Primary Mechanical Vibrators Mass-Spring System Equation Of

Primary Mechanical Vibrators Music 206: Mechanical Vibration • With a model of a wind instrument body, a model of the mechanical “primary” resonator is needed for a complete instrument. Tamara Smyth, [email protected] Department of Music, University of California, San Diego (UCSD) November 15, 2019 • Many sounds are produced by coupling the mechanical vibrations of a source to the resonance of an acoustic tube: – In vocal systems, air pressure from the lungs controls the oscillation of a membrane (vocal fold), creating a variable constriction through which air flows. – Similarly, blowing into the mouthpiece of a clarinet will cause the reed to vibrate, narrowing and widening the airflow aperture to the bore. <ul style="display: flex;"><li style="flex:1">1</li><li style="flex:1">Music 206: Mechanical Vibration </li><li style="flex:1">2</li></ul><ul style="display: flex;"><li style="flex:1">Mass-spring System </li><li style="flex:1">Equation of Motion </li></ul>• Before modeling this mechanical vibrating system, let’s review some acoustics, and mechanical vibration. • The force on an object having mass m and acceleration a may be determined using Newton’s second law of motion • The simplest oscillator is the mass-spring system: F = ma. – a horizontal spring fixed at one end with a mass connected to the other. • There is an elastic force restoring the mass to its equilibrium position, given by Hooke’s law mF = −Kx, Kxwhere K is a constant describing the stiffness of the spring. Figure 1: An ideal mass-spring system. • The spring force is equal and opposite to the force due to acceleration, yielding the equation of motion: • The motion of an object can be describe in terms of its d2x 1. displacement x(t) <ul style="display: flex;"><li style="flex:1">m</li><li style="flex:1">= −Kx. </li></ul>dt2 dx 2. velocity v(t) = dt dv d2x 3. acceleration a(t) = =<ul style="display: flex;"><li style="flex:1">dt </li><li style="flex:1">dt2 </li></ul><ul style="display: flex;"><li style="flex:1">Music 206: Mechanical Vibration </li><li style="flex:1">3</li><li style="flex:1">Music 206: Mechanical Vibration </li><li style="flex:1">4</li></ul><ul style="display: flex;"><li style="flex:1">Solution to Equation of Motion </li><li style="flex:1">Motion of the Mass Spring </li></ul><ul style="display: flex;"><li style="flex:1">• The solution to </li><li style="flex:1">• We can make the following inferences from the </li></ul>oscillatory motion (displacement) of the mass-spring system written as: d2x dt2 <ul style="display: flex;"><li style="flex:1">m</li><li style="flex:1">= −Kx, </li></ul>is a function proportional to its second derivative. • This condition is met by the sinusoid: x = A cos(ω0t + φ) x(t) = A cos(ω0t) – Energy is conserved, the oscillation never decays. – At the peaks: dx ∗ the spring is maximally compressed or stretched; ∗ the mass is momentarily stopped as it is changing direction; = −ω0A sin(ω0t + φ) dt d2x = −ω02A cos(ω0t + φ) dt2 – At zero crossings, • Substituting into the equation of motion yields: d2x ∗ the spring is momentarily relaxed (it is neither compressed nor stressed), and thus holds no PE ∗ all energy is in the form of KE. <ul style="display: flex;"><li style="flex:1">m</li><li style="flex:1">= −Kx, </li></ul>dt2 ✭<ul style="display: flex;"><li style="flex:1">✭</li><li style="flex:1">✭</li></ul>✭K✭<ul style="display: flex;"><li style="flex:1">✭</li><li style="flex:1">✭</li></ul>✭ ✭s(ω0t + φ), 20✭ <ul style="display: flex;"><li style="flex:1">✭</li><li style="flex:1">✭</li><li style="flex:1">✭</li></ul>✭– Since energy is conserved, the KE at zero crossings is exactly the amount needed to stretch the