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Условие Part A: mechanical properties (4 т.)Part B: electromechanical properties (including piezoelectric effect) (6 т.)Отговори Решение

IPhO 2003, Теория — Задача 2. A Piezoelectric Crystal Resonator under an Alternating Voltage

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Theoretical Question 2 — A Piezoelectric Crystal Resonator under an Alternating Voltage · 10 т.

Условие

Consider a uniform rod of unstressed length $\ell$ and cross-sectional area $A$ (Figure 2a). Its length changes by $\Delta\ell$ when equal and opposite forces of magnitude $F$ are applied to its ends faces normally. The stress $T$ on the end faces is defined to be $F/A$ . The fractional change in its length, i.e., $\Delta\ell/\ell$ , is called the strain $S$ of the rod. In terms of stress and strain, Hooke’s law may be expressed as

T = Y S \quad \text{or} \quad \frac{F}{A} = Y\frac{\Delta\ell}{\ell} \qquad (1)

where $Y$ is called the Young’s modulus of the rod material. Note that a compressive stress $T$ corresponds to $F < 0$ and a decrease in length (i.e., $\Delta\ell < 0$ ). Such a stress is thus negative in value and is related to the pressure $p$ by $T = -p$ .

For a uniform rod of density $\rho$ , the speed of propagation of longitudinal waves (i.e., sound speed) along the rod is given by

u = \sqrt{Y/\rho} \qquad (2)

Rectangular rod drawn in perspective along the x axis, with z and y axes at its left end; forces F pull outward at both ends; the cross-section is labelled A; dimension lines mark the length ℓ and the extension Δℓ.

Figure 2a

The effect of damping and dissipation can be ignored in answering the following questions.

Part A: mechanical properties (4 т.)

A uniform rod of semi-infinite length, extending from $x = 0$ to $\infty$ (see Figure 2b), has a density $\rho$ . It is initially stationary and unstressed. A piston then steadily exerts a small pressure $p$ on its left face at $x = 0$ for a very short time $\Delta t$ , causing a pressure wave to propagate with speed $u$ to the right.

Semi-infinite rod from x = 0 to ∞: a piston pushes with pressure p on the left face, which moves with velocity v; a darker compressed zone on the left, where pressure p acts, and a lighter unstressed zone to the right.

Figure 2b

Two strips of rod: the upper, unstressed, marks the position x of a cross section; the lower, labelled wave motion, shows the same cross section displaced by ξ, with pressure p and velocity v arrows.

Figure 2c

(a) If the piston causes the rod’s left face to move at a constant velocity $v$ (Figure 2b), what are the strain $S$ and pressure $p$ at the left face during the time $\Delta t$ ? Answers must be given in terms of $\rho$ , $u$ , and $v$ only. [1,6 т.]

(b) Consider a longitudinal wave traveling along the $x$ direction in the rod. For a cross section at $x$ when the rod is unstressed (Figure 2c), let $\xi(x, t)$ be its displacement at time $t$ and assume

\xi(x, t) = \xi_0 \sin k(x - u t) \qquad (3)

where $\xi_0$ and $k$ are constants. Determine the corresponding velocity $v(x, t)$ , strain $S(x, t)$ , and pressure $p(x, t)$ as a function of $x$ and $t$ . [2,4 т.]

Part B: electromechanical properties (including piezoelectric effect) (6 т.)

Consider a quartz crystal slab of length $b$ , thickness $h$ , and width $w$ (Figure 2d). Its length and thickness are along the $x$ -axis and $z$ -axis. Electrodes are formed by thin metallic coatings at its top and bottom surfaces. Electrical leads that also serve as mounting support (Figure 2e) are soldered to the electrode’s centers, which may be assumed to be stationary for longitudinal oscillations along the $x$ direction.

Quartz slab of length b, thickness h and width w drawn in perspective with axes x, y, z; leads from the centres of the top and bottom electrodes connect through switch K to an alternating voltage source V(t) with − and + terminals.

Figure 2d

Side view of the grey quartz slab of thickness h with electrodes on its top and bottom faces; leads are attached at the centre, halfway along the length (b/2 and b/2 marked); axes z and x; labels quartz and electrodes.

