Dr. Paul J. Angiolillo

Purpose

To develop an understanding of how light is refracted, to apply the concept of refraction to the study of how thin lenses form images, to learn to draw ray diagrams to assist in the predicting the locations of images formed by spherical thin lenses, and to determine the focal length of a converging lens using three different methods.

Hypothesis

When light hits the convex lens, it will refract and converge at a point in front of lens that will be construed from the rays diverged behind the lens. The distance between this point of convergence and the lens is the focal length, which should be similar in value to an indirect measurement of the focal length of the converging lens by varying the distance between the lens and an object. Using the distance between the lens and object and the lens and projected image, the focal length will be able to be determined using the thin lens equation 1/p + 1/q = 1/f. As the object is moved further past the curvature, the image will become smaller. The closer the object is moved to the lens between the curvature and focal length, the image will become larger. While the object is beyond the focal length, the image will be real and inverted. Once the object is placed between the focal length and lens, the image will become virtual and upright. Finally, the focal length interpreted using the conjugate method should also be similar to the previous focal length measurements.

Labeled Diagrams

See attached sheet.

Data

Part 1:

Part 2A:

f = 7.70 cm

Part 2B:

f_average = 7.40 cm

Part 2C:

f = 7.41 cm

Questions

1. What is the measured value of the focal length and how does it compare with the given value of focal length for the length for the lens?

The measured focal length is 7.70 cm which is fairly close to the given value of the focal length, 7.50 cm. The percent difference is 2.63%.

2. For each p and q in your data table above, calculate f. Record in your data table.

See data table.

3. Calculate and record the average value of your f’s.

The average value of the focal lengths is 7.40 cm.

4. How does the average focal length compare with the focal length printed on the lens? What is the percent difference?

The average focal length calculated is slightly less than the focal length printed on the lens. The percent difference is 1.34%.

5. Calculate the focal length of the lens using f = (D² – d²) / 4D where D is the distance between the object light and the image screen and d is the distance between the two conjugate positions.

D = 51.1 cm and d = 33.1 cm, so f = 7.41 cm.

6. How does this value for the focal length compare to the given value of the focal length for the lens? What is the percent difference?

This value for the focal length is again slightly less than the given focal length of the lens. The percent difference is 1.21%.

7. How do your focal lengths calculated from Parts 2A, 2B, and 2C compare? What is the percent difference between f from 2A to 2B? 2A to 2C? 2B and 2C?

The focal lengths from part 2B and 2C are nearly identical, while 2A is a tad larger than each of them. The percent difference between f from 2A to 2B is 3.97%, from 2A to 2C is 3.84%, and from 2B and 2C is 0.135%.

Conclusion

During part 1 of the experiment, ray diagrams of thin lenses were constructed for the following scenarios: an object beyond the curvature, an object at the curvature, an object between the curvature and focal length, an object at the focal length, and an object between the focal length and lens itself. The image size, orientation, location, and realness were predicted and recorded for each scenario. These predictions were used to aid with part 2B of the experiment.

During the half of the experiment, the focal length of a positive lens was measured in three different manners. The first of these methods, part 2A, used the application of a direct measurement to find the focal length. Equipment was set up on the optics bench so that light shone through a parallel ray lens, slit plate, and 75 mm convex lens onto a ray table. The parallel ray lens was first adjusted in order to align the light rays in a parallel manner before placing the 75 mm convex lens in front of the ray table. The light rays refracted through the lens and diverged onto the ray table. The outermost rays were traced on a piece of paper and were extended to meet at a focal point. The distance between this focal point and the lens was measured as 7.70 cm and recorded as the focal length. This was 2.63% difference from the given value of 7.50 cm as the focal length of the lens. This error can be attributed in part to difficulty keeping the tracing paper in place. It undoubtedly shifted some during tracing of the light rays. It was also difficult decipher the exact distance from that drawn point to the lens, as they were not on the same plane; the ray table was elevated and at an angle in relation to the lens. These attributions are most likely what led to that slight error.

During part 2B, the thin lens equation, 1/p + 1/q = 1/f, was used to measure the focal length. The crossed arrow target was placed at five positions beyond the measured focal length from of the lens part 2A. The distance from the lens to object (crossed arrow target) and from the lens to focused image were recorded for each instance. As the object was moved further away from the lens, the image became smaller, and as it was moved closer to the lens, the image became larger. The image was real and inverted in all cases. The average focal length from these recordings was calculated to be 7.40 cm, which was 1.34% different than the given focal length of the lens. The minor error could be attributed to difficulty locating the exact position where the image was truly focused; there was not definitive way to tell if the image was focused or not. There was likely some error in the interpretation of the positioning of the instruments as well, due to their thickness. Each measurement may have been misread by a couple tenths of a centimeter.

During part 2C, the conjugate method was used to measure the focal length. The method called for the crossed arrow target to be placed directly in front of the light source with the viewing screen at the opposite end of the optics bench, as far away as possible. The 75 mm convex lens was positioned between the object and viewing screen, adjacent to the crossed arrow target. The lens was slowly moved towards the viewing screen until a large focused image came into picture on the viewing screen. This positioning was noted and recorded as the first conjugate position. The lens was then moved towards the viewing screen until a minute focused image came into picture. This positioning was noted and recorded as the second conjugate position. The distance between these two positions was recorded as d and the distance between the crossed arrow target and viewing screen was recorded as D. These values were subbed into the equation f = (D² – d²) / 4D to find the focal length. It was calculated to be 7.41 cm using this method, which is 1.21% different than the given focal length. Error from this part of the experiment could again be attributed to possibly not getting the image in perfect focus, which would through off the positioning. Moreover, it was also difficult to read the exact positioning of the lens, object, and image as these apparatuses all had a thickness to them. As in part 2B, there may have been a couple tenths of a centimeter error in each reading. The measured and calculated focal lengths from all three methods were fairly consistent. The focal lengths from part 2B and 2C were nearly identical, while 2A was a small amount larger than each of them. The percent difference between f from 2A to 2B was 3.97%, from 2A to 2C was 3.84%, and from 2B and 2C was only 0.135%.

