular integral (any particular solution of the equation, obtained by a trial function based on the RHS). The complementary function is a free-oscillation 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 Ωt and cos Ωt. Example 15. Write down the general solution of the following differential equation: Mechanics Topic E (Oscillations) - 17 David Apsley

18 4.2 Static Load A useful comparison is with the displacement under a steady load of the same amplitude F 0, rather than a variable load. The steady-state displacement x s is given by the position of static equilibrium: whence (32) 4.3 Undamped Forced Oscillation In the special case of no frictional damping, the equation of motion is k x m F sin t 0 The form of the forcing function suggests a particular integral of the form substituting this in the equation of motion to find C, one finds that. By (33) where, as above, is the displacement under a static load of the same amplitude. Hence the magnitude of forced oscillations is Mx s, where the amplitude ratio or magnification factor M is given by (34) Key Points (1) The response of the system to external forcing depends on the ratio of the forcing frequency Ω to the natural frequency ω. (2) There is resonance (M ) if the forcing frequency approaches the natural frequency (Ω ω). (3) If Ω < ω the oscillations are in phase with the forcing (C has the same sign as F 0 ), because the system can respond fast enough. (4) If Ω > ω the oscillations are 180 out of phase with the forcing (C has the opposite sign to F 0 ) because the imposed oscillations are too fast for the system to follow. (5) If Ω >> ω (very fast oscillations) the system will barely move (C 0). Mechanics Topic E (Oscillations) - 18 David Apsley

19 4.4 Damped Forced Oscillation The complete analysis of oscillations for a SDOF system involves forced motion and damping. The equation of motion is Because of the dx/dt term, a particular integral must contain both and. Substituting a trial solution of the form produces, after a lot of algebra, a solution (optional exercise) of the form where, as in Section 3, the damping ratio is (35) and the displacement under static load is This is most conveniently written in the amplitude/phase-angle form where the amplitude ratio M is given by (36) (37) and the phase lag by (38) Key Points (1) The response of the system to forcing depends on both the ratio of forcing to natural frequencies (Ω/ω) and the damping ratio. (2) Damping prevents the complete blow-up (M ) as Ω ω. (3) The imposed frequency at which the maximum amplitude oscillations (resonance) occur is slightly less than the undamped natural frequency ω. In fact, Mechanics Topic E (Oscillations) - 19 David Apsley

20 (4) The phase lag varies from 0 (as Ω 0) to π (as Ω ). However, the phase lag is always π/2 when Ω = ω, irrespective of the level of damping. Example 15. (Meriam and Kraige, modified) The seismometer shown is attached to a structure which has a horizontal harmonic oscillation at 3 Hz. The instrument has a mass m = 0.5 kg, a spring stiffness k = 150 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 5 mm, determine the amplitude of the horizontal movement x B of the structure. x (t) B k x m c Mechanics Topic E (Oscillations) - 20 David Apsley

21 Numerical Answers to Examples in the Text Full worked answers are given in a separate document online. Example s Example 2. (a) ω = 3, A = 13, = tan 1 (12/5) = 1.18 radians = 67.4 (b) ω = 2, A = 5, = tan 1 ( 4/3) = 2.21 radians = Example 3. (a) k = 160 N m 1 ; (b) 4 rad s 1 ; (c) Hz; (d) 1.57 s; (e) 0.4 m s 1 Example 4. (a) 2.69 m s 1 ; 2.42 m s 2 ; (b); 111 N; 183 N; 147 N Example 5. (a) k = 400 N m 1 ; (b) m; (c) 1.26 s Example m; no (the string becomes unstretched). Example 7. Example 8. (a) ; (b) x = 1.1 m; (c) (i) s; (ii) 2.24 m s 1 ; 50 m s 2 Example 9. (a) 2.0 kg m 2 ; (b) 1.70 s Example s Example 11. Example 12. (a), ; (b) (c) ; (d) Example 13. Mechanics Topic E (Oscillations) - 21 David Apsley

22 (a) ; (b) 2.09 s (c) 2.22 s; Example 14. It does oscillate; = 0.179, period = 1.53 s; for critical damping, c = 335 N s m 1 Example 15. Example mm Mechanics Topic E (Oscillations) - 22 David Apsley