Figure 2e

The quartz crystal under consideration has a density $\rho$ of $2.65\times10^{3}\ \mathrm{kg/m^3}$ and Young’s modulus $Y$ of $7.87\times10^{10}\ \mathrm{N/m^2}$ . The length $b$ of the slab is 1.00 cm and the width $w$ and height $h$ of the slab are such that $h \ll w$ and $w \ll b$ . With switch $K$ left open, we assume only longitudinal modes of standing wave oscillation in the $x$ direction are excited in the quartz slab.

For a standing wave of frequency $f = \omega/2\pi$ , the displacement $\xi(x, t)$ at time $t$ of a cross section of the slab with equilibrium position $x$ may be written as

\xi(x, t) = 2\xi_0\, g(x)\cos\omega t, \qquad (0 \le x \le b) \qquad (4a)

where $\xi_0$ is a positive constant and the spatial function $g(x)$ is of the form

g(x) = B_1 \sin k\left(x - \frac{b}{2}\right) + B_2 \cos k\left(x - \frac{b}{2}\right) \qquad (4b)

$g(x)$ has the maximum value of one and $k = \omega/u$ . Keep in mind that the centers of the electrodes are stationary and the left and right faces of the slab are free and must have zero stress (or pressure).

(c) Determine the values of $B_1$ and $B_2$ in Eq. (4b) for a longitudinal standing wave in the quartz slab. [1,2 т.]

(d) What are the two lowest frequencies at which longitudinal standing waves may be excited in the quartz slab? [1,2 т.]

The piezoelectric effect is a special property of a quartz crystal. Compression or dilatation of the crystal generates an electric voltage across the crystal, and conversely, an external voltage applied across the crystal causes the crystal to expand or contract depending on the polarity of the voltage. Therefore, mechanical and electrical oscillations can be coupled and made to resonate through a quartz crystal.

To account for the piezoelectric effect, let the surface charge densities on the upper and lower electrodes be $-\sigma$ and $+\sigma$ , respectively, when the quartz slab is under an electric field $E$ in the $z$ direction. Denote the slab’s strain and stress in the $x$ direction by $S$ and $T$ , respectively. Then the piezoelectric effect of the quartz crystal can be described by the following set of equations:

S = (1/Y)T + d_p E \qquad (5a)

\sigma = d_p T + \varepsilon_T E \qquad (5b)

where $1/Y = 1.27\times10^{-11}\ \mathrm{m^2/N}$ is the elastic compliance (i.e., inverse of Young’s modulus) at constant electric field and $\varepsilon_T = 4.06\times10^{-11}\ \mathrm{F/m}$ is the permittivity at constant stress, while $d_p = 2.25\times10^{-12}\ \mathrm{m/V}$ is the piezoelectric coefficient.

Let switch $K$ in Fig. 2d be closed. The alternating voltage $V(t) = V_m\cos\omega t$ now acts across the electrodes and a uniform electric field $E(t) = V(t)/h$ in the $z$ direction appears in the quartz slab. When a steady state is reached, a longitudinal standing wave oscillation of angular frequency $\omega$ appears in the slab in the $x$ direction.

With $E$ being uniform, the wavelength $\lambda$ and the frequency $f$ of a longitudinal standing wave in the slab are still related by $\lambda = u/f$ with $u$ given by Eq. (2). But, as Eq. (5a) shows, $T = YS$ is no longer valid, although the definitions of strain and stress remain unchanged and the end faces of the slab remain free with zero stress.

(e) Taking Eqs. (5a) and (5b) into account, the surface charge density $\sigma$ on the lower electrode as a function of $x$ and $t$ is of the form,

\sigma(x,t) = \left[D_1 \cos k\left(x - \frac{b}{2}\right) + D_2\right]\frac{V(t)}{h}

where $k = \omega/u$ . Find the expressions for $D_1$ and $D_2$ . [2,2 т.]

(f) The total surface charge $Q(t)$ on the lower electrode is related to $V(t)$ by

Q(t) = \left[1 + \alpha^2\left(\frac{2}{kb}\tan\frac{kb}{2} - 1\right)\right]C_0 V(t) \qquad (6)

Find the expression for $C_0$ and the expression and numerical value of $\alpha^2$ . [1,4 т.]

