The goal of this lab is to understand the dynamic parameters behind a second order oscillatory system

The goal of this lab is to understand the dynamic parameters behind a second order oscillatory system. We will see how different inertias, spring coefficients and damping coefficients will affect the natural frequency and damping ratio of the system’s response. This lab will help us in verifying mathematical relationships developed using the data gathered from the ECP Torsional Dynamic System (Model 205a).

2 Determining the Dynamic Parameters of the Dynamic System

Objective of this exercise is to obtain the dynamic parameters that characterize the torsional dynamic system. Dynamic parameters desired include inertias, spring coefficients, and damping coefficients. This can be done by modeling the damped oscillatory movements of the torsional dynamic system into a ‘lightly damped second order system’, in which we can easily find those parameters.

2.1 Mathematical Analysis

Knowing the transfer function of a system tells much about its behavior. It is given that the torsional system can be modeled as a lightly damped second order system with the following equation:

Eq. 2.1-1

From the Laplace transform of this equation:

Eq. 2.1-2

Thus we define the transfer function of the system as:

Eq. 2.1-3

Note that the general second order transfer function of a system is as follows:

Eq. 2.1-4

2.1.1 Finding ?, ?, and b, k, J

The transfer function R(s) is an example of general second order transfer function G(s). Simply comparing the constants in R(s) and G(s) will yield two equations Eq.2.1.1-1 and Eq.2.1.1-2. (Consider kR(s) and G(s) to be equal.)

Eq. 2.1.1-1 Eq. 2.1.1-2

Once we find appropriate values for ?n and ?, the above equations will give values for bsystem, k, and J.

2.1.2 Inverse Transform of the System Transfer Function

Some useful relationships pop up when we perform inverse Laplace transform to G(s).

Eq. 2.1.2-1

Eq. 2.1.2-2

By the definition of damped natural frequency, we know that.

When we substitute t0 to G(t) and divide it with G(tn) we get

. Eq. 2.1.2-3

Since t0 and tn are chosen at the peak of the graph, where and reaches its maximum, we can simplify the above equation down to

Eq. 2.1.2-4

T0 is the fundamental period for the damped oscillation so . After substituting 2? and applying natural logarithm to the both side of the equation will yield

Eq. 2.1.2-5

Thus we get

Eq. 2.1.2-3

where Xn and X0 stands for G(tn) and G(t0) respectively.

2.2 Finding the Natural Frequency and the Damping Ratio

A step input is given to the system by offsetting the plate position by hand and letting go. This procedure is done to all 4 cases labeled d11 (disc 1 loaded), d12 (disc 1 unloaded), d31 (disc 3 loaded), d32 (disc 3 unloaded). The results for disc 3 are shown below, with results of disc 1 being very similar.

Notice that the system response to the input decays exponentially over time while maintaining a constant damped frequency ?d. Information extracted from the graphs is tabulated in Table 2.2-1.


Sampled cycle, n

Time at 1st peak (s)

Time at nth peak (s)

Amplitude at 1st peak (count)

Amplitude at nth peak (count)

























Table 2.2-1 Information extracted from graphs and data obtained experimentally

n is chosen so that the error is minimized. We used only the first few peaks of the graphs for two reasons. First, there are no peaks when the amplitude gets very close to zero, because the plate stops oscillating when friction is greater than the oscillation force. Second, when amplitude gets small, greater percentage of the encoder’s reading would be composed of error. Thus the reading is taken near the beginning.

2.2.1 Finding the Natural Frequency

By applying basic properties of oscillation, and , we can obtain the natural frequency of the system using the information collected in Table.2.2-1. The resulting data is presented below in Table 2.2-2. Take note that the damped natural frequency is considered to be equal to the natural frequency under the following assumption,

Eq. 2.2.1-1


Fundamental Period T0 (s)

Natural Frequency ?n (rad/s)













Table 2.2.1-1 Calculated To and ?n

2.2.2 Finding the Damping Ratio

From the relationship we found in 2.1.2, we can easily get ?. With an additional assumption that ? is a lot less than 1, we can write Eq. 2.1.2-3 as follows:

Eq. 2.2.2-1

Using Eq. 2.2.2-1 we can find the damping ratio for the two unloaded cases. As presented in Table 2.2.2-1.


Damping Ratio ?





Table 2.2.2-1 Damping ratio of unloaded cases

The damping ratio tends to be larger at the bottom plate. This is justified because it is connected to a DC motor, which brings more bearings and bands into the system to dampen the response down.

2.3 Finding the Inertia and Other Parameters of the System

Basic properties of rotation enable us to just use the parallel axis theorem to sum up the individual inertias of the brass weights and inertias of the rotational disks to get the total inertia of the rotational body. Thus

Eq. 2.3-1

where Ji (i=1,2,…) is the inertia of the rotational disk and Jm is the inertia invoked by the addition of the brass weights.

