Would a bungee jumper represent a damped system

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Bungee Jump: Simulate Multibody Systems

This is a simple model of a bungee jumper consisting of a mass attached to a platform by a spring and damper.

Automatic 3D Animation

Multibody systems have visualizers to show what a real-world system would look like.

jumper

To simulate the model and view a 3D animation of it, follow the steps below:

  • Click the button in the top-right corner.
  • When the build is finished, click the Simulate button .
  • Click the Animate button .
  • Use your mouse or trackpad to drag the animation to a good angle and zoom in with your scroll wheel or by using the trackpad. Then click the Play button to play the animation.

Instant Plotting

Explore the how the force on the cord varies with time by simulating and plotting the variable ElasticCord.f .

The variable will automatically be plotted when the model is simulated.

plotting

Change Parameters

Changing parameters for the simulation can be done rapidly in Simulation Center. Switch to the Parameters tab and enter a new value for the parameter you would like to vary.

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parameters

Simulate again to see the effects of your changes.

ShowHideTerms and Conditions of Use

This domain example is an informational resource made freely available by Wolfram Research.

Forced Vibrations of Damped Single Degree of Freedom Systems: Damped Spring Mass System

We have so far considered harmonic forcing functions acting on undamped systems. We will now extend our analysis to include systems which include viscous damping. We will still limit our analysis to harmonic forcing functions of the form

Figure 5.1: Damped spring–mass system subjected to harmonic forcing function

Consider a damped spring-mass system subjected to a harmonic forcing function as shown in Figure 5.1(a). The FBD/MAD for this system is shown in Figure 5.1(b) where is once again the displacement from the static equilibrium position. Applying Newton’s Laws we obtain

[+!!downarrowsum F = ma:qquad F_0 sin(omega t) -kx -c dot{x} = m ddot{x}]

(5.1)

Once again, the response will be composed of a homogeneous solution (the transient response) and a particular solution (the steady state response) as

(5.2)

We have previously found the homogeneous solution. For example, in the underdamped situation the homogeneous solution is given in equation (3.11) as

[x_H(t) = e^{-zeta p t} Bigl[ A sin(sqrt{1-zeta^2}pt) + B cos(sqrt{1-zeta^2}pt) Bigr]]

(5.3)

where and are arbitrary constants. To find the particular solution to equation (5.1), we will
assume a solution of the form

(5.4)

Substituting these into the equation of motion gives

[m Bigl[ -mathbb{X} , omega^2sinleft(omega t - phiright) Bigr]+c Bigl[ mathbb{X} , omega cosleft(omega t - phiright) Bigr]+k Bigl[ mathbb{X} , sinleft(omega t - phiright) Bigr] = F_0 sin (omega t)]

(5.5)

Using the identities

begin{align*} sin (omega t - phi) &= sin omega t , cos phi - cos omega t , sin phi \ cos (omega t - phi) &= cos omega t , cos phi + sin omega t , sin phiend{align}

[mathbb{X} bigl(k-momega^2bigr) Bigl[ sin omega t cos phi - cos omega t sin phi Bigr] +c mathbb{X} omega Bigl[ cos omega t cos phi + sin omega t sin phi Bigr] = F_0 sin omega t]

or, collecting the and terms,

[mathbb{X} Bigl[ bigl(k-momega^2bigr) cos phi + c omega sin phi Bigr] sin omega t -mathbb{X} Bigl[ bigl(k-momega^2bigr) sin phi - c omega cos phi Bigr] cos omega t = F_0 sin omega t]

Comparing the left and right hand sides leads to two equations

[ mathbb{X} Bigl[ bigl(k-momega^2bigr) cos phi + c omega sin phi Bigr] = F_0]

(5.6a)

[ mathbb{X} Bigl[ bigl(k-momega^2bigr) sin phi - c omega cos phi Bigr] = 0]

(5.6b)

From (5.6b) we see that

(5.7)

Now (5.6a) (5.6b) gives

[mathbb X bigl(k-momega^2bigr) bigr(cancelto{1}{cos^2 phi + sin^2 phi}bigr) = F_0 cos phi]

