Ch03 9
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1. Forced vibrations with damping are modeled by an equation of the form mü + γu̇ + ku = F0cos(ωt), where m is mass, γ is damping coefficient, k is spring constant, F0 and ω are the force amplitude and frequency. 2. The general solution has the form u(t) = uC(t) + U(t), where uC(t) is the transient solution and U(t) is the steady-state or forced response. 3. The amplitude R of the steady-state forced response depends on the driving frequency ω, with a maximum value of Rmax occurring at the resonant frequency ωmax, which is
![Undamped Equation: Beats (3 of 3)
Using the results of the previous slide, it follows that
When |ω0- ω| ≅ 0, ω0+ ω is much larger than ω0 - ω, and
sin[(ω0 + ω)t/2] oscillates more rapidly than sin[(ω0- ω)t/2].
Thus motion is a rapid oscillation with frequency (ω0+ ω)/2,
but with slowly varying sinusoidal amplitude given by
This phenomena is called a beat.
Beats occur with two tuning forks of
nearly equal frequency.
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Ch03 9
- 1. Ch 3.9: Forced Vibrations We continue the discussion of the last section, and now consider the presence of a periodic external force: tFtuktutum ωγ cos)()()( 0=+′+′′
- 2. Forced Vibrations with Damping Consider the equation below for damped motion and external forcing funcion F0cosωt. The general solution of this equation has the form where the general solution of the homogeneous equation is and the particular solution of the nonhomogeneous equation is tFtkututum ωγ cos)()()( 0=+′+′′ ( ) ( ) )()(sincos)()()( 2211 tUtutBtAtuctuctu C +=+++= ωω )()()( 2211 tuctuctuC += ( ) ( )tBtAtU ωω sincos)( +=
- 3. Homogeneous Solution The homogeneous solutions u1 and u2 depend on the roots r1 and r2 of the characteristic equation: Since m, γ, and k are are all positive constants, it follows that r1 and r2 are either real and negative, or complex conjugates with negative real part. In the first case, while in the second case Thus in either case, m mk rkrrmr 2 4 0 2 2 −±− =⇒=++ γγ γ 0)(lim = ∞→ tuC t ( ) ,0lim)(lim 21 21 =+= ∞→∞→ trtr t C t ecectu ( ) .0sincoslim)(lim 21 =+= ∞→∞→ tectectu tt t C t µµ λλ
- 4. Transient and Steady-State Solutions Thus for the following equation and its general solution, we have Thus uC(t) is called the transient solution. Note however that is a steady oscillation with same frequency as forcing function. For this reason, U(t) is called the steady-state solution, or forced response. ( ) ( )tBtAtU ωω sincos)( += ( ) 0)()(lim)(lim 2211 =+= ∞→∞→ tuctuctu t C t ( ) ( ),sincos)()()( cos)()()( )()( 2211 0 tUtu tBtAtuctuctu tFtkututum C ωω ωγ +++= =+′+′′
- 5. Transient Solution and Initial Conditions For the following equation and its general solution, the transient solution uC(t) enables us to satisfy whatever initial conditions might be imposed. With increasing time, the energy put into system by initial displacement and velocity is dissipated through damping force. The motion then becomes the response U(t) of the system to the external force F0cosωt. Without damping, the effect of the initial conditions would persist for all time. ( ) ( ) )()( 2211 0 sincos)()()( cos)()()( tUtu tBtAtuctuctu tFtkututum C ωω ωγ +++= =+′+′′
- 6. Rewriting Forced Response Using trigonometric identities, it can be shown that can be rewritten as It can also be shown that where ( )δω −= tRtU cos)( ( ) ( )tBtAtU ωω sincos)( += 22222 0 222222 0 2 22 0 22222 0 2 0 )( sin, )( )( cos , )( ωγωω ωγ δ ωγωω ωω δ ωγωω +− = +− − = +− = mm m m F R mk /2 0 =ω
- 7. Amplitude Analysis of Forced Response The amplitude R of the steady state solution depends on the driving frequency ω. For low-frequency excitation we have where we recall (ω0)2 = k /m. Note that F0 /k is the static displacement of the spring produced by force F0. For high frequency excitation, , )( 22222 0 2 0 ωγωω +− = m F R k F m F m F R 0 2 0 0 22222 0 2 0 00 )( limlim == +− = →→ ωωγωωωω 0 )( limlim 22222 0 2 0 = +− = ∞→∞→ ωγωωωω m F R