Отговори

Решение

Part A

(a) Refer to Figure A1. The left face of the rod moves a distance $v\Delta t$ while the pressure wave travels a distance $u\Delta t$ with $u = \sqrt{Y/\rho}$ . The strain at the left face is

S = \frac{\Delta\ell}{\ell} = \frac{-v\Delta t}{u\Delta t} = \frac{-v}{u} \qquad (A1a)

From Hooke’s law, the pressure at the left face is

p = -YS = Y\frac{v}{u} = \rho u v \qquad (A1b)

Three snapshots of the rod at t=0, Δt/2 and Δt: pressure p pushes the left face, a dark compressed zone grows to length uΔt (marked above by a dimension line and a dashed vertical line) while the left face has moved by vΔt (marked below).

Figure A1

(b) The velocity $v$ is related to the displacement $\xi$ as in a simple harmonic motion (or a uniform circular motion, as shown in Figure A2) of angular frequency $\omega = ku$ . Therefore, if $\xi(x,t) = \xi_0 \sin k(x - u t)$ , then

v(x,t) = -ku\xi_0 \cos k(x - u t). \qquad (A2)

The strain and pressure are related to velocity as in Problem (a). Hence,

S(x,t) = -v(x,t)/u = k\xi_0 \cos k(x - u t) \qquad (A3)

\begin{aligned} p(x,t) &= \rho u v(x,t) = -k\rho u^2 \xi_0 \cos k(x - u t) \\ &= -YS(x,t) = -kY\xi_0 \cos k(x - u t) \end{aligned} \qquad (A4)

Alternatively, the answers may be obtained by differentiations:

v(x,t) = \frac{\Delta\xi}{\Delta t} = -ku\xi_0 \cos k(x - u t),

S(x,t) = \frac{\Delta\xi}{\Delta x} = k\xi_0 \cos k(x - u t),

p(x,t) = -Y\frac{\Delta\xi}{\Delta x} = -kY\xi_0 \cos k(x - u t).

Reference circle of radius ξ0 with a vertical dashed x axis; the radius makes angle kx−ωt with the horizontal; its projection ξ on the vertical axis and the velocity projection v are drawn as downward arrows.

Figure A2

Part B

(c) Since the angular frequency $\omega$ and speed of propagation $u$ are given, the wavelength is given by $\lambda = 2\pi/k$ with $k = \omega/u$ . The spatial variation of the displacement $\xi$ is therefore described by

g(x) = B_1 \sin k\left(x - \frac{b}{2}\right) + B_2 \cos k\left(x - \frac{b}{2}\right) \qquad (B1)

Since the centers of the electrodes are assumed to be stationary, $g(b/2) = 0$ . This leads to $B_2 = 0$ . Given that the maximum of $g(x)$ is 1, we have $B_1 = \pm 1$ and

g(x) = \pm \sin \frac{\omega}{u}\left(x - \frac{b}{2}\right) \qquad (B2)

Thus, the displacement is

\xi(x,t) = \pm 2\xi_0 \sin \frac{\omega}{u}\left(x - \frac{b}{2}\right)\cos\omega t \qquad (B3)

(d) Since the pressure $p$ (or stress $T$ ) must vanish at the end faces of the quartz slab (i.e., $x = 0$ and $x = b$ ), the answer to this problem can be obtained, by analogy, from the resonant frequencies of sound waves in an open pipe of length $b$ . However, given that the centers of the electrodes are stationary, all even harmonics of the fundamental tone must be excluded because they have antinodes, rather than nodes, of displacement at the bisection plane of the slab.