(5.8a)

while (5.6a) – (5.6b) gives

[mathbb{X} c omega bigr(cancelto{1}{sin^2 phi + cos^2 phi} bigr) = F_0 sin phi]

(5.8b)

Squaring each of (5.8a) and (5.8b) and adding the results leads to

[F_0^2 = mathbb{X}^2 Bigl[ bigl(k-momega^2bigr)^2 + bigl(c omega bigr)^2 Bigr]]

(5.9)

Therefore, the particular solution (5.4) for this problem is

[x_P(t) = frac{F_0}{sqrt{bigl(k-momega^2bigr)^2 + bigl(c omega bigr)^2}} sin (omega t - phi)]

(5.10)

where is given by equation (5.7). The total solution (for the underdamped case) is

[x(t) = e^{-zeta p t} Bigl[ A sinleft(sqrt{1-zeta^2}ptright) + Bcosleft(sqrt{1-zeta^2}ptright) Bigr] + frac{F_0}{sqrt{bigl(k-momega^2bigr)^2 + bigl(c omega bigr)^2}} sin(omega t - phi)]

In many cases we are often primarily interested in the long term steady state response of a system. Since the transient response will eventually damp out as we have seen, we will often ignore the transient part and consider the solution to be simply given by the steady state response as

[x(t) = frac{F_0}{sqrt{bigl(k-momega^2bigr)^2 + bigl(c omega bigr)^2}} sin (omega t - phi)]

(5.11)

Note that this can be rewritten as

[x(t) = frac{dfrac{F_0}{k}}{sqrt{Bigl(1-dfrac{m}{k}omega^2Bigr)^2 + Bigl(dfrac{c omega}{k} Bigr)^2}} sin (omega t - phi)]

(5.12)

However, as we have already discussed,

[frac{comega}{k} = c , left(frac{2 mensuremath{p}}{c_mathrm{C}}right) , frac{omega}{k}= 2 , cancelto{zeta}{frac{c}{c_mathrm{C}}} , cancelto{frac{1}{ensuremath{p}^2}}{frac{m}{k}} , ensuremath{p} omega= 2 zetafrac{omega}{ensuremath{p}}]

As a result, (5.12) becomes

[x(t) = underbrace{frac{ensuremath{delta_{mathrm{ST}}}}{ensuremath{sqrt{left[1- left(ensuremath{frac{omega}{ensuremath{p}}}right)^2right]^2 + Bigl[2 zeta ensuremath{frac{omega}{ensuremath{p}}} Bigr]^2}}}}_{text{Amplitude }mathbb{X}} , sin (omega t - phi)]

(5.13)

Here we can see that the amplitude of the response is given by

[mathbb{X} = ensuremath{delta_{mathrm{ST}}} frac{1}{ensuremath{sqrt{left[1-left(ensuremath{frac{omega}{ensuremath{p}}}right)^2right]^2 + Bigl[2 zeta ensuremath{frac{omega}{ensuremath{p}}} Bigr]^2}}}]

[boxed{frac{mathbb{X}}{ensuremath{delta_{mathrm{ST}}}} = frac{1}{ensuremath{sqrt{left[1-left(ensuremath{frac{omega}{ensuremath{p}}}right)^2right]^2 + Bigl[2 zeta ensuremath{frac{omega}{ensuremath{p}}} Bigr]^2}}}}]

(5.14)

which represents the dynamic magnification factor in the damped situation. Similarly, since

[frac{comega}{k-momega^2} = frac{dfrac{c omega}{k}}{1-dfrac{m}{k} omega^2} =dfrac{2 zeta ensuremath{dfrac{omega}{ensuremath{p}}}}{1-ensuremath{dfrac{omega}{ensuremath{p}}}^2}]

equation (5.7) can be written as

(5.15)

Equations (5.14) and (5.15) are illustrated in Figures 5.2(a) and (b) respectively.