Equations

C = 2f

1/p + 1/q = 1/f

f = (D² – d²) / 4D

Percent Difference = |x₁ – x₂| / (x₁ + x₂)/2 x 100%

Purpose

To develop an understanding of the Law of Reflection, to apply the Law of Reflection to finding images formed by plane and spherical mirrors, and to learn to draw ray diagrams to assist in predicting the locations of images formed by spherical concave mirrors.

Hypothesis

According to the Law of Reflection, the angle of incidence will equal the angle of reflection when light is shone off a flat reflecting surface. When light is shone off a spherical mirror, it will converge at a focal point. Light will converge at a real focal point in front the concave mirror, and light will converge at a virtual focal point somewhere behind the convex mirror. An object placed beyond the curvature of a mirror will cast an inverted, shrunken, real image. An object placed at the curvature of a mirror will project and inverted, true to size, real image. Finally, an object placed between the curvature and focal point will project an inverted, magnified, real image.

Labeled Diagrams

See attached sheet.

Data

Part 1

Part 2

Part 3

Focal length of mirror f = 5.5 cm

Case 1

Case 2

Case 3

Graphs

Part 3

Questions

Part 1

1. What statement can you make regarding the relative positioning of the normal, the incident ray and the reflected ray?

The angle between the incident ray and the normal is equal to the angle between the reflected ray and the normal.

2. Do your observations validate the Law of Reflection?

Yes, the observations validate the Law of Reflection as θ_i = θ_r or the values are extremely close in all trials.

Part 3

3. Using your data above, create a graph in Graphical Analysis of pq vs. p + q. Your graph should appear linear. Perform a linear fit on the graph.

See graphs section.

4. There is an equation in geometrical optics called the mirror equation. It relates the object distance p and the image distance q to the focal length of the mirror f: 1/p + 1/q = 1/f. The mirror equation can be used to determine a mirror’s focal length. Solve the above equation algebraically for f.

In the instance of case 1, for trial 1 f = 4.0 cm, for trial 2 f = 4.1 cm, and for trial 3 f = 3.8 cm. In the instance of case 2, for trial 1 f = 4.1 cm. In the instance of case 3, for trial 1 f = 4.0 cm, for trial 2 f = 4.2 cm, and for trial 3 f = 4.3 cm. The average value for f is 4.1 cm.

5. How is your answer to Question 4 related to the slope of your graph from Question 3?

The slope of the graph from Question 3 is 4.0 cm, so these values are strikingly similar.

6. What is the percent difference between your slope and the focal length of the mirror that you measured?

The percent difference between the slope, 4.0 cm, and the focal length of the mirror that was measured, 5.5 cm, is 32%.

7. The magnification of the image of an object from a spherical mirror can also be expressed as the ratio –q/p. Calculate this ratio for each of your object and image distances and record in your data table.

See data table.

8. How does the ratio of –q/p compare to your calculated magnifications h_i/h_o for each entry? What is the percent difference?

In regards to case 1, the values for both h_i/h_o and –q/p are negative, but the values for h_i/h_oare more negative than that of –q/p. The percent difference for trial 1 is 126%, for trial 2 is 132%, and for trial 3 is 153%.

Case 2 shares the same characteristics of case 1. The percent difference is 112%.

In regards to case 3, the values are quite dissimilar because all h_i/h_ovalues are positive while all –q/p values are negative. The percent difference for trial 1 is 1200%, trial 2 is 700%, and trial 3 is 569%. It is thought that the images were recorded as upright when they were really inverted, which caused this error, but it cannot be validated by repeating the laboratory procedure at this time.

9. Do your data verify the prediction from your ray diagrams?

In regards to case 1, the values for –q/p verify the predictions made from the ray diagram, as when the object was moved further away from the mirror, the images became smaller. The values for h_i/h_o dot not support this claim however, as they say that the image way magnified, but in reality the projected image was smaller. The images were also inverted as told by the negative sign.

In regards to case 2, neither the value for –q/p nor h_i/h_o verifies the prediction made from the ray diagram. The magnification should have been 0.

In regards to case 3, the values for –q/p do verify the predictions made from the ray diagram, as when the object was moved closer to f, the images became more magnified. The images were recorded as being upright, but in reality were probably inverted as suggested by theory and the negative sign the –q/p value carries.

Conclusion

For part 1 of the experiment, the reflection of light from a plane mirror was measured. Equipment was set up on the optics bench so that light shone through a slit plate and slit mask onto a plane mirror. A ray table was used to measure the angle at which the line hit and reflected off the mirror. The ray table was rotated from 0^o to 90^o at 10^o intervals. The angle of incidence and angle of reflection were measured for each trial. The measured angles were identical or nearly identical in all trials, which seem to confirm the Law of Reflection. Any discrepancy in the measurements may be attributed to the ray optics mirror not being perfectly aligned on the ray table; without any way to secure it in place, it may have shifted slightly during some of the trials. This would have caused a difference in the angles of incidence and reflection.

For part 2 of the experiment, the focal points of a concave and convex mirror were measured. Equipment was set up on the optics bench so that light shone through a parallel ray lens and then through a slit plate and then onto the concave or convex mirror situated on a ray table. The parallel ray lens had to be adjusted to make the light rays project in a parallel fashion onto the mirror. Once parallel, the mirror was situated so that the centermost light ray would hit the center of the mirror perpendicularly. The light rays converged at a focal point which was measured and recorded. In the case of the convex mirror, a piece of paper was place underneath the mirror and the projected light rays were draw onto the piece of paper. The paper was then removed and the lines were extended to find the focal point which was located behind the mirror. In the case of the concave mirror, the focal point was in front of the mirror. The focal length of the concave mirror was 0.060 m and the focal point of the convex mirror was 0.058 m. This slight discrepancy could be attributed to difficulty tracing the lines projected by the convex mirror, but these values are rather close in value, which is expected.

