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

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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 ...
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