This image has an empty alt attribute; its file name is 5_2.png

Figure 5.2: Response of a damped SDOF system: (a) Dynamic Magnification Factor (DMF), (b)Phase Angle

(a) In the limiting case where , these results are the same as those obtained in the undamped case.

begin{equation*}frac{mathbb X}{delta_{text{ST}}} = frac{1}biggl {p})^2biggr |}, qquad phi =begin{cases}0,&frac{omega}{p}<1\ pi, &frac{omega}{p}></p><p>1end{cases} end{equation*}” width=”266″ height=”64″ /></p><p>(b) At very low frequency ratios,  , the amplitude of the motion is approximately the static deflection.</p><p>(c) At very high frequency ratios,  , the amplitude of the response is significantly less than the static deflection.</p><p>(d) In both (b) and (c) above, damping has very little effect. Therefore at these extremes, it is often appropriate to use the results for an undamped system (simply because they are easier to use).</p><p>(e) Between these two extremes, and particularly near resonance, damping has the effect of limiting the amplitude of vibration. At resonance an infinite amplitude will never be reached and a steady state value can be obtained.</p><p>(f) When the system is damped, there is no sudden transition from <em>in phase</em> to completely <em>out of phase</em>. The response is always somewhat out of phase with the forcing function (we usually say the response <em>lags</em> the forcing function by the phase angle).</p><div style=

(g) At resonance, the response lags the forcing function by radians, regardless of the amount of damping present.

Graphic Representation

Reconsider the response of the system in equation (5.5)

[m Bigl[ -mathbb{X} omega^2sin (omega t - phi) Bigr]+c Bigl[ mathbb{X} omega cos (omega t - phi) Bigr]+k Bigl[ mathbb{X} sin (omega t - phi) Bigr] = F_0 sin omega t,]

and rearrange the results slightly as

[underbrace{F_0 sin omega trule[-3mm]{0pt}{0pt}}_{text{Disturbing Force}}- underbrace{k Bigl[ mathbb{X} sin (omega t - phi) Bigr]}_{text{Spring Force}}- underbrace{c Bigl[ mathbb{X} omega cos (omega t - phi) Bigr]}_{text{Damping Force}}+ underbrace{m Bigl[ mathbb{X} omega^2sin (omega t - phi) Bigr]}_{text{``Inertia'' Force}}= 0]

(5.16)

We can interpret each of the terms in this equation as the vertical component of a vector rotating CCW about the origin at a rate of rad/s as shown in Figure 5.3 below.

However, from (5.16) we see that the sum of these components must be zero. It is therefore convenient to add these terms vectorially so that the resulting polygon closes as in Figure 5.4.

Figure 5.3: Graphic representation of terms in equation of motion as vertical components of rotating vectors

Figure 5.4: Vectorial addition of rotating vectors

From this figure we can see clearly that

[F_0^2 = Bigl[ k mathbb{X} - m mathbb{X} omega^2 Bigr]^2 + Bigl[ c mathbb{X} omega Bigr]^2 qquad text{and}qquadtan phi = frac{c mathbb{X} omega}{k mathbb{X} - m mathbb{X} omega^2}]

which leads directly to

which are the results obtained previously (but with less work involved here).

Using this representation, we can understand the effect of damping in each of the three situations

[ensuremath{frac{omega}{ensuremath{p}}} < 1,qquad ensuremath{frac{omega}{ensuremath{p}}} = 1,qquad ensuremath{frac{omega}{ensuremath{p}}} ></p><p> 1.]” width=”205″ height=”31″ /></p><p>In this case, the  term is smaller than the  term. As a result, the inertia force is smaller than the spring force. As a result, the phase angle (the angle by which the response lags the forcing function) is less than  .</p><p><img decoding=

Here the inertia force exactly balances the spring force, while the disturbing force exactly balances the damping force. As a result, we see that the phase angle is always radians.

This image has an empty alt attribute; its file name is 5_74_2.png

frac{omega}{p} ></p><p> 1}” width=”37″ height=”19″ /></p><p>Here, the inertia forces are larger than the spring forces. The result is a larger phase angle, somewhere between  and  radians.</p><p><img decoding=

The tool below demonstrates the rotating vector representation. In the tool is the natural frequency . Change the parameters to see the effect on the vector schematic.