spring to displacement −A or compress it to +A. ✭<ul style="display: flex;"><li style="flex:1">✭</li><li style="flex:1">✭</li></ul>s(ω0t + φ) = − A co <ul style="display: flex;"><li style="flex:1">✭</li><li style="flex:1">✭</li></ul>−ω A co <ul style="display: flex;"><li style="flex:1">✭</li><li style="flex:1">✭</li></ul><ul style="display: flex;"><li style="flex:1">✭</li><li style="flex:1">✭</li><li style="flex:1">✭</li><li style="flex:1">✭</li></ul>✭mshowing that the natural frequency of vibration is rKω0 = .m<ul style="display: flex;"><li style="flex:1">Music 206: Mechanical Vibration </li><li style="flex:1">5</li><li style="flex:1">Music 206: Mechanical Vibration </li><li style="flex:1">6</li></ul>PE and KE in the mass-spring oscillator Conservation of Energy • The total energy of the ideal mass-spring system is constant: • All vibrating systems consist of this interplay between an energy storing component and an energy carrying (“massy”) component. E = PE + KE 1<ul style="display: flex;"><li style="flex:1">ꢀ</li><li style="flex:1">ꢁ</li></ul><ul style="display: flex;"><li style="flex:1">=</li><li style="flex:1">KA2 sin2(ω0t + φ) + cos2(ω0t + φ) </li></ul>21• The potential energy PE of the ideal mass-spring system is equal to the work done1 stretching or compressing the spring: =KA2 2Potential Energy (PE) 1 0.8 0.6 0.4 0.2 01PE = Kx2, 21 =KA2 cos2(ω0t + φ). 2<ul style="display: flex;"><li style="flex:1">0</li><li style="flex:1">0.1 </li><li style="flex:1">0.2 </li><li style="flex:1">0.3 </li><li style="flex:1">0.4 </li><li style="flex:1">0.5 </li><li style="flex:1">0.6 </li><li style="flex:1">0.7 </li><li style="flex:1">0.8 </li><li style="flex:1">0.9 </li><li style="flex:1">1</li></ul>• The kinetic energy KE in the system is given by the motion of mass: Time (s) Kinetic Energy (KE) 1 0.8 0.6 0.4 0.2 01KE = mv2, 2K1✟✯ ✟✟ 2 <ul style="display: flex;"><li style="flex:1">2</li><li style="flex:1">2</li></ul>✟mω A sin (ω0t + φ) ✟==021<ul style="display: flex;"><li style="flex:1">0</li><li style="flex:1">0.1 </li><li style="flex:1">0.2 </li><li style="flex:1">0.3 </li><li style="flex:1">0.4 </li><li style="flex:1">0.5 </li></ul>Time (s) <ul style="display: flex;"><li style="flex:1">0.6 </li><li style="flex:1">0.7 </li><li style="flex:1">0.8 </li><li style="flex:1">0.9 </li><li style="flex:1">1</li></ul>KA2 sin2(ω0t + φ). 2Figure 2: The interplay between PE and KE. 1work is the product of the average force and the distance moved in the direction of the force <ul style="display: flex;"><li style="flex:1">Music 206: Mechanical Vibration </li><li style="flex:1">7</li><li style="flex:1">Music 206: Mechanical Vibration </li><li style="flex:1">8</li></ul><ul style="display: flex;"><li style="flex:1">Other Simple Oscillators—Pendulum </li><li style="flex:1">Other Simple Oscillators—Helmholtz </li></ul>Resonator • Besides the mass-spring, there are other examples of <ul style="display: flex;"><li style="flex:1">simple harmonic motion. </li><li style="flex:1">• A Helmholtz Resonator: The mass of air in the </li></ul>neck serves as a piston, and the larger volume as the spring. • A pendulum: a mass m attached to a string of length l vibrates in simple harmonic motion, provided that x ≪ l. S LmVlmx• The resonant frequency of the Helmholtz resonator is given by Figure 3: A simple pendulum. r<ul style="display: flex;"><li style="flex:1">c</li><li style="flex:1">a</li></ul>f0 = , 2π V L where a is the area of the neck, V is the volume, L is the length of the neck and c is the speed of sound. • The frequency of vibration is r1gl<ul style="display: flex;"><li style="flex:1">f = </li><li style="flex:1">,</li></ul>2π where g is the acceleration due to gravity. • According to this