Since the fundamental tone has a wavelength $\lambda = 2b$ , the fundamental frequency is given by $f_1 = u/(2b)$ . The speed of propagation $u$ is given by

u = \sqrt{\frac{Y}{\rho}} = \sqrt{\frac{7.87\times10^{10}}{2.65\times10^{3}}} = 5.45\times10^{3}\ \mathrm{m/s} \qquad (B4)

and, given that $b = 1.00\times10^{-2}\ \mathrm{m}$ , the two lowest standing wave frequencies are

f_1 = \frac{u}{2b} = 273\ (\mathrm{kHz}), \quad f_3 = 3f_1 = \frac{3u}{2b} = 818\ (\mathrm{kHz}) \qquad (B5)

[Alternative solution to Problems (c) and (d)]:

A longitudinal standing wave in the quartz slab has a displacement node at $x = b/2$ . It may be regarded as consisting of two waves traveling in opposite directions. Thus, its displacement and velocity must have the following form

\begin{aligned} \xi(x,t) &= \xi_m\left[\sin k\left(x - \frac{b}{2} - ut\right) + \sin k\left(x - \frac{b}{2} + ut\right)\right] \\ &= 2\xi_m \sin k\left(x - \frac{b}{2}\right)\cos\omega t \end{aligned} \qquad (B6)

\begin{aligned} v(x,t) &= -ku\xi_m\left[\cos k\left(x - \frac{b}{2} - ut\right) - \cos k\left(x - \frac{b}{2} + ut\right)\right] \\ &= -2\omega\xi_m \sin k\left(x - \frac{b}{2}\right)\sin\omega t \end{aligned} \qquad (B7)

where $\omega = ku$ and the first and second factors in the square brackets represent waves traveling along the $+x$ and $-x$ directions, respectively. Note that Eq. (B6) is identical to Eq. (B3) if we set $\xi_m = \pm\xi_0$ .

For a wave traveling along the $-x$ direction, the velocity $v$ must be replaced by $-v$ in Eqs. (A1a) and (A1b) so that we have

S = \frac{-v}{u} \quad \text{and} \quad p = \rho u v \qquad (\text{waves traveling along } +x) \qquad (B8)

S = \frac{v}{u} \quad \text{and} \quad p = -\rho u v \qquad (\text{waves traveling along } -x) \qquad (B9)

As in Problem (b), the strain and pressure are therefore given by

\begin{aligned} S(x,t) &= -k\xi_m\left[-\cos k\left(x - \frac{b}{2} - ut\right) - \cos k\left(x - \frac{b}{2} + ut\right)\right] \\ &= 2k\xi_m \cos k\left(x - \frac{b}{2}\right)\cos\omega t \end{aligned} \qquad (B10)

\begin{aligned} p(x,t) &= -\rho u\omega\xi_m\left[\cos k\left(x - \frac{b}{2} - ut\right) + \cos k\left(x - \frac{b}{2} + ut\right)\right] \\ &= -2\rho u\omega\xi_m \cos k\left(x - \frac{b}{2}\right)\cos\omega t \end{aligned} \qquad (B11)

Note that $v$ , $S$ , and $p$ may also be obtained by differentiating $\xi$ as in Problem (b).

The stress $T$ or pressure $p$ must be zero at both ends ( $x = 0$ and $x = b$ ) of the slab at all times because they are free. From Eq. (B11), this is possible only if $\cos(kb/2) = 0$ or

kb = \frac{\omega}{u}b = \frac{2\pi f}{\lambda f}b = n\pi, \qquad n = 1, 3, 5, \cdots \qquad (B12)

In terms of wavelength $\lambda$ , Eq. (B12) may be written as

\lambda = \frac{2b}{n}, \qquad n = 1, 3, 5, \cdots. \qquad (B13)

The frequency is given by

f = \frac{u}{\lambda} = \frac{nu}{2b} = \frac{n}{2b}\sqrt{\frac{Y}{\rho}}, \qquad n = 1, 3, 5, \cdots. \qquad (B14)

This is identical with the results given in Eqs. (B4) and (B5).

(e) From Eqs. (5a) and (5b) in the Question, the piezoelectric effect leads to the equations

T = Y(S - d_p E) \qquad (B15)

\sigma = Yd_p S + \varepsilon_T\left(1 - Y\frac{d_p^2}{\varepsilon_T}\right)E \qquad (B16)

Because $x = b/2$ must be a node of displacement for any longitudinal standing wave in the slab, the displacement $\xi$ and strain $S$ must have the form given in Eqs. (B6) and (B10), i.e., with $\omega = ku$ ,

\xi(x,t) = \xi_m \sin k\left(x - \frac{b}{2}\right)\cos(\omega t + \phi) \qquad (B17)

S(x,t) = k\xi_m \cos k\left(x - \frac{b}{2}\right)\cos(\omega t + \phi) \qquad (B18)

where a phase constant $\phi$ is now included in the time-dependent factors.

