In order access the oblong conductor website, use the following information, and choose
Study the E-field at a Point
We used the following word document as a lab procedure in the PHY112 and PHY131 classes.
Obtain the output of basic circuits using Bode plots
What are Bode Plots? Bode plots sketch the AC amplitude and phase at a chosen node in a circuit or control loop, as a function of frequency. They show, for example, the high or low frequency cutoffs of a filter circuit, or the positions of resonance peaks.
Why Bode Plots? Getting the response of simple networks is very easy and intuitive using this engineering technique. They provide a fast way to visualize the frequency response of basic networks with a minimum of complex algebra.
This post shows a simple progressive explanation of reactance and impedance. Step 1 is to plot reactances for basic circuit elements: resistors, capacitors, and inductors. Step 2 is to define and use the impedance concept. The final step in most textbooks is to rotate the impedance vector as a phasor. The discussion here departs from that approach and switches to s-space (as used in Laplace transforms). It is possible to do inverse transforms on s-space functions to obtain circuit response to transients, but that is not the purpose here. Rather Step 3 in this ‘lab experiment’ is to determine the poles and zeroes of the transfer function of the circuit. The result is an immediate sketch of the output amplitude and phase of any chosen node in the circuit as a function of frequency.
Simple Example
This simple example is a high pass filter consisting of an inductor and a resistor.
Step 1: The resistance is R and the reactance is just ωL, where ω = 2π*frequency. These two components of impedance can be plotted as perpendicular vectors on the complex plane:
In the complex plane, R is a real quantity, and the reactance of the inductor is a positive imaginary (often written jωL, where j =√(-1) ). If the impedance were purely real, as in the case of a resistance, then the current and voltage have the same phase. If the impedance were purely inductive – a positive imaginary – then the current would be 90 degrees out of phase with the voltage. In fact considering the voltage as a real V_{0 }cos (ωt), and dividing by a positive imaginary, jωL, the resulting current is a negative imaginary. Thus the current vector lags 90º behind the voltage vector as these two phasors rotate counterclockwise on the complex axes. Thus the voltage ‘leads’ current in a pure inductor. The current is –(V_{0} /ωL) sin (ωt).
Step 2: In this simple example, the net impedance Z = (R ^{2} + (ωL)^{2} )^{½} . The net impedance is indicated in the above graph by the dashed arrow. The phase angle between voltage and current is the angle between Z and the real axis in the figure, thus φ = tan^{-1} (ωL/R). The voltage leads the current by the angle φ.
Power is not dissipated in the reactance L because the current and voltage are 90º out of phase in the inductor. Therefore the entire power dissipation in the circuit is in the resistance R. Since the current in the circuit is I_{rms} = V_{rms}/|Z|, the power in the circuit, equal to the power in the resistor, is I_{rms}^{2} R = (V_{rms }/ |Z| )^{2} R = V_{rms} I_{rms} cosφ.
Step 3: The above circuit is a simple voltage divider, where the voltage V_{in} is divided by the network to give V_{out }= V_{in } jωL/Z . These are complex numbers, dependent on frequency. The Bode plot allows a quick graphical analysis of the high pass filtering effect. Divide by V_{in } and take the Laplace transform of both sides, substituting the variable ‘s’ for jω:
H(s) must be put in the form of a quotient of two polynomials in s, and preferably the polynomials are factored. Then directly determine the zeroes in the numerator: s = 0 in this case; and the zeroes in the denominator: s = –1/τ. Use this information to generate the Bode plots. Go to the Bode plotting website: Bode website
Suppose for example that R = 10 and L = 25 mH. Then τ = .0025 sec. The zero of H(s) is at s = 0, and the pole is at s = -400 radians/sec. Entering this info in the Bode website obtain the following result:
The Bode plot is the Asymptotic plot. It is a quick approximation to the Real plot, and there are simple rules for drawing it. It shows that the transmission of the filter is very low at small frequencies, and becomes flat for frequencies above ∼400 rad/sec. The phase plot is also available:
The phase plot shows there is a 90 degree phase shift at low frequencies due to the inductor. But at high frequency, where the impedance of the inductor becomes much greater than that of the resistor, the phase tends toward zero degrees as if the inductor were an open circuit.
We provided a few editable worksheets to download for students in PHY131 (Physics Science and Engineering students).
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In order access the dual current-loop website, use the following information:
We used the following word document as a lab procedure in the PHY112 class. However, it can be used with more advanced or engineering-level classes, because the concepts are significant. Document:
In order access the dual current-loop website, use the following information:
We used the following word document as a lab procedure in the PHY112 class. However, it can be used with more advanced or engineering-level classes, because the concepts are significant.
In order access the capacitor simulation website, use the following information:
We used the following word document as a lab procedure in the PHY112 class. However, it can be used possibly in PHY101 as well, because questions are qualitative.
In order access the electric dipole simulation website, use the following information:
We used the following word document as a lab procedure in the PHY112 class. However, it can be used possibly in PHY101 as well, because questions are qualitative.
Fingers held at arm’s length are calibrated to provide an angular measure. As the space station crosses each finger on the hand of an up-stretched arm, record the time on a stopwatch or cellphone stopwatch App. This gives the angular velocity. Combine this with the distance up to the space station to get the ISS velocity.