equation, would changing the mass change the frequency? <ul style="display: flex;"><li style="flex:1">Music 206: Mechanical Vibration </li><li style="flex:1">9</li><li style="flex:1">Music 206: Mechanical Vibration </li><li style="flex:1">10 </li></ul><ul style="display: flex;"><li style="flex:1">Damped Vibration </li><li style="flex:1">Mass-spring-damper system </li></ul>• In a real system, mechanical energy is lost due to friction and other mechanisms causing damping. • Damping of an oscillating system corresponds to a loss of energy or equivalently, a decrease in the amplitude of vibration. • Unless energy is reintroduced into the system, the amplitude of the vibrations with decrease with time. Km• A change in peak amplitude with time is called an envelope, and if the envelope decreases, the vibrating system is said to be damped. xFigure 5: A mass-spring-damper system. 1 0.8 • The damper is a mechanical resistance (or viscosity) and introduces a drag force Fr typically proportional to velocity, 0.6 0.4 0.2 Fr = −Rv 0dx = −R ,−0.2 −0.4 −0.6 −0.8 −1 dt where R is the mechanical resistance. <ul style="display: flex;"><li style="flex:1">0</li><li style="flex:1">0.1 </li><li style="flex:1">0.2 </li><li style="flex:1">0.3 </li><li style="flex:1">0.4 </li><li style="flex:1">0.5 </li><li style="flex:1">0.6 </li><li style="flex:1">0.7 </li><li style="flex:1">0.8 </li><li style="flex:1">0.9 </li><li style="flex:1">1</li></ul>Figure 4: A damped sinusoid. <ul style="display: flex;"><li style="flex:1">Music 206: Mechanical Vibration </li><li style="flex:1">11 </li><li style="flex:1">Music 206: Mechanical Vibration </li><li style="flex:1">12 </li></ul><ul style="display: flex;"><li style="flex:1">Damped Equation of Motion </li><li style="flex:1">The Solution to the Damped Vibrator </li></ul>• The equation of motion for the damped system is obtained by adding the drag force into the equation of motion: • The solution to the system equation dx2 dt2 dx + 2α + ω02x = 0 d2x dt2 dx dt m+ R + Kx = 0, has the form dt or alternatively x = A(t) cos(ωdt + φ). where A(t) is the amplitude envelope A(t) = e−t/τ = e−αt and the natural frequency ωd d2x dt2 dx + 2α + ω02x = 0, dt ,where α = R/2m and ω02 = K/m. • The damping in a system is often measured by the quantity τ, which is the time for the amplitude to decrease to 1/e: qωd = ω02 − α2, is lower than that of the ideal mass-spring system r12m τ = =.<ul style="display: flex;"><li style="flex:1">α</li><li style="flex:1">R</li></ul>Kω0 = .m• Peak A and φ are determined by the initial displacement and velocity. <ul style="display: flex;"><li style="flex:1">Music 206: Mechanical Vibration </li><li style="flex:1">13 </li><li style="flex:1">Music 206: Mechanical Vibration </li><li style="flex:1">14 </li></ul><ul style="display: flex;"><li style="flex:1">Systems with Several Masses </li><li style="flex:1">System Equations for Two </li></ul>Spring-Coupled Masses • When there is a single mass, its motion has only one degree of freedom and one natural mode of vibration. a<ul style="display: flex;"><li style="flex:1">x1 </li><li style="flex:1">x2 </li></ul><ul style="display: flex;"><li style="flex:1">m</li><li style="flex:1">m</li></ul><ul style="display: flex;"><li style="flex:1">K</li><li style="flex:1">K</li><li style="flex:1">K</li></ul>• Consider the system having 2 masses and 3 springs: Figure 7: A short section of a string. a<ul style="display: flex;"><li style="flex:1">x1 </li><li style="flex:1">x2 </li></ul><ul style="display: flex;"><li style="flex:1">m</li><li style="flex:1">m</li></ul>• The extensions of the left, middle and right springs are x1, x2 − x1, and −x2, respectively. <ul style="display: flex;"><li style="flex:1">K</li><li style="flex:1">K</li><li style="flex:1">K</li></ul>Figure 6: 2-mass 3-spring system. • When a spring is extended by x, the mass attached to <ul style="display: flex;"><li style="flex:1">the </li><li style="flex:1">• The system will have two “normal” independent </li></ul>modes of vibration: – left experiences a positive horizontal restoring force Fr = Kx; 1. one in which masses move in the same direction, with frequency – right experiences an equal and opposite force Fr = −Kx, r1Kmf1 = where K is the spring constant. 