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Primary Mechanical Vibrators

Music 206: Mechanical Vibration

Tamara Smyth, [email protected] • With a model of a wind instrument body, a model of Department of Music, the mechanical “primary” resonator is needed for a University of California, San Diego (UCSD) complete instrument. November 15, 2019 • Many are produced by coupling the mechanical vibrations of a source to the of an acoustic tube: – In vocal systems, air pressure from the lungs controls the 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.

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Mass-spring System Equation of

• Before modeling this mechanical vibrating system, • The on an object having m and let’s review some , and mechanical vibration. a may be determined using ’s second law of motion • The simplest oscillator is the mass-spring system: – a horizontal spring fixed at one end with a mass F = ma. connected to the other. • There is an elastic force restoring the mass to its equilibrium position, given by Hooke’s law m F = −Kx, K x where K is a constant describing the stiffness of the Figure 1: An ideal mass-spring system. spring.

• The motion of an object can be describe in terms of • The spring force is equal and opposite to the force its due to acceleration, yielding the equation of motion: 2 1. x(t) d x m 2 = −Kx. dx dt 2. v(t)= dt dv d2x 3. acceleration a(t)= = dt dt2

Music 206: Mechanical Vibration 3 Music 206: Mechanical Vibration 4 Solution to Equation of Motion Motion of the Mass Spring

• The solution to • We can make the following inferences from the d2x oscillatory motion (displacement) of the mass-spring m = −Kx, dt2 system written as:

is a function proportional to its second derivative. x(t)= A cos(ω0t) • This condition is met by the sinusoid: – is conserved, the oscillation never decays. – x = A cos(ω0t + φ) At the peaks: dx ∗ the spring is maximally compressed or stretched; = −ω0A sin(ω0t + φ) dt ∗ the mass is momentarily stopped as it is 2 d x 2 changing direction; 2 = −ω0A cos(ω0t + φ) dt – At zero crossings, • Substituting into the equation of motion yields: ∗ the spring is momentarily relaxed (it is neither d2x compressed nor stressed), and thus holds no PE m = −Kx, dt2 ∗ all energy is in the form of KE. ✭✭ ✭✭ ✭✭✭ K ✭✭✭ 2 ✭✭✭✭ ✭✭✭✭ – Since energy is conserved, the KE at zero crossings −ω0✭A✭cos(✭✭ ω0t + φ) = − ✭A✭cos(✭✭ ω0t + φ), m is exactly the amount needed to stretch the spring showing that the natural of vibration is to displacement −A or compress it to +A. K ω0 = . rm

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PE and KE in the mass-spring oscillator

• The total energy of the ideal mass-spring system is • All vibrating systems consist of this interplay between constant: an energy storing component and an energy carrying E = PE + KE (“massy”) component. 1 2 2 2 = KA sin (ω0t + φ)+cos (ω0t + φ) • The PE of the ideal mass-spring 2 1 system is equal to the done1 stretching or = KA2  compressing the spring: 2 Potential Energy (PE) 1 1 2 PE = Kx , 0.8 2 1 0.6 2 2 0.4 = KA cos (ω0t + φ). 2 0.2

0 0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1 • The KE in the system is given by (s) Kinetic Energy (KE) the motion of mass: 1

1 0.8 KE = mv2, 2 0.6 0.4 1 ✟✟✯K ✟✟ 2 2 2 = ✟mω0A sin (ω0t + φ) 0.2 2 0 0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1 1 2 2 Time (s) = KA sin (ω0t + φ). 2 Figure 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

Music 206: Mechanical Vibration 7 Music 206: Mechanical Vibration 8 Other Simple Oscillators— Other Simple Oscillators—Helmholtz Resonator

• Besides the mass-spring, there are other examples of . • A Helmholtz Resonator: The mass of air in the • A pendulum: a mass m attached to a string of neck serves as a piston, and the larger volume as the length l vibrates in simple harmonic motion, provided spring. that x ≪ l. S L m

l V

m x • The resonant frequency of the Helmholtz resonator is Figure 3: A simple pendulum. given by c a f = , 0 2π VL • The frequency of vibration is r where a is the area of the neck, V is the volume, L is 1 g f = , the length of the neck and c is the of . 2π l r where g is the acceleration due to . • According to this equation, would changing the mass change the frequency?

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Damped Vibration Mass-spring-damper system

• In a real system, mechanical energy is lost due to • of an oscillating system corresponds to a and other mechanisms causing damping. loss of energy or equivalently, a decrease in the • Unless energy is reintroduced into the system, the of vibration. amplitude of the vibrations with decrease with time. K m • A change in peak amplitude with time is called an envelope, and if the envelope decreases, the vibrating x system is said to be damped. Figure 5: A mass-spring-damper system.