For part 3 of the experiment, the cases of 3 ray diagrams were tested. Equipment was set up on the optics bench so that light shone through a crossed arrow target onto an angled spherical mirror which then reflected an image onto a viewing holder. The focal length of the mirror was first determined by placing the mirror as far away from the crossed arrow as target as possible. The viewing screen was moved to locate the point where the image of the target was focused, and that was designated as the focal point. In the case of this experiment, the focal length was 5.5 cm. The target was then placed at three positions beyond the curvature, directly on the curvature, and then at three positions between the curvature and the focal length. The viewing screen was situated in each trial to find the point where the projected image was focused. The distance from object to mirror, distance from image to mirror, height of the object, and height of the image were measured in each trial.

The results from this part of the experiment are not very consistent. In case 1, the values for both h_i/h_o and –q/p were both negative, but the values for h_i/h_owere more negative than that of –q/p. The percent difference for trial 1 was 126%, for trial 2 was 132%, and for trial 3 was 153%. The values for –q/p seems most reasonable as they predict that the image was shrunken and inverted, which was actually the case. The values for h_i/h_o suggest that the images were magnified and inverted, which was not what was observed. Case 2 shares the same characteristics of case 1, in that the value for both h_i/h_o and –q/p was negative, but the value for h_i/h_owas more negative than that of –q/p. The percent difference was 112%. During this case, it was predicted that the image would be inverted, but would be life size; not magnified or shrunken. In regards to case 3, the values were quite dissimilar because all h_i/h_ovalues were positive while all –q/p values were negative. The percent difference for trial 1 was 1200%, trial 2 was 700%, and trial 3 was 569%. It is thought that the images were recorded as upright when they were really inverted, which caused this discrepancy, but it cannot be validated by repeating the laboratory procedure at this time. The values for –q/p are most logical, as they suggest that the image was inverted and magnify, which is also what theory suggests.

The error from this part of the experiment came from the inability to distinguish when the image on the viewing screen was focused. Many times it was thought that the image was focused, but may not have truly been focused; there was not way to tell with certainty if it was focused or not. The viewing screen could be moved a few centimeters in either direction and the image would look about the same. All measurements for the height of the image are in question as well. The spherical mirror was placed at an angle in order to view the image, but this angle was never taken into consideration in any of the equations. The undoubtedly is what caused all the h_i/h_ovalues have such a stark difference from the –q/p values. It was not stated in the lab manual how to take that angle into consideration, and thus those values should most likely be thrown out. The –q/p values are most representative of the projected image, though the values for –q/p and h_i/h_o should have been equal.

Equations

θ_i = θ_r

C = 2f

1/p + 1/q = 1/f

Magnification = -q/p = h_i/h_o

Percent Difference = |x₁ – x₂| / (x₁ + x₂)/2 x 100%

Purpose

To understand the magnetic field generated by a long, straight, current-carrying conductor, the behavior of a conductor carrying a constant current in a magnetic field, and to verify the relationship associated with the force between two current-carrying conductors and the current in the wires.

Hypothesis

According to the theory that F = mg = μ_oLI² / 2πd_o, as the weight on the upper wire the increased, the current needed to return it to its original position will also increase as long as the length of the upper conductor and center-to-separation between the wires are kept constant. In addition, with the theory that d = rs / 2D and that d_o = d + r_upper + r_lower, iff the distance between the mirror and whiteboard were to increase, then less current would be needed to return the upper wire to its original position when a weight force is applied to it. Furthermore, the larger the value of s_o, the more current will be needed to for the upper wire to reach its original position when a weight force is applied to it. Finally, because the wires run in opposite directions, they will experience a force of repulsion when a current is run through them.

Labeled Diagrams

See attached sheet.

Data

First Separation

Second Separation

Graphs

Part 1

Part 2

Questions

Part 1

1. Plot a graph in Graphical Analysis of F vs. I (F is the weight force of the mass in the pan). What type of graph is it? What appears to be the relationships between force and current? Does your graph verify what is expected from theory?

It is a quadratic graph. The relationship between force and current appears to be that F is proportional to I² multiplied by some constants (μ_oL / 2πd_o). Yes, the graph verifies what is expected from theory.

2. Linearize your graph. To do this, you will need to perform some operation on the x-axis variable. Include both your graph of F vs. I and your linearized graph in your lab report.

See graphs section above.

3. What is the slope of your linearized graph?

For part one, the slope of the linearized graph is 1.19 x 10^-5 N/I². For part two, the slope of the linearized graph is 7.65 x 10^-6 N/I².

4. How does the slope of your linearized graph compare to μ_oL / 2πd_o? What is the percent difference?

For part one, μ_oL / 2πd_o = 5.99 x 10^-6 Tm / A, so the percent difference is 66.1%. For part two, μ_oL / 2πd_o = 9.65 x 10^-6 Tm / A, so the percent difference is 23.1%.

5. Assuming your linearized graph is a fairly good straight line and the slope is about right (see previous question), what physical relationships have you confirmed in this experiment?

The physical relationships that should have been confirmed are that F = mg = μ_oLI² / 2πd_o.

Conclusion

To begin the experiment, a current was connected to two parallel wires in a manner so that it would run in opposite directions. No current was actually run through the wires at this point. The length of the upper wire was measured, along with the radius of both the upper and lower wires, the distance from the mirror to the end of the upper conductor, and the distance from the mirror to the whiteboard. A He/Ne laser was turned on and reflected off the mirror onto a whiteboard, making sure that the upper and lower wires were touching. The initial position of the laser was marked on the board, and then the wires were allowed to repel to a distance of about 2 mm apart, using a counterweight to achieve this separation. The new position of the laser on the whiteboard was marked, and the distance between the two noted points was measured and recorded.

Masses were then added in increments of about 20 mg to the pan on the upper wire. At each weight, current was run through the wires to return the laser to its secondary position. After reaching the secondary position, the current was recorded and the next trial was performed, until six trials were completed. This whole process was repeated after using the counterweight to achieve an initial separation between the wires of about 4 to 5 mm.

Talk about results calculation number from questions in comparison to slope (percent difference).

