Cheng,+Halowell,+Huddleston,+Samani

flat

= Lab: Traverse Waves, Standing Waves on a String =

**Hypothesis:** -As tension increases the frequency must increase in order to make the same number of anti-nodes. This is due to inertia. With more tension on the string it takes more energy to move the string. We know this based on a few equations. so we can substitute that into another equation to come up with. This shows that the relationship will not be linear though because of the square root. -Frequency is inversely related to the wavelength. We know this from the equation where the velocity is constant. -We already know that as the frequency increases the wavelength decreases. When the wave lengths decrease there can be more antinodes on the string.


 * Materials:**

String, masses, wave generator, spreadsheet


 * Procedure:**


 * 1) Determine the amount of antinodes to look for.
 * 2) Adjust frequency until the desired amount of antinodes has been acquired.
 * 3) Fine-tune the frequency until it is at its maximum amplitude.
 * 4) Record Data.
 * 5) Repeat.

(for tension vs. frequency)


 * 1) Set a mass.
 * 2) Choose a constant for the amount of antinodes to look for.
 * 3) Adjust frequency until the desired amount of antinodes has been acquired.
 * 4) Fine-tune the frequency until it is at its maximum amplitude.
 * 5) Record Data.
 * 6) Repeat with different masses.

Photo of Set-Up:

Data:

Graphs: Experiment 1: As expected, there is a direct relationship between the number of antinodes and frequency. Increasing frequency decreases wavelength, because those two variables are indirectly related by the equation, wave speed = frequency * wavelength. A decreased wavelength then causes the number of antinodes to increase, because wavelength and (n) are inversely related by the equation, Length = (n)(0.5)(wavelength). The coefficient in from of the graph should be close to 20, becuase that reflects the number of antinodes times the wave speed divided by the quantity of length of the string and two).

Experiment 2: This graph demonstrates an exponential inverse relationship between frequency and wavelength. waves speed= frequency*wavelength. Therefore, frequency is equal to wave speed *(1/wavelength). The exponent should be very close to negative one, because the wavelength variable exists below the denominator. The coefficient is wave speed.

Experiment 3: Tension and frequency have a direct relationship. Using the equation for wave speed on a string and the equation for resonance, you realize that frequency is related to the square root of Tension, causing the variable x in the graph to have an exponent close to -0.5.

Calculations:

**//Discussion Questions: //** 1. Calculate the tension T that would be required to produce the n = 1 standing wave for the red braided string. 2. What would be the effect if the string stretched significantly as the tension increased? How would that have affected the data? This would increase the elastic property of the wave, which consequently increases wave speed. Increasing the wave speed affects the wavelength and frequency. Wavelength would decrease. Frequency would increase. 3. What is the effect of the type of string on the amount of hanging mass needed to create a set number of nodes? Explain this. The type of string affects the tension in the string and its elastic property. The stronger the string, the higher the mass needed to attain the same tension as a weaker string. 4. What is the effect of changing frequency on the number of nodes? Changing the frequency changes the interference of the two waves. If you change the frequency enough, the two waves will eventually have resonance again, but at different points than before. The places where a crest meets exactly with a trough are called nodes. 5. What factors affect the number of nodes in a standing wave? Based on the equation of a standing wave, length of the string and wavelength affect the number of antinodes. Wavelength is affected by wave speed. Increasing wave speed increases wavelength. The number of antinodes is also determined by the frequency, because frequency changes the pattern of interference. Our results also show that increasing tension also increases the frequency. The number of antinodes are related to the number of nodes on a standing wave.