1 • The damper is a mechanical resistance (or )

0.8 and introduces a drag force Fr typically proportional 0.6 to velocity, 0.4

0.2 Fr = −Rv 0 dx −0.2 = −R , dt −0.4

−0.6 where R is the mechanical resistance.

−0.8

−1 0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1

Figure 4: A damped sinusoid.

Music 206: Mechanical Vibration 11 Music 206: Mechanical Vibration 12 Damped Equation of Motion The Solution to the Damped Vibrator

• The equation of motion for the damped system is • The solution to the system equation obtained by adding the drag force into the equation of 2 dx dx 2 motion: +2α + ω0x =0 d2x dx dt2 dt m + R + Kx =0, dt2 dt has the form or alternatively x = A(t)cos(ωdt + φ). d2x dx +2α + ω2x =0, where A(t) is the amplitude envelope dt2 dt 0 −t/τ −αt 2 A(t)= e = e , where α = R/2m and ω0 = K/m. • The damping in a system is often measured by the and the natural frequency ωd quantity τ, which is the time for the amplitude to ω = ω2 − α2, decrease to 1/e: d 0 q 1 2m is lower than that of the ideal mass-spring system τ = = . α R K ω0 = . rm • Peak A and φ are determined by the initial displacement and velocity.

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Systems with Several System Equations for Two Spring-Coupled Masses

• When there is a single mass, its motion has only one a x x degree of freedom and one natural mode of vibration. 1 2 m m • Consider the system having 2 masses and 3 springs: K K K

a x1 x2 Figure 7: A short section of a string. m m K K K • The extensions of the left, middle and right springs are x1,x2 − x1, and −x2, respectively. Figure 6: 2-mass 3-spring system. • When a spring is extended by x, the mass attached to • The system will have two “normal” independent the modes of vibration: – left experiences a positive horizontal restoring 1. one in which masses move in the same direction, force Fr = Kx; with frequency – right experiences an equal and opposite force 1 K Fr = −Kx, f1 = where K is the spring constant. 2πrm 2. one in which masses move in different directions • The equation of motion for the displacement of the with frequency first mass: 1 3K mx¨1 = −Kx1 + K(x2 − x1), f2 = and the second mass, 2πr m (assuming equal masses and springs). mx¨2 = −K(x2 − x1)+ K(−x2).

Music 206: Mechanical Vibration 15 Music 206: Mechanical Vibration 16 Additional Modes Forced (Driven) Vibration

• When modes are independent, the system can vibrate • Getting they system to vibrate at any single mode of in one mode with minimal excitation of another. vibration requires exciting, or driving the system at • Unless constrained to one-dimension, the masses can the desired mode. also move transversely (at right-angles to the springs). • See Dan Russell’s site: The forced • An additional mass adds an additional mode of • When a simple harmonic oscillator is driven by an vibration. external force F (t), the equation of motion becomes • An N-mass system has N modes per degree of mx¨ + Rx˙ + Kx = F (t). freedom. • As N gets very large, it becomes convenient to view • The driving force may have harmonic time the system as a continuous string with a uniform dependence, it may be impulsive, or it can be a mass density and . random function of time ().

Figure 8: Increasing the number of masses (Science of Sound).

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Phase of Driven Vibration Resonance

• The force driving an oscillation can be illustrated by • At resonance, there is maximum transfer of energy holding a slinky in the vertical direction. between the hand and the mass-spring system.

• Move the hand up and down slowly: • Plotting amplitude A with respect to fh, shows a curve that is almost symmetrical about its peak – At low f , both the hand and the mass move in h bandwidth the same direction, and the spring hardly stretches Amax, i.e., it has a measured a distance at all. down from Amax. • Increasing f makes it harder to move mass, and it 0 h −5

lags behind the driving force. −10

−15

• At resonance fh = f0: −20

−25 Magnitude (dB)

– the mass is 1/4 cycle behind the hand. −30 – −35 the amplitude is at its maximum. −40 0 500 1000 1500 2000 2500 3000 Driving Frequency, Hz

• The higher the Q (quality factor), of the vibrating Figure 9: A Resonator shows peak amplitude when the driving force is equal to the system’s system, the more abrupt the transition in . 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.

Music 206: Mechanical Vibration 19 Music 206: Mechanical Vibration 20 Quality Factor and Bandwidth How to Discretize?

• The bandwidth of the resonance is typically described • The one-sided Laplace transform of a signal x(t) is f using the quantity Q = 0 , for quality factor. defined by ∆f ∞ ∆ ∆ −st • A high-Q has a sharp resonance, and a low-Q has a X(s) = Ls{x} = x(t)e dt 0 broad resonance curve. Z where 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 • The differentiation theorem for Laplace transforms states that resonance. d x(t) ↔ sX(s) • In terms of our mechanical system, given the dt R where x(t) is any differentiable function that resistance α = , the quality factor is: 2m approaches zero as t goes to infinity. ω Q = 0 . • The of an ideal differentiator is 2α H(s)= s, which can be viewed as the Laplace transform of the operator d/dt. • Given the equation of motion mx¨ + Rx˙ + Kx = F (t), the Laplace Transform is 2 2 s X(s)+2αsX(s)+ ω0X(s)= F (s).

