Physics 401, Spring 2016 Eugene V. Colla

Transcription

1 Physics 41, Spring 16 Eugene V. Colla.8 (rad) time (s)

2 1.Driven torsional oscillator. Equations.Setup. Kinematics 3.Resonance 4.Beats 5.Nonlinear effects 6.Comments 3/7/16

3 3/7/16 3

4 Tacoma (WA) Narrows Bridge Disaster 3/7/16 4

5 Tacoma (WA) Narrows Bridge, 194 3/7/16 5

6 Tacoma (WA) Narrows Bridge, 194 3/7/16 6

7 Tacoma (WA) Narrows Bridge, 194 3/7/16 7

8 Egyptian Bridge disaster. January195, St. Petersburg, Russia. 3/7/16 8

9 Egyptian Bridge disaster. January195, St. Petersburg, Russia. 3/7/16 9

10 Egyptian Bridge disaster. January195, St. Petersburg, Russia. 3/7/16 1

11 Dancing Bridge in Volgograd (Russia) (record from st long). May miles 3/7/16 11

12 In autumn 11, 1 semi-active tuned mass dampers were installed in the bridge. Each one consists of a mass 5, kg (11,5 lb), a set of compression springs and a magnethoreological damper. 3/7/16 1

13 3/7/16 13

14 The goals: (i) analyze the response of the damped driven harmonic oscillator to a sinusoidal drive. (ii) transient response and (iii) steady-state solution. wire M L 1 L R disk motor Angular displacement: θ cos(ωt); torque: Kλθ cos(ωt) L 1 λ = L 1 + L I ሷ θ + Kθ + R ሶ θ = m = Kθ cos(ωt) I is momentum of inertia, [kgm ] R is a damping constant [N m s]. K is the total spring constant [Nm] Viscous damping Torque by motor 3/7/16 14

15 Motor Pendulum 3/7/16 15

16 I ሷ θ + Kθ + R ሶ θ = m = Kθ cos(ωt) Solutions: sum of (1) Transient solution + () steady solution due to torque m (1) Transient solution (1 st week experiment) Iθ+ Rθ+ Kθ = -at θ t = A e cos ω t- a = R I ω = o K I a 1 o 1 The homogeneous equation of motion Transient solution Attenuation constant Natural (angular) frequency Damped (angular) frequency 3/7/16 16

17 at t ( t) A e cos( t ) 1 1 a Transient solution Once this response dies away in time the system response only on the frequency of drive Initially the system responds on the characteristic frequency 1 ( ) Re ( ) i t ss t e So the steady-state solution must have the similar time dependence as the drive I ሷ θ + Kθ + R ሶ θ = m = Kθ cos(ωt) Substituting ss (t) in equation of motion we will find the equations for ( ) 4 a e i( ) and 1 a ( ) tan

18 I ሷ θ + Kθ + R ሶ θ = m = Kθ cos(ωt) () steady solution t B cos t s B tan R R a I I o o o o Steady state solution Amplitude function Phase function Damping constant 3/7/16 18

19 time domain form for steady-state solution will be ss( t) cos( t ( )) 4 a Phase Amplitude B() General solution for equation of motion consist of the sum of sum of two components: (t) = t (t) + ss (t) at ( t) ( t) ( t) Ae cos( t ) Bcos( t ( )) t ss 1 Coefficients A and could be determined from initial conditions

20 Fitting function: (rad) 1 f =.5Hz (fitting) ( f ) A f f f f =pf; =a To create a new fitting function go Tools Fitting Function Builder or press F8 Model Equation Reduced Chi-Sqr.1 1 f d (Hz) Resonance1 (User) Adj. R-Square y=a*f^/sqrt((f^-x^)^+x^*gamma^) 3.E-4 Value Standard Error pend A pend f e-4 pend gamma E-4 Count Model Equation Reduced Chi-S qr Regular Residual of Sheet1 pend Gauss Fit Counts Gauss y=y + (A/(w*sqrt(PI/)))*exp(-*((x-xc)/w)^) Adj. R-Square Value Standard Error Counts y Counts xc 6.543E Counts w Counts A Counts sigma.1199 Counts FWHM.84 Counts Height Regular Residual of Sheet1 pend

21 (rad) 3 1 Phase.1 1 f d (Hz) Scanning the driving frequency we can measure the amplitude of the pendulum oscillating and the phase shift Both parameters Amplitude and phase can be defined by DAQ program or using Origin