The ISS will be visible only in the morning and the evening, because for most of the night, the Earth casts its shadow over the shuttle orbit. The shuttle travels from generally West to generally East; therefore it is very brightly illuminated soon after it rises in the morning, and a little less brilliant when it rises in the evening, due to the relative positions of the sun, the ISS, and the observer.
Hint: holding the hand absolutely still is difficult. It may help if there is a star or tree branch above the hand to help you steady your hand.
Let the finger width be w, and the distance from the eye to the finger be r. Then the angular displacement of any object whose image crosses the finger is Δθ = w/r. Let Δt be the time to cross over a finger. Then calculate the angular velocity as ω = Δθ/Δt = w/(rΔt).
To find the velocity of the ISS, multiply the angular velocity by the distance R from the eye to the ISS: v = ωR. A slight complication is that the distance from the observer to the ISS depends on the elevation angle. If the ISS passes directly overhead, so maximum elevation = 90ᵒ, then R is just the typical distance between the Earth and the space station, R = 249 +_ 5 mi = km. If the maximum elevation angle of the ISS is θ, then R = 249/sin θ.
On the morning of March 30, 2015, in Phoenix, AZ, USA, the table in the Spot the Station website shows that the ISS first became visible 10ᵒ above the horizon, reached a maximum elevation of 77ᵒ, and travelled for ≈ 6 minutes before disappearing at an elevation of 12ᵒ.
Holding my hand with fingers straight but slightly apart, so I could see narrowly between my fingers, my hand-width is 8.0 cm. Hence each finger averages 2.0 cm width. The average time to cross a finger was 3.0 s.
When I hold my bent hand above my head, the fingers are 33 cm above my eyes. Therefore the angular velocity of the space station appeared to me as ω = (2.0 / 33 * 3.0) = 0.0202 radians/sec. If the ISS had passed directly overhead, the velocity would just be ω*R, with R = 249 miles. But since the elevation was 77ᵒ instead of 90ᵒ, the perpendicular distance between the object and me (the observer) was actually the hypotenuse of a triangle whose height was 249 miles and whose angle was 77ᵒ. Therefore R = 249 / sin 77ᵒ = 256 miles. Converting to kilometers, R = 256 mi *1.61 km/mi = 411 km, and v = 0.0202 * 411 = 8.3 km/s. This is in reasonable agreement with look-up values – 7.6-7.7 km/s.
All low orbits — including the space station– move at about 5 miles (or 8 km) per second. Since the space station is about 250 miles above the Earth, the angular velocity of motion past an earthbound observer is about 5 /250 = .02 radians per second.
It is exciting to realize that while you are viewing the ISS for 6 minutes, it is moving across the country a distance of 300 s x 5 = 1500 miles or 2400 km. It is exciting to think that when the ISS is just going out of your view, it is passing over some far-away place that is 750 miles distant!
]]>Place a lightweight metalized ball on an electronic balance. The scale registers any changes in the force on the ball. A charged rod held over the ball induces an opposite charge in the ball, because the ball is grounded to the scale. It is not a bad approximation to assume the charges are equal and opposite point charges, separated by the distance between the top of the ball and the bottom of the charged rod. Use Coulomb’s law to calculate the amount of charge. Vary the distance to show the inverse square law.
F =kq^{2}/r^{2}
Solve the force equation for q. For example, if the scale reads -0.5 gram for a distance of 4 cm, this gives q = [(.0005 x 10 x .04^{2}/(9 x 10^{9} ) ]^{1/2} = 30 nC †. A few tens of nanocoulombs is about as much charge as one produces in the classroom, while the class van der Graaf machine accumulates a couple of hundred nC.
Students give the ball a consistent x-velocity by rolling it down an incline from a fixed height they choose. They figure out its x-velocity by seeing how far from the table edge the ball lands on the floor. Then they put the magnet in the path of the ball — at position M in the drawing — and repeat the process so that the ball and magnet hit and attach as they launch off the table. Again they find the new distance the pair goes before hitting the floor to measure the velocity after collision.
To measure the velocity of an object such as the steel ball just before rolling off the table alone, or the steel ball after sticking to the magnet, students must:
We used the following word document as a lab procedure in the PHY121 class. However, it can be used in PHY111 or possibly in PHY101 as well.
PHY121 Worksheet for Momentum Conservation Lab
Terryl Fender, Prof. of Physics, South Mountain Community College, Phoenix, AZ
]]>Smart phones are expensive, but almost every student has one – everywhere in the world! Neither the school nor the teacher need to buy this powerful piece of equipment to use for physics lab. In this experiment, students use inexpensive cell phone Apps to generate tones and to measure frequency.
When the air in a test tube vibrates, the air motion must be a maximum at the mouth of the tube, and zero at the closed end. The air motion for the fundamental (lowest possible) frequency is a ‘breathing’ mode, as indicated in the figure: The distance between a node (zero motion at the left end) and the next consecutive antinode (maximum motion at the right open end) is 1/4 of the wavelength. Therefore a full wavelength is 4× the length of the test tube.
Air motion in an open tube ‘breathes’ symmetrically at both ends, so the fundamental resonance looks like: The distance between a node and the next consecutive antinode is 1/4 of the wavelength, so the full wavelength of the fundamental wavelength is 2× the length of the open tube.
I will be attaching videos here to demonstrate the open and closed tube resonances. The procedure for the closed tube is:
c = f * λ = f * L/4
The procedure for the open tube is a little different:
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