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Finite Difference FDA of Equation of Motion

• The finite difference approximation (FDA) amounts to replacing derivatives by finite differences, or

d ∆ x(t) − x(t − δ) x(nT ) − x[(n − 1)T ] x(t) = lim ≈ . dt δ→0 δ T • The z transform of the first-order difference operator is (1 − z−1)/T . Thus, in the , 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 1 − z−1 2 s −→ c where c = . 1+ z−1 T  

Music 206: Mechanical Vibration 23 Music 206: Mechanical Vibration 24 Pressure Controlled Valves Classifying Pressure-Controlled Valves

• Returning now to our model of wind instruments.... • The method for simulating the reed typically depends • Blowing into an instrument mouthpiece creates a on whether an additional upstream or downstream pressure difference across the surface of the reed. pressure causes the corresponding side of the valve to open or close further. • When the reed oscillates, it creates and alternating opening and closure to the bore, allowing airflow • As per Fletcher, the couplet (σ1,σ2) may be used entry during the open phase and cutting it off during describe the upstream and downstream valve the closed phase. behaviour, respectively. • The effect, is often seen as a periodic train of • A σn value of pressure pulses into the bore. – +1 indicates an opening of the valve, • Sound sources of this kind are referred to as – -1 indicates a closing of the value, pressure-controlled valves and they have been on side n in response to a pressure increase. simulated in various ways to create musical synthesis models and vocal systems.

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Three simple configurations of PC valves Valve Displacement

    1) (−, +)  U   p2 p1 

S1 S2

   2) (+, −)  U  S3 p1   x S3 U  p2 p1 p2

S1 S2

3) (+, +)      p1  p2 Figure 11: Geometry of a blown open pressure-controlled valve showing effective areas  U S1,S2,S3.

Figure 10: Simplified models of three common configurations of pressure-controlled valves. • Consider the double reed in a blown open 1. (−, +): the valve is blown closed (as in woodwind configuration. instruments or reed-pipes of the pipe organ). • Surface S1 sees an upstream pressure p1, surface S2 2. (+, −): the valve is blown open (as in the simple sees the downstream pressure p2 (after flow lip-reed models for brass instruments, the human separation), and surface S3 sees the flow at the larynx, harmonicas and harmoniums). interior of the valve channel and the resulting 3. (+, +): the transverse (symmetric) model where the Bernoulli pressure. Bernoulli pressure causes the valve to close perpendicular to the direction of airflow.

Music 206: Mechanical Vibration 27 Music 206: Mechanical Vibration 28 Valve Driving Force Discretizing Valve Displacement

• The Laplace Transform of the valve displacement: 2 ms X(s)+ mgsX(s)+ KX(s) − Kx0 = F (s). • With these areas and the corresponding geometric couplet defined, the motion of the valve opening x(t) • The bilinear transform, defined by the substitution is governed by 1 − z−1 2 s −→ c where c = , d2x dx 1+ z−1 T m +2mγ +k(x−x )= σ p (S +S )+σ p S ,   dt2 dt 0 1 1 1 3 2 2 2 yields −1 −2 where γ is the damping coefficient, x0 the equilibrium X(z) 1+2z + z = −1 −2, position of the valve opening in the absence of flow, F (z)+ kx0 a0 + a1z + a2z K the valve stiffness, and m the reed mass. where • The couplet therefore, is very useful when evaluating 2 a0 = mc + mgc + k, the force driving a mode of the vibrating valve. 2 a1 = −2(mc − k), 2 a2 = mc − mgc + k.

• The corresponding difference equation is 1 x(n) = [Fk(n)+2Fk(n − 1) + Fk(n − 2)) − a0 a1x(n − 1) − a2x(n − 2)],

where Fk(n)= F (n)+ kx0.

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Volume Flow

 Volume Flow as a Function of Pressure Difference   400 threshold of oscillation pm Bore U pb 350 x    y 300 Lip Reed λ 250 Figure 12: The clarinet Reed. / s) 3 200

• The steady flow through a valve is determined based U (cm on input pressure p1 and the resulting output pressure 150 p2. 100 • The difference between these two pressure is denoted reed closed

∆p and is related to volume flow via the stationary 50 Bernoulli equation 0 0 10 20 30 40 50 60 70 80 90 100 2∆p p − p (hPa) U = A , 0 1 ρ s Figure 13: The reed table. where A is the cross section area of the air column (and dependent on the opening x).

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