PHY183-Lecture49

# PHY183-Lecture49 - Review Damped Oscillations Damping force...

This preview shows page 1. Sign up to view the full content.

This is the end of the preview. Sign up to access the rest of the document.

Unformatted text preview: Review: Damped Oscillations ! Damping force: Fd = !bv Physics for Scientists & Engineers 1 Spring Semester 2006 Lecture 49 ! d 2x dx m 2 = !b ! kx dt dt d 2 x b dx k + + x=0 dt 2 m dt m "# < " 0 or !"# t Small damping: x(t) = Ae !"# t b < 2 mk cos(" 't) + Be sin(" 't) k \$ b ' #& ) m % 2m ( 2 !" = April 21, 2006 Physics for Scientists&Engineers 1 1 April 21, 2006 b 2m ! 2 ! ' = ! 0 # ! "2 = !0 = k m 2 Physics for Scientists&Engineers 1 Small Damping d x b dx k + + x = 0; b < 2 mk dt 2 m dt m 2 Damping ! The equation of motion has different solutions, depending on the magnitude of the damping +C exp(!" # t) x(t) = C exp(!" # t)cos(" 't + \$ 0 ) Small damping: Large damping: b < 4mk b > 4mk b = 4mk !C exp(!" # t) Critical damping: x(t) = Ae April 21, 2006 !"# t cos(" 't) + Be !"# t sin(" 't) 3 April 21, 2006 Physics for Scientists&Engineers 1 4 Physics for Scientists&Engineers 1 Large Damping d 2 x b dx k + + x = 0; b > 4mk dt 2 m dt m Critical Damping d 2 x b dx k + + x = 0; b = 4mk dt 2 m dt m ! For b > 2 mk we have to try a different ansatz. The damping frequency !" is now larger than the frequency !0 of the un-damped oscillation ! No more oscillations; solutions are now two exponentials: ! Special case for b = 4mk can be obtained by a limiting procedure of either of the two previous ones ! New ansatz x(t ) = A exp("!1t ) + B exp("!2t ) 2 2 x(t) = A exp(!(" # + " #2 ! " 0 )t) + B exp(!(" # ! " #2 ! " 0 )t) x(t ) = A exp(#"! t ) + Bt exp(#"! t ) !" = b 2m ! The coefficients A and B can be determined from the initial conditions. ! Long-time behavior of this solution is governed by second term because it decays more slowly than the first term April 21, 2006 Physics for Scientists&Engineers 1 5 ! Again we find ! Coefficients A and B again found from initial conditions A = x0 , B = v0 + x0! " April 21, 2006 Physics for Scientists&Engineers 1 6 Comparison of 3 Cases ! Critical damping provides solution that returns to the origin (and stays there) quickest Over-damped Under-damped Critical damping Example, Critical Damping Question: ! A spring with a spring constant of k = 1.00 N/m, has a mass m = 1.00 kg attached to it, and moves in a medium with damping constant b = 2.00 kg/s. The mass is released from rest at a position of +5 cm from equilibrium. Where will it be after a time of 1.75 s? Answer: ! First find out which case applies: 2 mk = 2 (1 kg) ! (1 N/m) = 2 kg/s Same value as b => critical damping ! Damping frequency = ! " = b / 2m = 1.00 / s dx = v(t = 0) = v0 = 0 ! Initial conditions: x0 = x(t = 0) = +5 cm dt t = 0 ! Use solution for critical damping: x(t) = A exp(!" t) + Bt exp(!" t) # # 7 April 21, 2006 Physics for Scientists&Engineers 1 8 Example: shock absorber April 21, 2006 Physics for Scientists&Engineers 1 Example, Critical Damping (2) Answer (cont.): dx = !" # A exp(!" # t) + B(1 ! " # t)exp(!" # t) ! Take derivative: v = dt ! Use initial conditions (t = 0): x(0) = A exp(!" # 0) + B0 exp(!" # 0) = A = 5 cm Recall: Forced Harmonic Motion ! Example: someone on a swing undergoes ~harmonic pendulum motion; pushing someone in regular intervals allows that person to swing higher. ! Motion with periodic driving force of the kind v(0) = !" # A exp(!" # 0) + B(1 ! " # 0)exp(!" # 0) = !" # A + B = 0 ! Complete solution: \$ B = A" # Fdriving (t ) = Fd cos(!d t ) with Fd and !d constants. x(t) = (5 cm)(1 + ! " t)exp(#! " t) = (5 cm)(1 + (1.00/s)t)exp(#(1.00/s)t) ! Position after 1.75 s: x(1.75 s) = (5 cm)(1 + 1.75)exp(!1.75) = 2.39 cm April 21, 2006 Physics for Scientists&Engineers 1 9 April 21, 2006 Physics for Scientists&Engineers 1 10 Un-damped Forced Oscillation ! Start again with r r r Fk + Fdriving = ma Resonance ! Steady-state solution (after a reasonably long transition time) ! Differential equation (x-axis) x(t) = C cos(! d t) d2x !kx + Fd cos(" d t) = m 2 dt ! Reorder to get to standard form: ! Amplitude C depends on how far away from the intrinsic frequency !0 is the driving frequency !d is: C= d2x k F + x ! d cos(" d t) = 0 dt 2 m m Fd 2 m(! " ! d ) 2 0 ! When " d # " 0 , what happens? ! Resonance! 11 April 21, 2006 Physics for Scientists&Engineers 1 12 April 21, 2006 Physics for Scientists&Engineers 1 ! Driven Oscillation with Damping r r ! With damping F = !bv , the functional form of the solution is still Resonant Amplitude !" = 0.3 Hz, !" = 0.5 Hz, !" = 0.7 Hz x(t ) = C! cos("d t \$ #! ) Fd 2 2 2 m (" 0 # " d )2 + 4" d " !2 ! But now the amplitude is C! = ! Amplitude cannot be infinite, even if ! d = ! 0 There the amplitude has the value of Fd /(2m"0"! ) (approximately the maximum amplitude) April 21, 2006 Physics for Scientists&Engineers 1 13 Note: maximum amplitude reached slightly below !0 April 21, 2006 Physics for Scientists&Engineers 1 14 Phase between force and response ! Phase between driving force and oscillation 2 % " 2 ' "d & # ' arctan ( 0 ( 2" " ) ) 2 d ! + * Phase Space ! Typically plot displacement as a function of time (or velocity vs. time, or acceleration vs. time) ! Now plot velocity vs. displacement => phase space plot ! First: No damping ! Different amplitudes of the oscillation result in ellipses of different size ! Periodic motion => closed contours 15 April 21, 2006 Physics for Scientists&Engineers 1 16 \$! = ! In phase for !d << !0 ! for !d = !0 2 ! Out of phase by ! Out of phase by " for " d >> " 0 ! Video of unwanted resonance phenomenon at end of class April 21, 2006 Physics for Scientists&Engineers 1 ! ! Phase Space (2) ! Plot of damped motion: (k = 11 N/m, m = 1.8 kg, b = 0.5 kg/s, C = 5 cm, #0 = 1.6) Same plot in phase space Point attractor! April 21, 2006 Physics for Scientists&Engineers 1 17 ...
View Full Document

{[ snackBarMessage ]}

Ask a homework question - tutors are online