Theoretical_Exam_Solution_2_English

# Theoretical_Exam_Solution_2_English - IPHO 2003 Theoretical...

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Unformatted text preview: IPHO 2003 Theoretical Question 2 ( Solution ) (2003/07/17) p.1/5 Solution- Theoretical Question 2 A Piezoelectric Crystal Resonator under an Alternating Voltage Part A (a) Refer to Figure A1. The left face of the rod moves a distance v ∆ t while the pressure wave travels a distance u ∆ t with ρ / Y u = . The strain at the left face is u v t u t v S- =- = = ∆ ∆ ∆ (A1a)* 1 From Hooke’s law, the pressure at the left face is uv u v Y YS p ρ = =- = (A1b)* (b) The velocity v is related to the displacement ξ as in a simple harmonic motion (or a uniform circular motion, as shown in Figure A2) of angular frequency ku = ϖ . Therefore, if ) ( sin ) , ( t u x k t x- = ξ ξ , then ) ( cos ) , ( t u x k ku t x v-- = ξ . (A2)* The strain and pressure are related to velocity as in Problem (a). Hence, ) ( cos / ) , ( ) , ( t u x k k u t x v t x S- =- = ξ (A3)* ) ( cos ) , ( ) ( cos ) , ( ) , ( 2 t u x k kY t x YS t u x k u k t x uv t x p-- =- =-- = = ξ ξ ρ ρ (A4)*------------------------------------------------------------------------------------------------- Alternatively, the answers may be obtained by differentiations: ) ( cos ) , ( t u x k ku t t x v-- = = ξ ∆ ξ ∆ , ) ( cos ) , ( t u x k k x t x S- = = ξ ∆ ξ ∆ , 1 An equations marked with an asterisk contains answer to the problem. p u ∆ t t=0 ∆ t/2 p p Figure A1 v ∆ t ∆ t p p ξ kx-ϖ t v x ξ Figure A2 IPHO 2003 Theoretical Question 2 ( Solution ) (2003/07/17) p.2/5 ) ( cos ) , ( t u x k kY x Y t x p-- =- = ξ ∆ ξ ∆ .------------------------------------------------------------------------- Part B (c) Since the angular frequency ϖ and speed of propagation u are given, the wavelength is given by λ = 2 π / k with k = ϖ / u . The spatial variation of the displacement ξ is therefore described by...
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Theoretical_Exam_Solution_2_English - IPHO 2003 Theoretical...

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