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Physics 42200 Waves & Oscillations Lecture 9 – French, Chapter 4 Spring 2014 Semester Matthew Jones Comments on the Assignment • Question 1 – Primarily concerned with the case when ≪ (weak damping). – In this case ≈ and the total energy changes much more slowly than the position, . • Question 2 – A graph will help a lot… ( ) Forced Oscillators and Resonance + + = • Natural oscillation frequency: = • Amplitude of steady-state oscillations: ⁄ = − cos = + • Phase difference: = tan$% − − 2 Resonance Phenomena • Change of variables: = • Natural oscillation frequency: = • = Amplitude of steady-state oscillations: = • Strong damping force Strong damping force & = − / − Phase difference: = tan $% 1/& − 2 = 1 + & 1− %/ 1 2& large small & Resonance Phenomena • Steady state amplitude: ( ⁄ ) Q=5 Q=4 Q=3 Q=2 Q=1 Peak is near ⁄ / ≈ 1. The peak occurs at exactly ⁄ = 1. Average Power • The rate at which the oscillator absorbs energy is: )*( ) = )*( ) /2 2 & − 1 1 + & & = 10 Full-Width-at-Half-Max: &=5 ./0 = &=3 &=1 / & = Examples 4 • Resonant RLC circuit: 2(3) 1 5 ℰ cos 89 7 +9 8 1 = :+ < 9 ; 8 The transformer does not play a role in the analysis of the circuit. It is just a convenient way to isolate the driving voltage source from the part of the circuit that oscillates. Resonant Circuit 89 1 = 7 + 9 : + < 9 8 = ℰ cos 8 ; • Differentiate once: 8 9 89 1 7 +: + 9 = −ℰ sin 8 8 ; • Redefine what we man by “ = 0”: 8 9 89 1 7 +: + 9 = ℰ cos 8 ; 8 • Change of variables: 8 9 89 ℰ + + 9 = cos 8 8 7 Resonant Circuit 8 9 89 ℰ + + 9 = cos 8 8 7 • Amplitude of steady state current oscillations: / =ℰ ; %/ 1 − + & • Voltage across the capacitor: 9 = 9 ?@ = ; • Amplitude of voltage oscillations measured across C: / =ℰ %/ 1 − + & Actual Data Voltage • If we know that 7 = 235A/, can we estimate : and ;? 7 = 235A/ Peak voltage is at BD EF = 290 /H D EF = 2C BD EF B = /2C [Mhz] Energy % • Energy stored on a capacitor is I = ; • The graph of the stored energy is proportional to the square of the voltage graph. • If we defined 2∆ as the FWHM on the graph of power vs frequency, then it will correspond to 1/ 2 of the peak voltage. Voltage Resonant Circuit 2∆B = 110 /H = 2C ∆B = 6.91 × 10M N $% 7 = 235A/ But B= /2C Peak position is [Mhz] D EF = = :/7 so we can find :: : = 7 = 6.91 × 10M N $% 235A/ = 162Ω % P@ − QR ; = 1.20nF ;= P % R YR STUVW X R Lifetime of Oscillations • Amplitude of a damped harmonic oscillator: = [ $Q=/ cos • Maximum potential energy: 1 \= ∝ [ $Q= 2 • After time = 1/ , the energy is reduced by the factor 1/[. • We call ^ = 1/ the “lifetime” of the oscillator. Other Resonant Systems Other Resonant Systems Other Resonant Systems Wire Bond Resonance • Wire bonds in a magnetic field: _ f ` Lorentz force is a = b c 8ℓ × e The tiny wire is like a spring. A periodic current produces the driving force. Wire Bond Resonance Resonance in Nuclear Physics • A proton accelerated through a potential difference gains kinetic energy g = [ : i Phys. Rev. 75, 246 (1949). Resonance in Nuclear Physics • In quantum mechanics, energy and frequency are proportional: I=ℏ • A given energy corresponds to a driving force with frequency . • When a nucleus resonates at this frequency, the proton energy is easily absorbed. Nuclear Resonance “Lifetime” is defined in terms of the width of the resonance. Resonance • Resonances are the main way we observe fundamental particles.