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Driven Oscillations Lab |
last updated Nov. 3, 2002
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The pre-lab and the instructions for the pre-lab are located at http://www2.hawaii.edu/~jmcfatri/labs/RLCdrivoscprelab.html .
The goal of this lab is
- to observe the effect of placing an AC voltage source in an RLC circuit.
- to plot the "response curve", or the amplitude of the oscillations versus frequency,
- to verify the equation for the response curve
The applet you have run in the pre-lab is the exact mechanical analog of the RLC driven damped oscillator circuit we will build today.This lab is a continuation of the previous lab. We will have the same RLC Circuit we had in the natural oscillation experiment, but now we will drive the circuit using a frequency generator. You should refer to you previous notes if you have questions about the theory behind the RLC circuit. I will assume everyone is comfortable with what we learned last time.
- The position of the ball is analogous to the amount charge on the capacitor plate. The velocity is analogous to current. The acceleration is analogous to the change in current per time.
- The mass of the spring system resists a change in velocity. In the circuit the inductor resists the change in current. mass --> inductor
- The damping force (the water) resisted the motion of the ball. This force was proportional to the velocity of the ball. In the circuit a resistor provides a resistive force to the charges in the wire which is proportional to the current. damping --> resistor
- The motor provided a driving force. In the circuit the AC power supply will supply the driving force. motor--> AC power supply
- The spring provided a force which was based upon the position of the ball F = -kx. The capacitor provides an electrical force based upon how much charge is on the plate. k --> 1/C
Hence, I will only state the results, which are similar to last time:
Using Kirchoff's Rules, for the RLC circuit + AC voltage source, we have the equation:L dI/dt + RI + Q/C = V0 sin(wt) This equation (again I will only state the results) has the solution:Q(t) = Q(w) sin (wt + phase) This last line means that as I change the driving frequency, the amplitude of the resulting sine wave will change. In this experiemtn we wish to determine what factors determine this amplitude.
where Q(w) is a frequency-dependant amplitude
Part I The Response Curve: Amplitude vs. frequency
1) Construct the circuit shown below. Note that the thick wires on the transformer are connected to the frequency generator, and the thin wires are connected to the RLC circuit. We will be using a new type of frequency generator.
2) Write down the values of R, L, and C, which are given.3) You will be working with frequencies in the 2000 Hz range. Set your frequency generator for this output range.
4) Find Vmax , the maximum amplitude. Record the frequency at which this maximum amplitude occurs.
5) Calculate V max / sqrt(2). Find two frequencies which have this amplitude. There is one above and one below the frequency of the maximum. Record these frequencies.
6) Record at least three more data points close to the peak.
Part II: Width vs. Resistance
1) Repeat part I for different resistances: R1, R2, R1 series R2, and R1 parallel R2
2) Determine the total resistance in each case.
3) Plot the half power width versus the resistance of the circuit.
L = ________________ +/- __________
(if you do not have a label on your solenoid, assume L = 4.5 +/- 0.5 mH)
R1 = 5.1 Ohms +/- 10% (Green Brown Gold = 5.1 Ohms)
R2 = 3.6 Ohms +/- 10% (Orange Blue Gold = 3.6 Ohms)
C = _______________ +/- __________
w0 = ______________ +/- __________ (recall w0 = 1/sqrt(LC) )
Plot 1: R1
Voltage frequency
setting
(V)
(rad/s)V max V max/sqrt(2) w1= V max/sqrt(2) w2= V 3 V 4 V 5 half-power width = |w2-w1| = _______________ +/- ___________
Plot 2: R2
Voltage frequency
setting
(V)
(rad/s)V max V max/sqrt(2) w1= V max/sqrt(2) w2= V 3 V 4 V 5 half-power width = |w2-w1| = _______________ +/- ___________
Plot 3: R1 series R2
Voltage frequency
setting
(V)
(rad/s)V max V max/sqrt(2) w1= V max/sqrt(2) w2= V 3 V 4 V 5 half-power width = |w2-w1| = _______________ +/- ___________
Plot 4: R1 parallel R2
Voltage frequency
setting
(V)
(rad/s)Vmax Vmax/sqrt(2) w1= Vmax/sqrt(2) w2= V3 V4 V5 half-power width = |w2-w1| = _______________ +/- ___________
How to do the fit:
The graph you get should give you good values for A, d, and w. I will only check your w value. It should be equal to your calculated natural frequency, within the calculated error.
1) Run Graphical Analysis (GA)
2) Enter data as usual
3) Double click on the graph window to get rid of connecting lines and add error bars.
4) Click on the graph window to select that window
5) Go the the menu Analyze and select Automatic Fit.
6) In the formula field, enter:
y = A / ( (w^2 - x^2)^2 + (d^2)*(x^2) )^0.5
and click OK. A is the value V0/LC, d is the R/L value, and w is the natural frequency.
7) When the numbers stop changing, the fit is done.
8) If you are satisified with the fit, then select that fit; otherwise, click the "try new fit" button.
1) How does increased
resistance affect the width of the V vs. w curve? Justify this by comparing
the widths of your four graphs.
2) How do your RLC graphs compare with the graphs you made for the mechanical driven damped oscillator (in the applet)? Does it appear to follow the same general fit?