22 Amplitude () t ss 4 a At resonance = ss() t Q a Combination of high initial amplitude, and high quality Q or low damping factor a could be result of the destruction of the mechanical system 3/7/16

23 For correct representation of the resonance curve take care about choosing of the step size in frequency. 3/7/16 3

24 There are two parameters used to measure the rate at which the oscillations of a system are damped out. One parameter is the logarithmic decrement d, and the other is the quality factor, Q. d, is defined by 6 (t max ) (t max +T 1 ) 8.49 d ln (rad) time (s) Q ~ 1.8

25 It can be shown that Q can f 1 =.496Hz be calculated as 1 / or Amplitude 1 f=.66hz f 1 /f. is bandwidth of the resonance curve on the θ half power level or max for amplitude graph f d (Hz) Here Q~7.9

26 Consider sum of two harmonic signals of frequencies 1 and y 1 =Asin( 1 t+ 1 ); y =Bsin( t+ ) In case A=B y=y 1 +y =Asin ω 1+ω β 1 = φ 1+φ ; β = φ 1 φ If 1 ω 1+ω = and y= Acos Wt + β sin t + β 1 ω 1 ω t + β 1 =W cos ω 1 ω t + β ; 1 =.78 =.94 1 y time (s) f (Hz)

27 More general case A B 1 and y 1 =Asin( t); y =Bsin(( +)t) y=y 1 +y =Csin( + )t where C = A + B + ABcos(at) β = tan 1 B sin(αt) A + B cos αt if A + Bcos( αt) + ቊ π if A + Bcos αt < A + B + AB A + B AB 1 1 = = 1 y time (s).8.3 f (Hz)

28 pend (rad) Amplitude 6 4 Two peaks corresponding and t (sec) Time domain trace Frequency (Hz) Beating spectrum Use Origin to analyze the frequency spectrum!

29 at ( t) ( t) ( t) Ae cos( t ) Bcos( t ( )) t ss 1 t (t) 4 Beats dying in time. How fast it depends on damping. When you will work on resonance data wait until you will see the steady state oscillations. (rad) -4 4 time (s) 4 (rad) (rad) time(s) time(s)

30 at ( t) ( t) ( t) Ae cos( t ) Bcos( t ( )) t ss (rad) (rad) time (s) time (s) t (t) This can be seen well from envelope plot Origin 8.6: Analysis Signal Processing Envelope

31 at ( t) ( t) ( t) Ae cos( t ) B cos( t ( )) C t ss 1 First let we apply FFT to find 1 and Result: =3.14rad -1 and =.898 rad -1

32 t ss t t ( t) ( t) ( t) Ae cos( t ) Bcos( t ( )) C 1 8 fitting parameters From fitting A.651 t B C Result from FFT: =3.14rad -1 and =.898 rad -1

33 FFT Compare with original pendulum spectrum Pendulum Possible origin of extra peaks: (i) Nonlinear behavior of pendulum (ii) Not a single frequency driving force provided by motor (iii) Not ideal fitting function Residuals

34 at ( t) ( t) ( t) Ae cos( t ) Bcos( t ( )) t ss 1 4 t (t) (rad) We also can analyze the decrease of the amplitude of the 1 component by analyzing the spectrum as a function of time -4 4 time (s) motor.4 First 55 sec 1 s1 1 R (V) First 55 sec Last 55 sec Origin 9.: Analysis Signal Processing FFT f (Hz) Last 55 sec

35 .1 C VC. -.1 L 5 time ( s) 1 5 V (V) V(t) VC (V) R -5-1 : Graph 5 time ( s) 1 1

36 Find peaks.1 R V(t) C L V C VC (V) time (s) Envelope FFT.1 31kHz 1 34kHz VC (V). Magnitude time (s) f (khz)

37 .4 d~ /3 1-1 f =.4891Hz f d =.163 Hz f d =.163 f -f d f (rad). Amplitude f +f d time (s) Frequency (Hz)

38 In the case of driving frequency f d =f1/n where N is integer we can observe more complicated motion of the pendulum f d f =.4891Hz f 1-1 f d =.44 Hz (rad).3. d~ Amplitude f d +f time (s) Frequency (Hz)

39 1 7 Magnitude Frequency (Hz) Detailed analyzes* shows that even if = sin ωt the driving torque contains several harmonics of *P. Debevec (UIUC, Department of Physics)