Sources of error (elaborate on these a little bit and add any more you can think of): Inaccurate s_o measurement (dots marked were not extremely precise). The laser was pointed at an angle at the mirror, not head on. This would cause the calculation of d to be slightly thrown off. Hard to match laser perfection with marking because fluctuation in reaching equilibrium. Any other inaccuracies in measurements. Overall the perfect differences are probably not too bad considering all the possible sources of error.

Equations

B = μ_oI’ / 2πd_o

F = ILB = IL μ_oI’ / 2πd_o = μ_oLI² / 2πd_o

F = mg = μ_oLI² / 2πd_o

d = rs / 2D

d_o = d + r_upper + r_lower

Percent Difference = |x₁ – x₂| / (x₁ + x₂)/2 x 100%

Purpose

To measure and determine the relationship between a magnetic field generated by a line of current and a radial distance from a conductor, and to measure and determine the relationship between a magnetic field at the center of a coil and the number of turns in a coil.

Hypothesis

As the distance from the center of the conductor increases, the magnetic field strength will decrease in accordance with the equation B = μ_oI / 2πR. As the number of turns in the coil increases, the magnetic field strength will increase, in accordance with the equation B = Nμ_oI / 2R.

Labeled Diagrams

See attached sheet.

Data

Part 1

Current = 7 A

Part 2

Current = 7 A

Diameter of Cylindrical support = 0.130 m

Graphs

Part 1

Part 2

Questions

Part 1

1. Theory states that the magnetic field produced by a long straight current-carrying wire decreases in strength as you get further from the wire. The exact dependence of the magnetic field strength B on radial distance from the wire R is B = μ_oI / 2πR where μ_o is the permeability of free space and has a value of 4π x 10^-7 Tm/A. For your data from Part 1, plot a graph in Graphical Analysis of B vs. 1/R. Your graph should have a linear trend. Perform a Linear Fit on your graph. What is the value of the slope?

The slope is 9.2 x 10^-7 Tm.

2. Calculate your value of μ_oI / 2π. How does it compare to the slope of your graph? What is the percent difference?

The value is 1.4 x10^-6 Tm, which is larger in value in comparison to the slope. The percent difference is 41%.

Part 2

3. Theory states that the magnetic field produced by a circular loop of current- carrying wire increases in strength as the number of turns of wire is increased. The exact dependence of the magnetic field strength B on the number of turns N is B = Nμ_oI/ 2R where R is the radius of the loop of wire. For your data from Part 2, plot a graph in Graphical Analysis of B vs. N. Your graph should have a linear trend. Perform a Linear Fit on your graph. What is the value of the slope?

The graph of the slope is 6.53 x 10^-5 T/turn.

4. Calculate your value of μ_oI / 2R. How does it compare to the slope of your graph? What is the percent difference?

The value is 6.7 x 10^-5 T/turn, which is very similar yet slightly large in value than the slope. The percent difference is 2.6%.

Conclusion

During part one of the experiment, magnetic field strength was measured as a function of radial distance from a conductor. First, a piece of polar graph paper with concentric circles starting at a diameter of 0.5 cm increasing in increments of 0.5 cm to 10 cm was punched through a rigid aluminum conductor at the center of the concentric circles. The paper was placed on the plastic table of the apparatus and was aligned using a compass so that the parallel lines on the sheet were pointing north. The paper was then secured to the apparatus using tape. The high amperage DC power supply was connected in series with a high power resistor and the aluminum wire at the side and on top of the apparatus. The magnetic field sensor was zeroed and the DC power supply was set to 7 A. With the current on and kept constant, the magnetic field strength was recorded at each circle on the polar graph paper by holding the sensor in line with the parallel lines on the sheet and so that the white dot on the sensor was on the left and at a 90^o angle with the parallel lines pointing north. The magnetic field strength was recorded using Vernier Lab Pro. This value was recorded along with the respective radius.

Graphical Analysis was used to plot B vs. 1/R and perform a linear trend. The resulting slope was 9.2 x 10^-7 Tm and the correlation was 0.9734. It was somewhat difficult to get accurate readings at the smaller radii, which negatively affected this correlation. The slope from this plot in comparison to the value of μ_oI / 2π, 1.4 x10^-6 Tm, yielded a percent difference of 41%. The data followed the expected trend of decreasing in magnetic field strength as the radius increased. Possible sources of error include the difficulty of aligning the sensor perfectly along the radius of each circle, and it was also a challenge to get an accurate reading from the sensor as the readings kept jumping around. Additionally, if the sheet was not perfectly aligned northward, there would be interference from the earth’s magnetic pull which would have affected the readings.

During part two of the experiment, magnetic field strength was measured as a function of the number of turns in a wire. First, the power supply was turned off and the center aluminum wire was removed along with the top table of the tangential galvanometer. The magnetic field sensor was inserted into the side of the cylindrical support by screwing it into the side of the top table support using a threaded plastic bushing. The white dot of the sensor was aligned to be in the center of the cylindrical support with the white dot facing upward. The sensor was zeroed and the power supply was turn on to 7 A. A piece of flexible wire was wound around the outside of this support to complete one full turn. The magnetic field sensor measured B using “Events with Entry” on Lab Pro. The magnetic field strength was measured for six full turns of the wire. The diameter of the cylindrical support was also measured, being 0.013 m.

Graphical Analysis was used to plot B vs N and perform a linear trend. The resulting slope was 6.53 x 10^-5 T/turn and the correlation was 0.9943. The data followed the expected trend of increasing in magnetic field strength as the number of turns increased. The slope from this plot in comparison to the value of μ_oI / 2R, 6.7 x 10^-5 T/turn, yielded a percent difference of only 2.6%. Possible error could have resulted from the earth’s magnetic field. Additionally, the wire was not perfectly wrapped around the base, which would throw the readings off as it would not have been an exact turn of the wire. Finally, it was difficult to align the sensor perfectly in the middle of the support, so again this would contribute to some error in the readings.