By assumption, the electric field $E$ between the electrodes is uniform and depends only on time:

E(x,t) = \frac{V(t)}{h} = \frac{V_m\cos\omega t}{h} \qquad (B19)

Substituting Eqs. (B18) and (B19) into Eq. (B15), we have

T = Y\left[k\xi_m \cos k\left(x - \frac{b}{2}\right)\cos(\omega t + \phi) - \frac{d_p}{h}V_m\cos\omega t\right] \qquad (B20)

The stress $T$ must be zero at both ends ( $x = 0$ and $x = b$ ) of the slab at all times because they are free. This is possible only if $\phi = 0$ and

k\xi_m\cos\frac{kb}{2} = d_p\frac{V_m}{h} \qquad (B21)

Since $\phi = 0$ , Eqs. (B16), (B18), and (B19) imply that the surface charge density must have the same dependence on time $t$ and may be expressed as

\sigma(x,t) = \sigma(x)\cos\omega t \qquad (B22)

with the dependence on $x$ given by

\begin{aligned} \sigma(x) &= Yd_p k\xi_m \cos k\left(x - \frac{b}{2}\right) + \varepsilon_T\left(1 - Y\frac{d_p^2}{\varepsilon_T}\right)\frac{V_m}{h} \\ &= \left[Y\frac{d_p^2}{\cos\frac{kb}{2}}\cos k\left(x - \frac{b}{2}\right) + \varepsilon_T\left(1 - Y\frac{d_p^2}{\varepsilon_T}\right)\right]\frac{V_m}{h} \end{aligned} \qquad (B23)

(f) At time $t$ , the total surface charge $Q(t)$ on the lower electrode is obtained by integrating $\sigma(x,t)$ in Eq. (B22) over the surface of the electrode. The result is

\begin{aligned} \frac{Q(t)}{V(t)} &= \frac{1}{V(t)}\int_0^b \sigma(x,t)\,w\,dx = \frac{1}{V_m}\int_0^b \sigma(x)\,w\,dx \\ &= \frac{w}{h}\int_0^b \Big[Y\frac{d_p^2}{\cos\frac{kb}{2}}\cos k\Big(x - \frac{b}{2}\Big) + \varepsilon_T\Big(1 - Y\frac{d_p^2}{\varepsilon_T}\Big)\Big]dx \\ &= \left(\varepsilon_T\frac{bw}{h}\right)\left[Y\frac{d_p^2}{\varepsilon_T}\left(\frac{2}{kb}\tan\frac{kb}{2}\right) + \left(1 - Y\frac{d_p^2}{\varepsilon_T}\right)\right] \\ &= C_0\left[\alpha^2\left(\frac{2}{kb}\tan\frac{kb}{2}\right) + (1 - \alpha^2)\right] \end{aligned} \qquad (B24)

where

C_0 = \varepsilon_T\frac{bw}{h}, \qquad \alpha^2 = Y\frac{d_p^2}{\varepsilon_T} = \frac{(2.25)^2\times10^{-2}}{1.27\times4.06} = 9.82\times10^{-3} \qquad (B25)

(The constant $\alpha$ is called the electromechanical coupling coefficient.)

Note: The result $C_0 = \varepsilon_T bw/h$ can readily be seen by considering the static limit $k = 0$ of Eq. (5) in the Question. Since $\tan x \approx x$ when $x \ll 1$ , we have

\lim_{k\to 0} Q(t)/V(t) \approx C_0[\alpha^2 + (1 - \alpha^2)] = C_0 \qquad (B26)

Evidently, the constant $C_0$ is the capacitance of the parallel-plate capacitor formed by the electrodes (of area $bw$ ) with the quartz slab (of thickness $h$ and permittivity $\varepsilon_T$ ) serving as the dielectric medium. It is therefore given by $\varepsilon_T bw/h$ .