- 8. Maximum Amplitude of Forced Response Thus At an intermediate value of ω, the amplitude R may have a maximum value. To find this frequency ω, differentiate R and set the result equal to zero. Solving for ωmax, we obtain where (ω0)2 = k /m. Note ωmax < ω0, and ωmax is close to ω0 for small γ. The maximum value of R is )4(1 2 0 0 max mk F R γγω − = 0lim,lim 0 0 == ∞→→ RkFR ωω −=−= mkm 2 1 2 2 2 02 2 2 0 2 max γ ω γ ωω
- 9. Maximum Amplitude for Imaginary ωmax We have and where the last expression is an approximation for small γ. If γ 2 /(mk) > 2, then ωmax is imaginary. In this case, Rmax= F0 /k, which occurs at ω = 0, and R is a monotone decreasing function of ω. Recall from Section 3.8 that critical damping occurs when γ 2 /(mk) = 4. +≅ − = mk F mk F R 8 1 )4(1 2 0 0 2 0 0 max γ γωγγω −= mk2 1 2 2 0 2 max γ ωω
- 10. Resonance From the expression we see that Rmax≅ F0 /(γ ω0) for small γ. Thus for lightly damped systems, the amplitude R of the forced response is large for ω near ω0, since ωmax ≅ ω0 for small γ. This is true even for relatively small external forces, and the smaller the γ the greater the effect. This phenomena is known as resonance. Resonance can be either good or bad, depending on circumstances; for example, when building bridges or designing seismographs. +≅ − = mk F mk F R 8 1 )4(1 2 0 0 2 0 0 max γ γωγγω
- 11. Graphical Analysis of Quantities To get a better understanding of the quantities we have been examining, we graph the ratios R/(F0/k) vs. ω/ω0 for several values of Γ = γ 2 /(mk), as shown below. Note that the peaks tend to get higher as damping decreases. As damping decreases to zero, the values of R/(F0/k) become asymptotic to ω = ω0. Also, if γ 2 /(mk) > 2, then Rmax= F0 /k, which occurs at ω = 0.
- 12. Analysis of Phase Angle Recall that the phase angle δ given in the forced response is characterized by the equations If ω ≅ 0, then cosδ ≅ 1, sinδ ≅ 0, and hence δ ≅ 0. Thus the response is nearly in phase with the excitation. If ω = ω0, then cosδ = 0, sinδ = 1, and hence δ ≅ π /2. Thus response lags behind excitation by nearly π /2 radians. If ω large, then cosδ ≅ -1, sinδ = 0, and hence δ ≅ π . Thus response lags behind excitation by nearly π radians, and hence they are nearly out of phase with each other. 22222 0 222222 0 2 22 0 )( sin, )( )( cos ωγωω ωγ δ ωγωω ωω δ +− = +− − = mm m ( )δω −= tRtU cos)(
- 13. Example 1: Forced Vibrations with Damping (1 of 4) Consider the initial value problem Then ω0 = 1, F0 = 3, and Γ = γ 2 /(mk) = 1/64 = 0.015625. The unforced motion of this system was discussed in Ch 3.8, with the graph of the solution given below, along with the graph of the ratios R/(F0/k) vs. ω/ω0 for different values of Γ. 0)0(,2)0(,2cos3)()(125.0)( =′==+′+′′ uuttututu
- 14. Example 1: Forced Vibrations with Damping (2 of 4) Recall that ω0 = 1, F0 = 3, and Γ = γ 2 /(mk) = 1/64 = 0.015625. The solution for the low frequency case ω = 0.3 is graphed below, along with the forcing function. After the transient response is substantially damped out, the steady-state response is essentially in phase with excitation, and response amplitude is larger than static displacement. Specifically, R ≅ 3.2939 > F0/k = 3, and δ ≅ 0.041185.
- 15. Example 1: Forced Vibrations with Damping (3 of 4) Recall that ω0 = 1, F0 = 3, and Γ = γ 2 /(mk) = 1/64 = 0.015625. The solution for the resonant case ω = 1 is graphed below, along with the forcing function. The steady-state response amplitude is eight times the static displacement, and the response lags excitation by π /2 radians, as predicted. Specifically, R = 24 > F0/k = 3, and δ = π /2.