Equations

Percent Difference = |E2 – E1| / [ (E1 + E2) / 2 ] x 100%

B x R = μ_oI / 2π

B / N = μ_oI / 2R

Purpose

To investigate the current flow and voltages in series and parallel circuits, and also to use Ohm’s law to calculate equivalent resistances of series and parallel circuits.

Hypothesis

The calculated equivalent resistances for the series circuits will abide by the equation R_eq = R₁ + R₂ and for the parallel circuits the value will be similar to 1/R_eq = 1/R₁ + 1/R₂. The current flow is expected to be uniform throughout the series circuits, but will be stronger through the smaller resistor in the parallel circuits. The voltages across each resistor should add up to V_TOT in the series circuits, and the voltages should be uniform throughout the parallel circuits.

Labeled Diagrams

See attached sheet.

Data

Part 1: Series Circuits

Power supply: 2.5 V

Part 2: Parallel Circuits

Power supply: 2.5 V

Part 3: Currents

Power supply: 5 V

Graphs

Part 3: Currents

Trial 1:

Average current through 10 Ω: 0.0655 A

Average current through 51 Ω: 0.0656 A

Trial 2:

Average current through 68 Ω: 0.0648 A

Average current through 51 Ω: 0.0925 A

Questions

Part 1:

1. Examine the results of Part 1. What is the relationship between the three voltage readings: V₁, V₂, and V_TOT?

V₁ plus V₂ is about equal to V_TOT.

2. Using the measurements you have made above and your knowledge of Ohm’s law, calculate the equivalent resistance (R_eq) of the circuit for each of the three series circuits you tested.

See data table.

3. Study the equivalent resistance readings for the series circuits. For each of the three series circuits, compare the experimental results with the resistance calculated using the rule for calculating equivalent resistance outlined in the Theory section. In evaluating your results, consider the tolerance of each resistor by using the minimum and maximum values in your calculations.

The range for the 10 Ω resistor is 9.5 Ω to 10.5 Ω. This gives a theoretical R_eq range of 19.0 Ω to 21.0 Ω for trial 1. The calculated value of 19.0 Ω falls just on the edge of this range. For trial 2, the 10 Ω resistor has the same range and the 51 Ω resistor has a range of 48.45 Ω to 53.55 Ω. This gives a theoretical R_eq range of 57.95 Ω to 64.05 Ω. The calculated value of 58.2 Ω again falls just on the edge of that range. For trial 3, the theoretical R_eq range is 96.9 Ω to 107.1 Ω, and the calculated R_eq value of 106 yet again falls within that range.

Part 2:

4. Using the measurements you have made above and your knowledge of Ohm’s law, calculate the equivalent resistance (R_eq) of the circuit for each of the three parallel circuits you tested.

See data table.

5. Study the equivalent resistance readings for the parallel circuits. Do your results verify what is expected for R_eq from the Theory section?

Yes, the calculated equivalent resistances have about the same value as 1/R_eq = 1/R₁ + 1/R₂.

6. Examine the results of Part 2. What do you notice about the relationship between the three voltage readings V₁, V₂, and V_TOT in parallel circuits?

V₁, V₂, and V_TOT are all about equal.

Part 3:

7. What did you discover about the current flow in a series circuit in Part 3?

The current flow through each resistor is equal.

8. What did you discover about the current flow in a parallel circuit in Part 3?

The current flow through each resistor differs.

9. If the two measured currents in your parallel circuit were not the same, which resistor had the large current going through it? Why?

The smaller resistor had the largest current going through it because current prefers to go through the path of least resistance.

Conclusion

During part 1 of the experiment….procedure (what was done), results, expectations and sources of error.

During part 2 of the experiment….procedure (what was done), results, expectations and sources of error.

During part 3 of the experiment…procedure (what was done), results, expectations and sources of error.

Equations

R = V / I

R_eq = R₁ + R₂ + R₃ + …

1/R_eq = 1/R₁ + 1/R₂ + 1/R₃ + …

V_TOT = V₁ + V₂ + V₃ + …

Purpose

To determine the mathematical relationship between current, potential difference, and resistance in a simple circuit, and also to compare the potential versus current behavior of a resistor to that that of a light bulb.

Hypothesis

The graph of potential difference versus current for a resistor should result in a linear plot, in accordance with the equation R = V / I. To extract even more from this equation, the slope of said linear plot will be equal to the resistance. The potential versus current behavior for that of a light bulb is predicted to not follow Ohm’s law, as light bulbs give off heat and light. It is not expected to have the same linear plot of current versus time that a resistor is characteristic of.

Labeled Diagrams

See attached sheet.

Data

Graphs

Resistor 10 Ω

Resistor 51 Ω

Light Bulb First 3 Points

Light Bulb Last 10 Points

Questions

1. As the potential across the resistor increased, the current through the resistor increased. If the change in current is proportional to voltage, the data should be in a straight line and it should go through zero. In these two examples how close is the y-intercept to zero? Is there a proportional relationship between voltage and current? If so, write the equation for each run in the form potential = constant x current. (Use a numerical value for the constant.)

The y-intercept is extremely close to zero for both the 10 Ω and 51 Ω resistor, being 0.0004256 V and -0.009978 V respectively. Yes, there is a proportional relationship between voltage and current. The equation for the first run (10 Ω resistor) is U = 9.655 I and the equation for the second run (51 Ω resistor) is U = 48.94 I.

2. Compare the constant in each of the above equations to the resistance of each resistor.

The constants are nearly equal to the resistance of their respective resistor.

3. The constant you determined in each equation should be similar to the resistance of each resistor. However, resistors are manufactured such that their actual value is within a tolerance. For most resistors used in this lab, the tolerance is 5% or 10%. Check with your instructor to determine the tolerance of the resistors you are using. Calculate the range of values for each resistor. Does the constant in each equation fit within the appropriate range of values for each resistor?

The range for the 10 Ω resistor is 9.5 Ω to 10.5 Ω. The determined constant of 9.655 fits into this range. The 51 Ω resistor has a range of 48.45 Ω and 53.55 Ω. The determined constant of 48.94 fits within that range.