For one part of the experiment we hypothesized that as the tension increases so does the frequency when trying to keep the number of antinodes constant. We did need keep the number of antinodes constant and rather we kept the tension constant. Our hypothesis was proved correct because since the tension remained constant as the frequency increased so did the number of antinodes. For example when the frequency was 19.3Hz there was one antinode and when the frequency was increased to 38.2Hz there were 2 antinodes. For the second part of the experiment we looked at the relationship between frequency and wavelength. We were right in our prediction that as the frequency increased the wavelength would decrease. This is because as was previously proved true as the frequency increases the number of antinodes increases. As the string is divided into more antinodes each wavelength is decreased. We can see this in our data for example when the frequency was 19.3Hz the wavelength was 3.2m and when the frequency increased to 38.2Hz the wavelength decreased to 1.6m. For the final part of this lab we were observing the relationship between wavelengths and antinodes. We predicted that as the wavelength decreased the number of antinodes increased. We found that the wavelength was definitely dependent on the number of antinodes. In experiment 3 we kept the same number of antinodes and changed the frequency and the wavelength stayed the same. Looking at the data from experiment 2 we can see that when there was 1 antinode the wavelength was 3.2m and when there were 2 antinodes the wavelength went down to 1.6m. The lab results came out as hypothesized. When the tension increased, the frequency increased. This can be observed in the table for the third experiment. In experiment two, it can be observed that the wavelength decreased as the frequency increased, as predicted in the hypothesis. And as the frequency increased, the number of antinodes increased in a direct relationship. However, there were sources of error, as with any experiment. Our error was very small as shown by the R^2 values, but there was still error regardless. For example, the smallest increment someone could adjust the wave maker was by .1 Hz. If we were able to use a wave maker that allows the adjustment of the Hz in smaller increments, the results would have been more exact. Also, if the string had any imperfections, cuts, tears, etc., the lab data could have been changed. This would yield less accurate data. These patterns of interference are important in understanding the behavior of waves. Waves are constantly colliding in our world, and their frequencies and amplitudes can cause different effects of destructive and constructive interference. The applet demonstrated that even in simple examples, such as a pebble thrown into a pool of water, interference affects the waves.
 * Conclusion **

=Lab: What is the relationship between the mass on a spring and its period of oscillation? =  Sam Cheng, Chris Halowell, Maddie Huddleston, and Bret Pontillo <span style="font-family: Arial,Helvetica,sans-serif;">**Hypothesis** <span style="font-family: Arial,Helvetica,sans-serif;"> Mass is directly related to period, but with a square root exponent. Because T=2pi*sqrt(m/k), increasing m, increases the period, but not in a linear manner.

<span style="font-family: Arial,Helvetica,sans-serif;">**Objectives** <span style="font-family: Arial,Helvetica,sans-serif;"> -To directly determine the spring constant k of a spring by measuring the elongation of the spring for specific applied forces. <span style="font-family: Arial,Helvetica,sans-serif;"> -To indirectly determine the spring constant k from measuring of the variation of the period T of oscillation for different values of mass on the end of the spring. <span style="font-family: Arial,Helvetica,sans-serif;"> -To compare the two values of spring constant k.

<span style="font-family: Arial,Helvetica,sans-serif;">**Materials** springs, tape, clamps and rods, masses, balance, stopwatch, meterstick, ruler

<span style="font-family: Arial,Helvetica,sans-serif;">**Procedure** <span style="font-family: Arial,Helvetica,sans-serif;"> 1. Set up the spring by hanging it with a clamp. Attach a meal rod to the end of the spring, to attach masses.

<span style="font-family: Arial,Helvetica,sans-serif;"> Hooke's Law <span style="font-family: Arial,Helvetica,sans-serif;"> 1. Measure the weight hanging from the string prior to each trial. <span style="font-family: Arial,Helvetica,sans-serif;"> 2. Using a ruler, measure the amplitude (distance) the spring moves out of equilibrium. <span style="font-family: Arial,Helvetica,sans-serif;"> 3. Because F=-kx, solve for k, using the force (weight) and and distance (amplitude). <span style="font-family: Arial,Helvetica,sans-serif;"> 4. Repeat for several trials, using different masses. <span style="font-family: Arial,Helvetica,sans-serif;"> 5. Graph the effect of force on the distance.

<span style="font-family: Arial,Helvetica,sans-serif;"> Period <span style="font-family: Arial,Helvetica,sans-serif;"> 1. Record the weight hanging on the spring before each trial. <span style="font-family: Arial,Helvetica,sans-serif;"> 2. Using a stopwatch, measure the amount of time for four trials. Then, divide the number of seconds by four to find out the period of that particular trial. <span style="font-family: Arial,Helvetica,sans-serif;"> 3. Using the equation, T=2*pi*sqrt(m/k), solve for the spring constant. <span style="font-family: Arial,Helvetica,sans-serif;"> 4. Repeat for several trials, using different masses.

