HW3, MAE 200A. Fall 2005. UCI

Nasser Abbasi

Problem 1

Due: Oct. 13 at beginning of class

Analyze the forced response of the pendulum. The forcing, as demonstrated in class, is the sinusoidal side to side movement of the attachment point.

Assume the lateral motion of the attachment point is of the form MATH

is the maximum displacement. The linearized dynamics about the equilibrium at $\theta =0$ are

MATH

Construct a physical pendulum. It can be as simple as the one I used in class, although a rigid rod would be preferable to an elastic one. Estimate the values of , and and use these values in your calculations. = length of the rod; = mass of the pendulum bob; =damping coefficient
Determine the natural frequency $\omega _{n}\$ and damping ratio $\zeta ~$ of your pendulum.
Determine the general solution (homogeneous plus particular) for the linear pendulum model. You can use the result I gave in lecture if you know how to obtain it. If you don't know how to obtain it, this would be a good opportunity to learn how to.
Using the particular solution, construct the Bode plots theta (amplitude and phase angle as functions of the forcing frequency $\omega$ ).
Experiment with forcing your pendulum at different frequencies and convince yourself that what you see corresponds to the predictions from the Bode plots.
If you move the attachment point according to where, relative to the natural frequency $\omega _{n}$ of the pendulum, $\omega _{1}$ is much less than $\omega _{n}$ , $\omega _{2}$ is close to $\omega _{n}$ , and $\omega _{3}$ is greater than $\omega _{n}$ , describe qualitatively what the $\theta$ response will look like after the homogeneous solution has died out. Your answer should be based on theory not experiment.

Please turn in all your work except for your pendulum. Bring your pendulum to class if you want to show it off, but this is optional.

Answer

Answer parts 1,2

A simple pendulum was used for this experiment.

The Mass was weighted and found out to be about

The length was measured to be 80 cm or

To determine the damping coefficient, since MATH I will first find $\zeta$ .

The method of logarithmic decrement was used. The logarithmic decrement $\Lambda$ is the natural logarithm of the ratio of any two successive amplitudes in the same direction. The pendulum mass is held initially at an angle $\theta _{0}=45^{0}$ in the positive direction, and then released. The mass will then make one full cycle by swinging to left and then back to the same side as it started and stop before starting its second cycle and so on. The angle the mass reach at the end of its first cycle was estimated to be $\theta _{1}=40^{0}$ and the angle it reached at the end of it second full cycle was estimated to be $\theta _{2}=35^{0}$

Hence MATH

The following diagram showing the process and the derivation of the above equation

To derive the equation MATH We utilize the above plot of for a damped second order system.

In the above diagram MATH and MATH where $\tau _{\nu }$ is the time it takes to make one full swing between these 2 successive oscillations.

MATH

But MATH where $\omega _{d}$ is the damped natural frequency. Hence hence

MATH

But hence the above equation becomes

MATH

Now that this equation is derived, it can be used to estimate $\zeta$ , Once $\zeta$ is found then can be easily found. MATH

square each side and solve for $\zeta$

MATH

Next, since Hence, then MATH

and since , then MATH

Answer part 3

First we find an analytical solution .

MATH

where is the homogenous solution (due to initial conditions) and is the particular solution (due to the forcing function). To find , looking at the ODE

MATH

We see this is a standard second order system. Let , , hence we can write the above as MATH the solution is MATH

where the damped natural frequency be hence the above can be written as

MATH

Now we can find the particular solution. Since the forcing function is a sinusoidal, we can try MATH

This particular solution will take care of the case when the forcing function is out of phase with the response, that is why both a $\sin$ and a $\cos$ function are present. Substitute $\theta _{p}$ into the dynamic equation that represents the linearized pendulum given by MATH

and since MATH then hence the above equation becomes

MATH

Now MATH

Hence (4) becomes

MATH

compare coefficients of we get

MATH

and MATH

Need to solve (5) and (6) for $G_{1},G_{2}\ ,$ first divide (5) and (6) by $\omega _{n}^{2}$ and call the ratio which is the response ratio, to obtain new expressions for and (6)

MATH

from (6a)

MATH

Plug (7) into (5a) we obtain

MATH

Substitute (8) into (7) to solve for $G_{1}$

MATH

Hence MATH

Where $G_{1},G_{2}$ are as given above. Hence

MATH

Hence the general solution is

MATH

Where , can now be determined from initial conditions. However, since we are interested in steady state solution (not the transient solution), we do not need to find these constants for the purpose of this solution.

The first term in the solution (the ) term, will damp down quickly with time since it has the inverse exponential term in it. What is left then is the particular solution.

Answer part 4

Since

MATH

Hence in the above equation we can write it as

MATH

Where amplitude MATH and phase MATH

To make amplitude in degrees instead of radians, convert the above by multiplying by $180/\pi$

and ,, and for let meters.

Now we are ready to generate the needed plots. I will show the following plot where the x-axis will show the input frequency $\omega$ in rad/sec, and the y-axis will show the amplitude in angles.

Part 5

Now I changed the forcing frequency to be close to the natural frequency of the pendulum (3.5 radians per second), and observed largest oscillation at that frequency. (note: input force is lateral, at the joint, so input frequency needs to be converted from cycles per second to radiance per second).

This is in agreement with the above plot.

Part 6

Now assume

For a linear system, the total response will be the same as the sum of the response to each of the above signals individually. (at each instance of time).

In addition, the response will have the same frequency but different amplitude and phase.

Hence the steady state amplitude will be the sum of the individual amplitudes, and the steady state phase shift will be the sum of the individual phase shifts. (at each instance of time).

This can be answered by looking at the analytical solution found above for (steady state solution) and taking the limit of as it approaches or or $\infty$ and see what happens to the amplitude and the phase in each case.

Since we found

MATH

Where

Hence for the case when , then , hence the limit of the solution will be , i.e. in stead state, the displacement goes to zero. For the phase we see that the phase will also go to zero, hence response will be in-phase with the input.

For the case when , then we see $\beta$ becomes very large. We see that the phase term goes to zero since it goes like $\frac{1}{\beta }$ in the limit. Hence the response in steady state when the input frequency is much larger than $\omega _{n}$ will be in phase. To see what happens to the amplitude, take the limit of the amplitude as

MATH

So when forcing frequency is much larger than $\omega _{n}$ , the steady state response will be fixed and will be proportional to the maximum amplitude . This agrees with the plot shown above where we see the response is steady at large $\omega .$ For the phase we see that the phase will also go to zero, hence response will be in-phase with the input. This agrees with the phase plot shown.

For the case when , then , hence

MATH

Hence the smaller the damping ratio $\zeta$ , the larger the amplitude. This means the smaller the damping coefficient the larger the amplitude. This is called resonance. For the phase, we see that hence at resonance, the phase is 90 $^{0}$ from the input. This agrees with the plot shown.

Appendix

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