4. Do your resistors follow Ohm’s law? Base your answer on your experimental data.

Yes, they follow Ohm’s law because the correlation for linear fit for each trial is equal to 1.000.

5. Describe what happened to the current through the light bulb as the potential increased. Was the change linear? Since the slope of the linear regression line is a measure of resistance, describe what happened to the resistance as the voltage increased. Since the bulb gets brighter as it gets hotter, how does the resistance vary with temperature?

The change was more curved at low voltage, but became more linear as the voltage was raised higher. Resistance increased as the voltage increased, as the slope was 3.261 V/A for the first 3 points and 37.66 V/A for the last 10 points. Since the bulb gets brighter as it gets hotter, that means resistance increases with temperature.

6. Does your light bulb follow Ohm’s law? Base your answer on your experimental data.

The light bulb seemed to follow Ohm’s law for the first few points, but the data began to curve into a steeper linear plot as voltage was increased. This is evident by the change in slop from 3.261 V/A for the first 3 points and 37.66 V/A for the last 10 points. The y-intercept for the last 10 points is also non-zero, which goes against Ohm’s law.

Purpose

To determine the spring constant of a spring by measuring its stretch versus applied force, to determine the spring constant of a spring by measuring the period of oscillation for different masses, and also to investigate the dependence of period of oscillation on the value of mass and amplitude of motion.

Hypothesis

If the applied force (mass) is to remain the same while the vertical displacement is increased, there period will remain the same. However, if the vertical displacement is held constant while the applied force in increased, the period will increase. This is in accordance with the derived equation T = 2π (M / k)^0.5.

Labeled Diagrams

See attached sheet.

Data

Part I

k = 3.53 N/m

Part II

Part III

m_s = 0.00940 kg

Slope = 10.5 s²/kg

Y-Intercept = 0.0687 N

k = 3.75 N/m

C = 0.694

% difference = 6.04%

Graphs

Part I

Part II

Part III

Questions

1. Do the data from Part 1 verify Hooke’s Law? State clearly the evidence for your answer.

The data correlate close to Hooke’s Law, but not quite. The law states that F = -ky, where F is in this case Mg and y equals the negative displacement. After graphing forces versus displacement, a value of 3.53 N/m was determined as the spring constant. However, when applying this value to the equation and using recorded displacement values, the calculated force come up less than the actual for used. For example, in the first trial y = -0.055 m. Multiplying that value by the extrapolated spring constant gives a theoretical force of 0.194 N, but the actual force used was 0.244 N. All other trials yield a similar lowball theoretical force.

2. How is the period T expected to depend upon the amplitude A? Do your data confirm this expectation?

Period is not expected to depend upon amplitude, as suggested by the equation T = 2π (M / k)^0.5, where amplitude is absent as a variable. The data confirms this expectation, as the period was nearly the same for each trial.

3. Consider the value you obtained for C. If you were to express C as a whole number fraction, which of the following would best fit your data (1/2, 1/3, 1/4, 1/5)?

The obtained value of C is 0.694, which is closest to 1/3.

4. Calculate T predicted by the equation T = 2π (M / k)^0.5 for M = 0.050 kg. Calculate T predicted by T = 2π ( [M + Cm_s] / k )^0.5 for M = 0.050 kg and your value of C. What is the percent difference between them? Repeat for a value for M of 1.000 kg. Is there a difference in the percent differences? If so, which is greater and why?

T predicted using the first equation and M = 0.050 kg is 0.726 s. T predicted using the second equation and M = 0.050 kg is 0.771 s. This is a percent difference of 6.01%.

T predicted using the first equation and M = 1.000 kg is 3.24 s. T predicted using the second equation and M = 1.000 kg is 3.26 s. This is a percent difference of only 0.615%. The greater percent difference occurs at the lower weight because the weight of the spring is almost insignificant at higher weight. The proportion of the mass to the spring is so great that it has almost no effect on the calculation.

Conclusion

During part one of the experiment, the vertical displacement of a spring was measured as a function of force applied to it. The starting position of the spring was recorded using a stretch indicator. Mass was added to the spring, and the displacement was recorded. This was repeated with various amounts of mass. From these data, a graph of force versus displacement was plotted, and a linear fit slope revealed the spring constant. In this endeavor, the spring constant was valued at 3.53 N/m.

However, when applying this spring constant to the recorded displacements in the Hooke’s Law equation, the calculated forces are lower than the recorded forces. In similar manner, when rearranging Hooke’s Law to solve for displacement, the calculated displacements are larger than the actual recorded displacements. This means there was human error, most likely in terms of not being precise with the displacement readings because the recordings for the masses used were accurate. Because such small masses were used, any error in displacement readings was augmented. The spring used may also not have been perfect.

During part two of the experiment, the period of the spring was measured as amplitude changed while mass remained constant. The period remained nearly the same throughout every trial, which was to be expected. Any differences in period may be accounted to inadequate stopwatch usage and inaccurate starting displacements throughout the trials. It should be noted that at amplitude of 0.100 m, the hook lost contact with the spring for a split second at the apex of oscillation, which accounts for its oddity in period. This was something that could not be avoided; at that amplitude the spring pulled the hook up too quickly which caused the loss of contact. The resulted graph of period versus amplitude yielded a linear fit slope of close to 0 (-0.247 s/m), which was predicted.

During part three of the experiment, the period of the spring was measured as mass was varied while amplitude remained constant. As the mass was increased, the period also increased. This was not surprising considering the given equations. The square of the period versus mass for each trial was plotted and a linear fit was taken. The slope and y-intercept of this line was then used to determine the spring constant and C, the fraction of the spring’s mass that should be taken into account for the equation T = 2π ( [M + Cm_s] / k )^0.5. The calculated value of k was 3.75 N/m, which is only 6.04% different from the value determined earlier of 3.53 N/m. The value of C was determined to be 0.694, which is closest to the whole number fraction of 1/3. Any error during parts two and three can be attributed to inaccurate stopwatch recordings and slight variance in displacement and release of the masses at each amplitude.