- 16. Example 1: Forced Vibrations with Damping (4 of 4) Recall that ω0 = 1, F0 = 3, and Γ = γ 2 /(mk) = 1/64 = 0.015625. The solution for the relatively high frequency case ω = 2 is graphed below, along with the forcing function. The steady-state response is out of phase with excitation, and response amplitude is about one third the static displacement. Specifically, R ≅ 0.99655 ≅ F0/k = 3, and δ ≅ 3.0585 ≅ π.
- 17. Undamped Equation: General Solution for the Case ω0 ≠ ω Suppose there is no damping term. Then our equation is Assuming ω0 ≠ ω, then the method of undetermined coefficients can be use to show that the general solution is tFtkutum ωcos)()( 0=+′′ t m F tctctu ω ωω ωω cos )( sincos)( 22 0 0 0201 − ++=
- 18. Undamped Equation: Mass Initially at Rest (1 of 3) If the mass is initially at rest, then the corresponding initial value problem is Recall that the general solution to the differential equation is Using the initial conditions to solve for c1and c2, we obtain Hence 0)0(,0)0(,cos)()( 0 =′==+′′ uutFtkutum ω 0, )( 222 0 0 1 = − −= c m F c ωω ( )tt m F tu 022 0 0 coscos )( )( ωω ωω − − = t m F tctctu ω ωω ωω cos )( sincos)( 22 0 0 0201 − ++=
- 19. Undamped Equation: Solution to Initial Value Problem (2 of 3) Thus our solution is To simplify the solution even further, let A = (ω0 + ω)/2 and B = (ω0 - ω)/2. Then A + B = ω0t and A - B = ωt. Using the trigonometric identity it follows that and hence ( )tt m F tu 022 0 0 coscos )( )( ωω ωω − − = ,sinsincoscos)cos( BABABA =± BABAt BABAt sinsincoscoscos sinsincoscoscos 0 −= += ω ω BAtt sinsin2coscos 0 =− ωω
- 20. Undamped Equation: Beats (3 of 3) Using the results of the previous slide, it follows that When |ω0- ω| ≅ 0, ω0+ ω is much larger than ω0 - ω, and sin[(ω0 + ω)t/2] oscillates more rapidly than sin[(ω0- ω)t/2]. Thus motion is a rapid oscillation with frequency (ω0+ ω)/2, but with slowly varying sinusoidal amplitude given by This phenomena is called a beat. Beats occur with two tuning forks of nearly equal frequency. ( ) 2 sin 2 0 22 0 0 t m F ωω ωω − − ( ) ( ) 2 sin 2 sin )( 2 )( 00 22 0 0 tt m F tu ωωωω ωω + − − =
- 21. Example 2: Undamped Equation, Mass Initially at Rest (1 of 2) Consider the initial value problem Then ω0 = 1, ω = 0.8, and F0 = 0.5, and hence the solution is The displacement of the spring–mass system oscillates with a frequency of 0.9, slightly less than natural frequency ω0 = 1. The amplitude variation has a slow frequency of 0.1 and period of 20π. A half-period of 10π corresponds to a single cycle of increasing and then decreasing amplitude. 0)0(,0)0(,8.0cos5.0)()( =′==+′′ uuttutu ( )( )tttu 9.0sin1.0sin77778.2)( =
- 22. Example 2: Increased Frequency (2 of 2) Recall our initial value problem If driving frequency ω is increased to ω = 0.9, then the slow frequency is halved to 0.05 with half-period doubled to 20π. The multiplier 2.77778 is increased to 5.2632, and the fast frequency only marginally increased, to 0.095. 0)0(,0)0(,8.0cos5.0)()( =′==+′′ uuttutu
- 23. Undamped Equation: General Solution for the Case ω0 = ω (1 of 2) Recall our equation for the undamped case: If forcing frequency equals natural frequency of system, i.e., ω = ω0 , then nonhomogeneous term F0cosωt is a solution of homogeneous equation. It can then be shown that Thus solution u becomes unbounded as t → ∞. Note: Model invalid when u gets large, since we assume small oscillations u. tt m F tctctu 0 0 0 0201 sin 2 sincos)( ω ω ωω ++= tFtkutum ωcos)()( 0=+′′
- 24. Undamped Equation: Resonance (2 of 2) If forcing frequency equals natural frequency of system, i.e., ω = ω0 , then our solution is Motion u remains bounded if damping present. However, response u to input F0cosωt may be large if damping is small and |ω0 - ω| ≅ 0, in which case we have resonance. tt m F tctctu 0 0 0 0201 sin 2 sincos)( ω ω ωω ++=