2.3.1 Inertia Invoked by the Brass Weights

We know that inertia of the single three-dimensional brass weight having the center of mass as its axis of rotation is

Eq. 2.3.1-1

with r = 0.025m and m = 0.5kg. Thus Jm0 is found to be 0.00015625 kg�m2. Since the system is designed to have four brass weights connected at a distance R from the axis of rotation, by applying parallel axis theorem, the inertia these four brass weights bring to the system can be formulated as

Eq. 2.3.1-2

with R = 0.09m and m = 0.5kg, which gives the total inertia Jm to be 0.016825 kg�m2.

2.3.2 Inertia of the Rotational Disks, Spring Coefficients and Damping Coefficients.

Jm is found in section 2.3.1, so we can just substitute Jm, along with all the values for ?n and ? from Table 2.2.1-1 and Table 2.2.2-1, into the Eq. 2.1.1-1 and 2.1.1-2. This gives six equations and six unknowns – b1, b3, k1, k3, J1, and J3. Using some algebra we then acquired the values for these dynamic parameters, as presented in Table 2.3.2-1. Note that for spring 1 and spring 2 are connected in series.






0.00015625 kg�m2




0.016825 kg�m2

b2 = b3



0.00226573253 kg m2



J2 = J3

0.00194013807 kg m2





Table 2.3.2-1 Final calculated results of the Dynamic Parameters

3 Second Order Damped Oscillator in Free Motion

3.1 Effects of Inertia and Stiffness

To study the effects of changing inertia, spring constant, and damping constant on the motion of the second order damped oscillating system, configurations that varied the inertia and/or the stiffness of the dynamic system is used. The graphs for configuration 1 and 2 are shown in Fig. 3.1-1 and Fig. 3.1-2:

3.1.1 The Effects of the Reduction in Inertia

Configuration 1 and d11 in section 2 have the exact same set up except that the configuration 1 has 2 brass weights with a new R that will give half the inertia of d11. So by comparing the results of Configuration 1 with those of the loaded lower disk (d11), we can find the effects for the reduction in inertia.

First, we need to determine a new R that will give half the lower disk inertia using only two brass weights. In section 2, the lower disc inertia with 4 brass weights was:

Eq. 3.1.1-1

Since we want a new R (rnew) that will give half the inertia of the above using 2 brass weights, we can set

Eq. 3.1.1-2

Then rnew will be

Eq. 3.1.1-3

By calculating the above equation, rnew is calculated to be 0.083469m. Utilizing the data obtained (as in Fig. 3.1-1) and using method similar to section 2.2.1, we find that the corresponding ?new is 17.395 rad/s. This is greater than ?nd11 that we had found to be 11.8283 rad/s in section 2.2.1 by a factor of 1.471. This is inline with the observation we can make by looking at Eq. 2.1.1-2. From the equation, we can see that if the total inertia J is reduced by a factor of 2, ?n should be reduced by a factor of .This is very close to the factor of 1.471 that we got.

3.1.2 The Effects of Change in Stiffness

The change stiffness is the increase in spring constant. This is achieved by replacing the clamp from the middle disk to the upper disk and removal of the middle disk. Having twice doubled the length of the thin steel rod (torsional spring) in Configuration #2, the stiffness decreased. By comparing the results from Configuration 1 with Configuration 2, we can find out how the change in stiffness affects the measured frequency of oscillation.

k12 was calculated in section 2.3.2 and it corresponds to the spring constant of Configuration 2. For Configuration 1, k1 would be the corresponding spring constant.


Sampled cycle n

Time at 1st peak to (s)

Time at nth peak tn (s)

Fundamental period To (s)

Natural freq.

?n (rad/s)

# 1






# 2






Table 3.1.2-1 experimental data of Configuration 1 and Configuration 2

The above table shows that Configuration 1 has a higher ?n than the Configuration 2 (17.395 > 12.393). Thus, we see that if the spring constant decreases (from k1 to k12 ) then the natural frequency also decreases.

In order to compare the differences quantitatively, based on Eq. 2.1.1-2 again, we can divide the equation for Configuration 1 by the equation for Configuration 2 and get ratio formula as follows:

Eq. 3.1.2-1

The ratio of ?c1 to ?c2 resulted from our data was 1.40361 and the resulting ratio from Eq.3.1.2-1 was calculated to be 1.41321. Ignoring discrepancy caused by non-ideal environment and apparatus, we can say that the two resulting ratios are essentially the same, verifying our theoretical finding with our experimental observation.

3.1.3 The Effects of Combined Change in Inertia and Stiffness

The natural frequency ?nd11 from the d11 case of section 2 was measured to be 11.828 rad/s. On the other hand the natural frequency ?nc1 from Configuration 2 was measured to be 12.393 rad/s. In order to find out any differences between measurements and theory, we can use the following two equations derived from Eq. 3.1.2-1.

Eq. 3.1.3-1 Eq. 3.1.3-2

Thus, using the above formulas we can calculate the theoretical value of ? after the variation in both inertia and spring constant. Combining the two formulas above we get

Eq. 3.1.3-3

Letting ?a be the natural frequency of d11 in section 2 and ?b be the natural frequency of Configuration 2, Eq. 3.1.3-3 gives us a ?b of 11.836 rad/s. It is fairly close to the experimental value of 12.393 rad/s as shown in table 2.2.1-1. The discrepancy is acceptable due to the large possible error during the placement of the mass to their new R of 0.083469m.