Equations

T = 2π (M / k)^0.5

T = 2π ( [M + Cm_s] / k )^0.5

slope = 4π² / k

y-intercept = 4π²Cm_s / k

Purpose

To compare the moments of inertia calculated using two different methods, and to verify that angular momentum is conserved in an interaction between a rotating disk and a ring dropped onto the disk.

Hypothesis

If a weighted ring is added to the disk, the moment of inertia will be the same as the disk without the weighted ring. The angular momentum before the ring is dropped on the disk during part two will be greater than the angular momentum after the ring is dropped.

Labeled Diagrams

See attached sheet.

Data

Part 1

Mass of disk (M): 1.500 kg
Radius of disk (R): 0.114 m
Radius of shaft (r): 0.006 m
Mass of ring (m): 1.420 kg
Inner radius of ring (R₁): 0.054 m
Outer radius of ring (R₂): 0.064 m

Disk Alone

Disk plus ring

Part 2

Graphs

Part One (Disk Alone)

Part One (Disk Plus Ring)

Part Two

Questions

Part 1

1. In your data table in Part 1, you have two values for the moment of inertia. One is found from the theoretical equation for moment of inertia that is introduced in the Theory section and other is an experimental value obtained using Newton’s 2^nd law for rotational motion, τ = Iα, in conjunction with the definition of torque, τ =rF. How well do your two values agree with each other? What is the percent difference? Which do you think is likely a better way to calculate a value for moment of inertia?

The values are extremely close, as the percent difference for the disk alone is 9.5% and the disk plus the ring is 0% (they are of equal value). I think the better way to calculate the moment of inertia is to use I = ½ MR², as it is a more elegant equations that takes into account less variables. The other equation takes more variables into account, mainly for calculation torque, which I feel leads to increased error.

Part 2

1. How do your values for the angular momentum before and after the ring is dropped onto the disk compare? What is the percent difference?

The angular momentum before the ring is dropped onto the disk is greater than the angular momentum after the ring is dropped onto the disk. The percent difference is 11.7%.

2. Does there appear to be an inverse relationship between moment of inertia and angular velocity?

No, there appears to be a direct relationship between moment of inertia and angular velocity. As the angular velocity decreased, so did the moment of inertia.

3. How well do your results support the theory of conservation of momentum? What are the limitations of the experimental setup?

The results somewhat support the theory of conservation of momentum. The percent difference is 11.7%, which I suppose isn’t a huge discrepancy, but it could be better. The limitations of the experimental setup were that it is difficult to drop the ring on the spinning disk perfectly. We were able to drop the ring into the grooves of the disk, but there was still some wiggle room in those grooves. The ring would need to fit in the grooves like a puzzle piece in order to be positioned dead center to yield the least amount of error.

Conclusion

Lab Summarized

During the first part of the lab, the moments of inertia for a spinning disk with and without a weighted ring on top were calculated using two different methods. The force coercing the disk to spin was a 300 g weight attached to the shaft of the disk using a string a pulley system. The weight was allowed to free fall and the resulting graph of velocity versus time was used to find the final angular velocity by taking the mean of the segment after which the string had completely unraveled from the shaft. The angular acceleration was found from the slope of this graph up to that point.

These values, along with the force of kinetic friction, found by determining the minimum force needed to get the disk spinning, were used to find the moment of inertia. The moment of inertia was also calculated a second way, using the radii and masses of the disk and ring.

The second part of the experiment was performed much like part one of the experiment using the disk alone, only this time shortly after the string had unraveled, the ring was dropped onto the spinning disk. Using the angular velocities and moment of inertias, determined much like they were in part one, the angular moments before and after the ring were dropped were calculated and compared.

The percent differences between the two different calculations of the moments of inertia in part one were quite low. Using the disk alone, the percent difference was 9.5% and with the disk plus the ring, the percent difference as 0%. Under perfect conditions, the values should have been equal. The angular acceleration and final velocity for the disk along were greater than that of the measurement for the disk plus the ring, which could be expected, due to the extra mass. The percent difference between the angular momentums before and after the ring was dropped in part two was 11.7%. They should have been equal under ideal conditions.

As stated previously in the questions, some of this error is most likely due from the ring not being placed dead center around the spinning disk. If the disk was dropped at and angle and did not make complete contact with the disk at the same instant, this could have also caused error. If the shaft was not properly lubricated, this would have caused error throughout the experiment. Lastly, if the string ever caught a snag while unraveling, this would have also contributed to the error.

Equations

I =τ/α
I = ½ MR2
L = I_fω_f
I_iω_i = I_fω_f

Purpose

To utilize two different methods of determining the initial velocity of a fired ball, namely a ballistic pendulum and treating the ball as a projectile, and then compare these two calculated values. The loss of kinetic energy from firing the ball into the pendulum is also an area of interest.

Hypothesis

The initial velocity determined by firing the ball into the ballistic pendulum should theoretically be equal to the initial velocity determined by firing the ball as a projectile.

Labeled Diagrams

See attached sheet.

Data

Part One

v= 6.23 m/s α_v=0.0187 m/s

Part Two

Y=1.003 m v= 8.75 m/s α_v=0.120 m/s

Questions

1. Compare the two different values of v average. Calculate the percent difference between them. State whether the two measurements agree within the combined standard errors of the two values of v average.

The average initial velocity for the ballistic pendulum was 6.23 m/s while the average initial velocity for the projectile determination was 8.75 m/s. This is a percent difference of 33.6%. It should have been expected that these two values would be equal. The two measurements also do not agree within the combined standard errors of the two values for v average, as the standard errors only total 0.1387 m/s, and the average velocities fall out of that range.

2. Calculate the loss in kinetic energy when the ball collides with the pendulum as the difference between ½ mv² (the kinetic energy before) and ½ (m + M)V² (the kinetic energy immediately after the collision). What is the fractional loss in kinetic energy? Calculate by dividing the loss by the original kinetic energy.

The average kinetic energy before the collision is 1.35 J and the average kinetic energy immediately after the collision is 0.272 J, so the loss of kinetic energy is 1.08 J. The fractional loss in kinetic energy is 0.8.