Transmissibility in Forced Damped Vibrations

The force transmitted between a machine and its supporting structure is an important consideration in the vibration isolation of machinery. Typically a machine is vibrating due to some internal disturbing force and we would like to isolate the supporting structure from these vibrations. (The reverse situation also applies.)

Figure 5.5: Vibration isolation

A typical situation is shown in Figure 5.5. The goal is often to reduce the maximum force transmitted to the supporting structure. When we previously considered the undamped situation, the only force transmitted to the support was through the spring. Now, however, force can be transmitted through both the spring and the damper. Further, we can not simply added these two forces as scalars because they do not reach their maximum values at the same time (i.e. they are not in phase). However, as we have seen these two forces are always 90 out of phase

[F_text{S} = k x(t) = k mathbb{X} sin (omega t - phi),qquad F_text{C} = c dot{x}(t) = c mathbb{X} omega cos (omega t - phi)]

so the resulting magnitude of the sum of these two forces is

(5.17)

(This force can be seen schematically in the diagram in Figure 5.6 based on the graphical interpretation discussed earlier.) However, we have already determined the amplitude of the response in this situation (equation (5.9)) to be

(5.18)

Figure 5.6: Maximum force transmitted to the supporting structure

Combining (5.17) and (5.18) gives

[ensuremath{F_{T_{max}}}} = cancel k cdotfrac{frac{F_0}{cancel k}}{sqrt{Bigl[1-frac{m}{k}omega^2Bigr]^2 +Bigl[frac{c omega}{k} Bigr]^2}}cdotsqrt{1 + Bigl( frac{c omega}{k}Bigr)^2}]

[frac{F_{T_{max}}}{F_0} = frac{sqrt{1 + Bigl( dfrac{c omega}{k}Bigr)^2}}{sqrt{Bigl[1-dfrac{m}{k}omega^2Bigr]^2 +Bigl[dfrac{c omega}{k} Bigr]^2}}]

[frac{m}{k}omega^2 = Bigl(ensuremath{frac{omega}{ensuremath{p}}}Bigr)^2 qquadtext{and}qquadfrac{comega}{k} = 2zetaensuremath{frac{omega}{ensuremath{p}}}]

we get the result

[boxed{ensuremath{TR} = frac{F_{T_{max}}}{F_0} = frac{sqrt{1 + Bigl( 2 zeta ensuremath{dfrac{omega}{ensuremath{p}}} Bigr)^2}}{ensuremath{sqrt{left[1-ensuremath{left(frac{omega}{ensuremath{p}}}right)^2right]^2 + Bigl[2 zeta ensuremath{frac{omega}{ensuremath{p}}} Bigr]^2}}}}]

(5.19)

This is the transmissibility in the damped situation which represents the ratio of the maximum force transmitted to the supporting structure to the maximum disturbing force (which here is simply ).

Figure 5.7: Transmissibility in a damped spring–mass system

This relationship is illustrated in Figure 5.7. As this figure shows, the transmissibility is always greater than one if . This means that more force is transmitted to the supporting structure than if the machine were rigidly attached to the structure. Near resonance, even small amounts of damping can significantly reduce the force transmitted to the structure.

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ensuremath{frac{omega}{ensuremath{p}}} ></p><p>For sqrt{2}” width=”51″ height=”22″ />, the force transmitted to the structure is less than the disturbing force. In this region, however, the addition of damping increases the transmissibility compared to the undamped situation. If a system is operating in this region, an undamped support offers better transmissibility characteristics than one with damping. However, some damping is often still desirable if the system must pass through resonance to reach its operating state.</p><h4><span class=EXAMPLE

A machine with a mass of 100 kg is supported on springs of total stiffness 700 kN/m and has an unbalanced rotating element which results in a disturbing force of 350 N at a speed of 3000 RPM. Assuming the system has a damping ratio of , determine

  1. the amplitude of the motion due to the rotating imbalance,
  2. the transmissibility, and
  3. the maximum force transmitted to the supporting structure.