Marking Scheme

Theoretical Question 2

A Piezoelectric Crystal Resonator under an Alternating Voltage

Total Scores	Sub Scores	Marking Scheme for Answers to the Problem
Part A	(a)	The strain $S$ and pressure $p$ on the left face.
		➢ 0.4 for $\lvert\Delta\ell\rvert = v\Delta t$ and $\ell = u\Delta t$ .
4.0 pts.	1.6	➢ 0.4 for $S = -v/u$ . (0.1 for sign)
		➢ 0.4 for relating $p$ to $S$ as $p = -YS$ . (0.1 for sign)
		➢ 0.4 for $p = \rho u v$ . (0.1 for sign)
	(b)	The velocity $v(x, t)$ , strain $S(x, t)$ , and pressure $p(x, t)$ .
	2.4	➢ 0.3×3 sinusoidal variation with correct phase constant. (0.2 for phase constant.)
		➢ 0.3×3 for amplitude.
		➢ 0.2×3 for dependence on $x$ and $t$ as ( $kx$ - $ku\,t$ ).
Part B	(c)	The function $g(x)$ for a standing wave of angular frequency $\omega$ .
		➢ 0.4 for $g(b/2) = 0$ .
6.0 pts	1.2	➢ 0.3+0.1 for $B_1 = \pm 1$ (0.1 for both signs)
		➢ 0.4 for $B_2 = 0$
	(d)	The two lowest standing wave frequencies.
		➢ 0.2 for wavelength of fundamental tone $\lambda = 2b$ .
	1.2	➢ 0.2 for excluding even harmonics.
		➢ (0.3+0.1) for $f_1 = u/2b = 273$ kHz. (0.1 for value)
		➢ (0.3+0.1) for $f_3 = 3u/2b = 818$ kHz. (0.1 for value)
	(e)	The surface charge density $\sigma$ as a function of $x$ and $t$ .
		➢ 0.1×2 for $\xi$ and $S$ , each a separable function of $x$ and $t$ .
	2.2	➢ 0.1×2 for $\xi$ and $S$ , each depends on time as $\cos\omega t$ with $\phi = 0$ .
		➢ 0.3 for spatial part $\xi(x) = \xi_m \sin k(x - b/2)$ .
		➢ 0.3 for spatial part $S(x) = k\xi_m \cos k(x - b/2)$ .
		➢ 0.3 for $T(x) = [k\xi_m \cos k(x - b/2) - d_p V_m/h]Y$ .
		➢ 0.3 for $k\xi_m\cos(kb/2) = d_p V_m/h$ .
		➢ 0.6 for $D_1$ (0.3) and $D_2$ (0.3) in $\sigma(x)$ .
	(f)	The constants $C_0$ and $\alpha^2$ .
	1.4	➢ 0.2 for relation between $\sigma$ and $Q$ as $Q(t) = (\int_0^b \sigma(x)w\,dx)\cos\omega t$ .
		➢ 0.3 for noting $Q(t)/V(t) \approx C_0$ as $k \to 0$ .
		➢ 0.4 for $C_0 = \varepsilon_T bw/h$ .
		➢ 0.4+0.1 for $\alpha^2 = Yd_p^2/\varepsilon_T = 9.82\times10^{-3}$ . (0.1 for value)

[Answer Sheet] Theoretical Question 2

A Piezoelectric Crystal Resonator under an Alternating Voltage

Wherever requested, give each answer as analytical expressions followed by numerical values and units. For example: area of a circle $A = \pi r^2 = 1.23\ \mathrm{m^2}$ .

(a) The strain $S$ and pressure $p$ at the left face are (in terms of $\rho$ , $u$ , and $v$ )

$S =$
$p =$

(b) The velocity $v(x, t)$ , strain $S(x, t)$ , and pressure $p(x, t)$ are

$v(x, t) =$
$S(x, t) =$
$p(x, t) =$

$B_1 =$
$B_2 =$

(d) The lowest two frequencies of standing waves are (expression and value)

The Lowest
The Second Lowest

(e) The expressions of $D_1$ and $D_2$ are

$D_1 =$
$D_2 =$

(f) The constants $\alpha^2$ (expression and value) and $C_0$ are (expression only)

$\alpha^2 =$
$C_0 =$

Оригинал в Архива: 2.pdf · официални решения: 2_sol.pdf · допълнителни данни: IPhO_2003_Theoretical Question 2.pdf