2π • The equation of motion for the displacement of the first mass: 2. one in which masses move in different directions with frequency rmx¨1 = −Kx1 + K(x2 − x1), and the second mass, 13K f2 = <ul style="display: flex;"><li style="flex:1">2π </li><li style="flex:1">m</li></ul>mx¨2 = −K(x2 − x1) + K(−x2). (assuming equal masses and springs). <ul style="display: flex;"><li style="flex:1">Music 206: Mechanical Vibration </li><li style="flex:1">15 </li><li style="flex:1">Music 206: Mechanical Vibration </li><li style="flex:1">16 </li></ul>Additional Modes Forced (Driven) Vibration • When modes are independent, the system can vibrate in one mode with minimal excitation of another. • Getting they system to vibrate at any single mode of vibration requires exciting, or driving the system at the desired mode. • Unless constrained to one-dimension, the masses can also move transversely (at right-angles to the springs). • See Dan Russell’s site: The forced harmonic oscillator • An additional mass adds an additional mode of vibration. • When a simple harmonic oscillator is driven by an external force F(t), the equation of motion becomes • An N-mass system has N modes per degree of freedom. mx¨ + Rx˙ + Kx = F(t). • The driving force may have harmonic time dependence, it may be impulsive, or it can be a random function of time (noise). • As N gets very large, it becomes convenient to view the system as a continuous string with a uniform mass density and tension. Figure 8: Increasing the number of masses (Science of Sound). <ul style="display: flex;"><li style="flex:1">Music 206: Mechanical Vibration </li><li style="flex:1">17 </li><li style="flex:1">Music 206: Mechanical Vibration </li><li style="flex:1">18 </li></ul><ul style="display: flex;"><li style="flex:1">Phase of Driven Vibration </li><li style="flex:1">Resonance </li></ul>• At resonance, there is maximum transfer of energy between the hand and the mass-spring system. • The force driving an oscillation can be illustrated by holding a slinky in the vertical direction. • Plotting amplitude A with respect to fh, shows a curve that is almost symmetrical about its peak • Move the hand up and down slowly: – At low fh, both the hand and the mass move in the same direction, and the spring hardly stretches at all. Amax, i.e., it has a bandwidth measured a distance <ul style="display: flex;"><li style="flex:1">down from Amax </li><li style="flex:1">.</li></ul>0−5 • Increasing fh makes it harder to move mass, and it lags behind the driving force. −10 −15 −20 −25 −30 −35 • At resonance fh = f0: – the mass is 1/4 cycle behind the hand. – the amplitude is at its maximum. −40 <ul style="display: flex;"><li style="flex:1">0</li><li style="flex:1">500 </li><li style="flex:1">1000 </li><li style="flex:1">1500 </li><li style="flex:1">2000 </li><li style="flex:1">2500 </li><li style="flex:1">3000 </li></ul>Driving Frequency, Hz • The higher the Q (quality factor), of the vibrating system, the more abrupt the transition in phase. Figure 9: A Resonator shows peak amplitude when the driving force is equal to the system’s natural frequency. • Both the peak Amax the bandwidth ∆f depend on the damping in the system: 1. heavily damped: ∆f is large and Amax is small. 