3.2 The “Feel” of Viscous Damping

Varying the damping coefficient, bsystem, from 0.05N-m/(rad/s) to 0.50N-m/(rad/s), we observed that the larger the damping coefficient, the greater the force we must apply to turn the lower disk. In another word, larger damping coefficient increases the disk impedance, thus reducing the initial acceleration caused by any input force. For constant torque, increase in b will result in both a decrease in angular displacement and a decrease in steady state velocity for the disk, vice versa. These phenomenons are represented by equations as shown below:

Eq. 3.2-1

Eq. 3.2-2

3.3 Combined Effects of Inertia, Stiffness and Damping

In four different cases, the damping ratio ? and the natural frequency ?n vary from the change of b, k, and J. Case 1 to Case 3 tests the step response of the system from effect of changing total inertia (J) while holding k and b constant. On the other hand, Case 3 and Case 4 tests combined effect of k and b, with J kept constant.

Rearrange Eq. 2.1.1-1 from section 2, we can obtain the damping ratio ? as follows

Eq. 3.3-1

The natural frequency also can be calculated a rearranged version of Eq. 2.1.1-2 as below

Eq. 3.3-2

Setting up according to cases stated in the lab instruction, experimental data of these four different cases are collected and presented in Table 3.3-1


Damping coefficient, b (N-m/(rad/s))

Total disk inertia, J (kg-m2)

Spring constant, k (N-m/rad)

Damping ratio, ? *calculated

?n *calculated (rad/s)

























Table 3.3-1: Dynamic parameters for combined effects of inertia, stiffness, and damping

The responses of the four cases are shown in Figure 2.3.1 to Figure 2.3.4 as below:

Fig. 3.3-1 System Response of Case 1

Fig. 3.3-2 System Response of Case 2

Fig. 3.3-3 System Response of Case 3

Fig. 3.3-4 System Response of Case 4

From the graph of Case 1, we see that it has fast overshoot and oscillation while settling fast. The graph of Case 2 shows that it has the least overshoot and oscillation, but settles very fast. Case 3’s graph shows fast overshoot and oscillation, however with relatively slow settling. Finally Case 4 shows the most overshoot and oscillation but settles again relatively slow.

3.3.1 Effect on Damping Ratio ? with Respect to Change of b, k and J

Combining 3.3-1 and 3.3-2, we can obtain a function of ? in terms of b, k, and J as shown in Eq. 3.3.1-1.

Eq. 3.3.1-1

Thus, damping ratio ? is linearly proportional to the damping constant b, and inversely proportional to the square root of both the total inertia J and the spring constant k. This is consistence with results obtained in Case 1 and 2 as seen in Table 3.3-1, with b and k constant, a smaller J obtains a larger ?, vice versa.

3.3.2 The Various Response Shapes and Speed of Response Relate to ?n and ?

We first determine how the change of damping ratio ? will affect the shape of response by percent overshoot. The percent overshoot (%OS) as a function of ? is shown in Eq. 3.3.2-1

, Eq. 3.3.2-1

Higher percent overshoot indicates more oscillation in the response; where as lower percent overshoot means less oscillation. From Eq. 3.3.2-1, we observe that the percent overshoot is inversely proportional to the damping ratio ?. Comparing to Case 1 and Case 4 as seen in Table 3.3-1, we can see that this is true. The natural frequency ?n of the two cases are really close, but with ?4 smaller than ?1, the graph of Case 4 should have more cycles of oscillation than Case 1.This is exact what we observed when comparing Fig. 3.3.1 to Fig. 3.3.4. Case 2 holds the largest value of ?. This implies that less oscillation would show up in Figure 2.3.2.

In second order under damped system the time constant ? allow us to picture how steep the “folding function” is and how fast the system reaches steady state. The time constant ? is relate to the damping ratio and natural frequency as shown in the relationship of Eq. 3.3.2-2

Eq. 3.3.2-2

The larger ?n and ? gets, the smaller the resulting ?. This indicates a larger natural frequency and damping ratio will result in a system that reaches steady state sooner, vice versa. Case 2 has relatively the largest ?n and ? , thus it should reach steady state the soonest. We can see that the actual experimental data is in line with our theoretical analysis in Fig. 3.3-2.

Comparing Cases 3 and 4 allows us to find the relationship of how the change of ? and ?n affects the speed of the response. The period of response from the inverse Laplace transfer Eq. 2.1.2-2 shown below

Eq. 3.3.2-3

The larger the natural frequency and the smaller the damping ratio end up with a smaller period of respond and thus a faster response. Case 4 has a smaller damping ratio and a larger natural frequency compared to Case 3, therefore Case 4 responses faster than Case 3. Moreover, the time constants of both cases are very close. Thus, in the transient state, we can observer more oscillation in Case 4 since it has a shorter period.

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