3. Calculate the ratio M / (m + M) for the values of m and M in Part 1. Compare this ratio with the ratio calculated in the previous question. Express the fractional loss of kinetic energy in symbol form and use equations from the Theory section to show it should equal M / (m + M).

The ratio M / (m + M) is equal to 0.8. This ratio is exactly the same as the fractional loss of kinetic energy.

The fractional loss of kinetic energy equals ( ½ mv² – ½ (m + M)V² ) / ( ½ mv² ).

Conclusion

During part one of the experiment, a ball was fired into a ballistic pendulum to ultimately determine its initial velocity. This process was repeated five times in order to obtain average values to work with in order to eliminate error. By massing the ball and the pendulum, recording the initial and final heights, the values for V and finally v could be calculated. It was found that the average initial velocity of the ball was 6.23 m/s.

During part two of the experiment, the same ball was fired as a projectile instead of into a ballistic pendulum. The ball was fired from a table horizontally to the ground. A piece of carbon paper was used to capture the spot where the ball first struck the ground. Height and horizontal distance the ball traveled were then measured in order to determine the initial velocity of the ball. The average initial velocity of the ball was 8.75 m/s.

As far as the accuracy of the results from the lab, the percent difference between the average velocities calculated is 33.6%. This is a fairly significant difference, which suggests that there sources of error during the procedure. The notched part of the ballistic setup could have had finer groves to yield more accurate measurements. The major contributor of error, however, was most likely from the distance measurements from the projectile part of the lab. One positive to come from the results was that the fractional loss in kinetic energy was identical to the mass ratios from the ballistic pendulum setup, which is theoretically expected.

Equations

∆KE = ½ mv²

½ (m + M)V² = (m + M)gh

mv = (m + M)V

V = (2gh)^0.5 v = (m + M) (2gh)^0.5 / m v = ∆x / (2∆y / g)^0.5

Purpose

To determine the relationship between force, displacement, potential energy, kinetic energy, and work by using a force sensor to pull a spring and also to push a cart.

Hypothesis

The work done on the spring will be greatest at its furthest displacement and that the greater the work done on the cart, the greater its acceleration will be.

Labeled Diagrams

See attached sheet.

Data

Part One

Stretch

Part Two

Weight of cart: 5.29 N

Mass of cart: 0.54 kg

Graphs

See attached sheets.

Questions

1. In Part 1 you did work to stretch the spring. The graph of force vs. distance depends on the particular spring you used, but for most springs will be a straight line. This corresponds to Hooke’s law, of F=-kx, where F is the force applied by the spring when it is stretched a distance x. k is the spring constant, measured in N/m. What is the spring constant of the spring? From your graph, does the spring follow Hooke’s law? Do you think that it would always follow Hooke’s law, no matter how far you stretched it? Why is the slope of your graph positive, while Hooke’s law has a minus sign?

The spring constant is 81.668 N/m. From the graph, it does appear that the spring follows Hooke’s law as it produced a fairly straight line. I think the spring would follow Hooke’s law until it is all the way stretched out and cannot be stretched any more, or if breaks. The slope of the graph is positive as it is showing force applied on the spring. Hooke’s law shows the force applied by the spring, so that would be in the opposite direction in which it is pulled, thus being negative.

2. The elastic potential energy stored by a spring is given by ∆PE = ½ kx², where x is the distance. Compare the work you measured to stretch the spring to 10 cm, 20 cm, and the maximum stretch to the stored potential energy predicted by this expression. Should they be similar?

The ∆PE for my intervals was 0.20 J at 7 cm, 0.80 J at 14 cm, and 2.16 J at 23 cm. The ∆PE increased as the displacement is increased. This should be expected, as the spring becomes harder to pull the more it is stretched out.

3. In Part 2 you did work to accelerate the cart. In this case the work went to changing the kinetic energy. Since no spring was involved and the cart moved along a level surface, there is no change in potential energy. How does the work you did compare to the change in kinetic energy. Here, since the initial velocity is zero, ∆KE = ½ mv² where m is the total mass of the cart and any added weights, and v is the final velocity. Record you values in the data table.

The work done, 0.04796 J, is greater than the ∆KE, which is 0.027 J.

Conclusion

Lab Summarized

The overall goal of the lab was to investigate the relationship between work, potential energy, and kinetic energy. The goal was achieved using a spring and force sensor along with a motion detector to determine the work done on the spring when pulling it. Using the motion detector, displacement was determined, which could then be used to determine the spring constant from Hooke’s law. The force and acceleration were also collected using the force sensor and motion detector. Graphs produced of force versus position could be integrated to find the work done on the spring over certain intervals. The slope of the linear fit of this graph could also be used to produce the spring constant. Finally, the elastic potential energy stored in the spring could be determined from the aforementioned data.

During the second part of the experiment, the force sensor was used to push and thus accelerate a cart on the frictionless track toward a motion detector. The measured weight of the cart and final velocity could then be used to determine the change in kinetic energy of the cart. The work applied could again be determined by taking an integral over the time period in which the cart was pushed and accelerated with the force sensor.

The data collected for part one seems fairly conclusive. The determined spring constant of 81.668 N/m is comparable to known spring constants. The values for work extrapolated by integrating the graphs of force versus position are extremely close to the calculated values for work, or potential energy. The values of 0.1779 and 0.20 J, 0.8080 and 0.80 J, and 2.152 and 2.16 J are nearly identical, which shows part one of the experiment was performed rather well (or luckily). In regards to part two of the experiment, the work done on the cart found by integrating the graph of force versus position, 0.04796 J, is almost twice as large as the calculated change in kinetic energy, 0.027 J. The only way that this could be accounted for is if the wrong interval was used for the integral on the graph. Looking at the graph, it seems like the integral taken should have been from the start to the top of the peak, and not the whole peak.

Equations

W = F * s

W = ∆PE + ∆KE

∆PE = ½ kx²

∆KE = ½ mv²

F = -kx