2. little damping: ∆f is small and Amax is large, creating a sharp resonance. <ul style="display: flex;"><li style="flex:1">Music 206: Mechanical Vibration </li><li style="flex:1">19 </li><li style="flex:1">Music 206: Mechanical Vibration </li><li style="flex:1">20 </li></ul><ul style="display: flex;"><li style="flex:1">Quality Factor and Bandwidth </li><li style="flex:1">How to Discretize? </li></ul>• The bandwidth of the resonance is typically described f0 • The one-sided Laplace transform of a signal x(t) is defined by using the quantity Q = , for quality factor. Z∆f ∞<ul style="display: flex;"><li style="flex:1">∆</li><li style="flex:1">∆</li></ul>X(s) = Ls{x} = x(t)e−stdt • A high-Q has a sharp resonance, and a low-Q has a broad resonance curve. 0where t is real and s = σ + jω is a complex variable. • For a vibrator set into motion and left to vibrate freely, its decay time is proportional to the Q of its resonance. • The differentiation theorem for Laplace transforms states that dx(t) ↔ sX(s) dt where x(t) is any differentiable function that approaches zero as t goes to infinity. • In terms of our mechanical system, given the R<ul style="display: flex;"><li style="flex:1">resistance α = </li><li style="flex:1">, the quality factor is: </li></ul>2m ω0 • The transfer function of an ideal differentiator is H(s) = s, which can be viewed as the Laplace transform of the operator d/dt. <ul style="display: flex;"><li style="flex:1">Q = </li><li style="flex:1">.</li></ul>2α • Given the equation of motion mx¨ + Rx˙ + Kx = F(t), the Laplace Transform is s2X(s) + 2αsX(s) + ω02X(s) = F(s). <ul style="display: flex;"><li style="flex:1">Music 206: Mechanical Vibration </li><li style="flex:1">21 </li><li style="flex:1">Music 206: Mechanical Vibration </li><li style="flex:1">22 </li></ul><ul style="display: flex;"><li style="flex:1">Finite Difference </li><li style="flex:1">FDA of Equation of Motion </li></ul>• The finite difference approximation (FDA) amounts to replacing derivatives by finite differences, or dx(t) − x(t − δ) x(nT) − x[(n − 1)T] ∆x(t) = lim δ→0 ≈.<ul style="display: flex;"><li style="flex:1">dt </li><li style="flex:1">δ</li><li style="flex:1">T</li></ul>• The z transform of the first-order difference operator is (1 − z−1)/T. Thus, in the frequency domain, the finite-difference approximation may be performed by making the substitution 1 − z−1 s → T• The first-order difference is first-order error accurate in T. Better performance can be obtained using the bilinear transform, defined by the substitution <ul style="display: flex;"><li style="flex:1">ꢂ</li><li style="flex:1">ꢃ</li></ul>1 − z−1 1 + z−1 2s −→ c where c = .T<ul style="display: flex;"><li style="flex:1">Music 206: Mechanical Vibration </li><li style="flex:1">23 </li><li style="flex:1">Music 206: Mechanical Vibration </li><li style="flex:1">24 </li></ul><ul style="display: flex;"><li style="flex:1">Pressure Controlled Valves </li><li style="flex:1">Classifying Pressure-Controlled Valves </li></ul><ul style="display: flex;"><li style="flex:1">• Returning now to our model of wind instruments.... </li><li style="flex:1">• The method for simulating the reed typically depends </li></ul>on whether an additional upstream or downstream pressure causes the corresponding side of the valve to open or close further. • Blowing into an instrument mouthpiece creates a pressure difference across the surface of the reed.

Primary Mechanical Vibrators Mass-Spring System Equation Of

Details

Download

Copyright

We respect the copyrights and intellectual property rights of all users. All uploaded documents are either original works of the uploader or authorized works of the rightful owners.

Support