EXAMPLE

A vibrating system, together with its inertia base, is originally mounted on a set of springs. The operating frequency of the machine is 230 RPM. The system weighs 1000 N and has an effective spring modulus 4000 N/m. Shock absorbers are to be added to the system to reduce the transmissibility at resonance to 3. Additionally, the transmissibility at the normal operating speed should be kept below 0.2.

Forced Vibrations Due to a Rotating Imbalance

A common cause of forced vibrations in machinery is a rotating imbalance, shown schematically in Figure 5.8.

Figure 5.8: Schematic of a rotating imbalance

Here a machine with total mass is constrained to move in a vertical direction. is a rotating mass (which is included in the total mass ) with eccentricity that is rotating at a constant speed . During the resulting motion the machine body (the part that is not the eccentric mass) vibrates vertically, with a position described by the cooordinate . The eccentric mass moves relative to the machine body, and its total vertical displacement is given by .

Figure 5.9: FBD/MAD for rotating imbalance

Figure 5.9 shows a FBD/MAD for this situtation. Applying Newton’s Laws in the vertical direction gives

 begin{align*} +!!uparrowsum F = ma:qquad -k x - c dot{x} &= bigl( M-ensuremath{tilde{m}} bigr) ddot{x} + ensuremath{tilde{m}} dfrac{d^2}{d,t^2} bigl( x + e sin omega t bigr) \ &= bigl( M - ensuremath{tilde{m}} bigr) ddot{x} + ensuremath{tilde{m}} bigl( ddot{x} - e omega^2sin omega t bigr) end{align*}

[M ddot{x} + cdot{x} + k x = underbrace{ensuremath{tilde{m} e omega^2}rule[-1mm]{0pt}{0pt}}_{F_0} sin omega t.]

(5.20)

This is the same equation of motion we obtained previously (equation (5.1)) with

As a result we know the solution to the steady response is

[mathbb{X} = frac{dfrac{ensuremath{tilde{m} e omega^2}}{k}}{ensuremath{sqrt{Bigl[1-ensuremath{Bigl(dfrac{omega}{ensuremath{p}}Bigr)}^2Bigr]^2 + Bigl[2 zeta ensuremath{dfrac{omega}{ensuremath{p}}} Bigr]^2}}}, qquadphi = tan^{-1} Biggl[ frac{2 zeta ensuremath{frac{omega}{ensuremath{p}}}}{1-left(ensuremath{frac{omega}{ensuremath{p}}}right)^2} rule[-1cm]{0pt}{0pt}Biggr]]

(5.21)

As discussed previously,

so that equation (5.21) can be rearranged to give

[boxed{frac{Mmathbb{X}}{ensuremath{tilde{m}} e} = frac{left(ensuremath{dfrac{omega}{ensuremath{p}}}right)^2}{ensuremath{sqrt{Bigl[1-ensuremath{Bigl(dfrac{omega}{ensuremath{p}}Bigr)}^2Bigr]^2 + Bigl[2 zeta ensuremath{dfrac{omega}{ensuremath{p}}} Bigr]^2}}}}]

(5.22)

This result is shown in Figure 5.10 for a variety of damping ratios.

Figure 5.10: Response amplitude for a rotating imbalance

[frac{Mmathbb{X}}{ensuremath{tilde{m}} e} = frac{left(ensuremath{dfrac{omega}{ensuremath{p}}}right)^2}ensuremath{Bigl{ensuremath{p}}Bigr)^2Bigr|}}, qquad phi =begin{cases}0,& ensuremath{dfrac{omega}{ensuremath{p}}} < 1, \[6mm] pi,& ensuremath{dfrac{omega}{ensuremath{p}}} ></p><p> 1.end{cases}]” width=”289″ height=”83″ /></p><h4><span class=EXAMPLE

A counter-rotating eccentric weight exciter is used to produce the forced oscillation of a spring and damper supported mass as shown below. By varying the rotation speed, a resonant amplitude of 0.60 cm was recorded. When the speed of rotation was increased considerably beyond the natural frequency, the amplitude appeared to approach a constant value of 0.08 cm.

Determine the damping ratio for the system.

Video Lecture: Forced Vibration of Damped SDOF Systems Section 5.1

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