<span style="font-family: Arial,Helvetica,sans-serif;"> Application of the experimental k values <span style="font-family: Arial,Helvetica,sans-serif;"> 1. Average the k values found experimentally in the two procedures above. <span style="font-family: Arial,Helvetica,sans-serif;"> 2. Apply a recorded amount of mass on the spring. <span style="font-family: Arial,Helvetica,sans-serif;"> 3. Measure amplitude. <span style="font-family: Arial,Helvetica,sans-serif;"> 4. Using the equation for period, solve for the theoretical period of the each trial. <span style="font-family: Arial,Helvetica,sans-serif;"> 5. Record the experimental period for each trial. <span style="font-family: Arial,Helvetica,sans-serif;"> 6. The experimental and theoretical periods should be very close. <span style="font-family: Arial,Helvetica,sans-serif;"> 7. A graph can also be obtained from the relationship between mass and period. =<span style="font-family: Arial,Helvetica,sans-serif;"> =

<span style="font-family: Arial,Helvetica,sans-serif;">**Data** <span style="font-family: Arial,Helvetica,sans-serif;"> <span style="font-family: Arial,Helvetica,sans-serif;">We also found a theoretical period based on the average of the k's from the first two subprocedures. The experimental periods were different, and therefore, yielded a different k value. <span style="font-family: Arial,Helvetica,sans-serif;"> <span style="font-family: Arial,Helvetica,sans-serif;">

<span style="font-family: Arial,Helvetica,sans-serif;">**Calculations**

<span style="font-family: Arial,Helvetica,sans-serif;"> Solving for Spring Force Constant:

<span style="font-family: Arial,Helvetica,sans-serif;">Getting k from the period equation: <span style="font-family: Arial,Helvetica,sans-serif;">

<span style="font-family: Arial,Helvetica,sans-serif;">Getting k from Hooke's Law: <span style="font-family: Arial,Helvetica,sans-serif;">

<span style="font-family: Arial,Helvetica,sans-serif;">Getting k from the graph:

<span style="font-family: Arial,Helvetica,sans-serif;">**Discussion questions** <span style="font-family: Arial,Helvetica,sans-serif;">4. <span style="font-family: Arial,Helvetica,sans-serif;">Generate the equations and the corresponding graphs for <span style="font-family: Arial,Helvetica,sans-serif; line-height: 150%; margin-left: 1.5in; text-indent: -1.5in;"> i. position with respect to time. <span style="font-family: Arial,Helvetica,sans-serif; line-height: 150%; margin-left: 1.5in; text-indent: -1.5in;">Trial 1 <span style="font-family: Arial,Helvetica,sans-serif; line-height: 150%; margin-left: 1.5in; text-indent: -1.5in;">
 * 1) <span style="font-family: Arial,Helvetica,sans-serif;">**Do the data for the displacement of the spring versus the applied force indicate that the data for the spring constant is indeed constant for the range of forces?**Yes, the data does indicate that the spring constant is indeed constant because we obtained similar results from each.
 * 2) <span style="font-family: Arial,Helvetica,sans-serif;">**Why is time for more than one period measured?**We measured the time for more than one period to minimize the effect the stopwatch had on our error. When using a stopwatch for timing it is difficult to get exact results due to reaction time. By timing for more periods at a time the error does not have as large an impact.
 * 3) <span style="font-family: Arial,Helvetica,sans-serif;">**Discuss the agreement between the k values derived from the two graphs. Which is more accurate?**The K value derived from the graph showing force vs. displacement is more accurate because there is no error from the stopwatch included. The K value in the more accurate graph is 28N/m while the k value found through the less accurate method was 25N/m. Though they are different they are similar.

<span style="font-family: Arial,Helvetica,sans-serif; line-height: 150%; margin-left: 1.5in; text-indent: -1.5in;"> ii. velocity with respect to time. <span style="font-family: Arial,Helvetica,sans-serif; line-height: 150%; margin-left: 1.5in; text-indent: -1.5in;">

<span style="font-family: Arial,Helvetica,sans-serif; line-height: 150%; margin-left: 1.5in; text-indent: -1.5in;"> iii. acceleration with respect to time.



<span style="font-family: Arial,Helvetica,sans-serif;">**5. A spring constant k=8.25N/m. If the spring is displaced -0.150m from its equilibrium position, what is the force that the spring exerts?** <span style="font-family: Arial,Helvetica,sans-serif;"> <span style="font-family: Arial,Helvetica,sans-serif;"> **6****. A massless spring has a spring constant of K=7.85N/m. A mass m=0.425 kg is placed on the spring, and it is allowed to oscillate. What is the period T of oscillation?** <span style="font-family: Arial,Helvetica,sans-serif;"> <span style="font-family: Arial,Helvetica,sans-serif;"> 7. **We neglected to take into account the mass of the spring itself. Are your results any better when using the more accurate relationship (where m is the hanging mass and ms is the mass of the spring)? Redo graph #2 using, and explain these results.** <span style="font-family: Arial,Helvetica,sans-serif;"> <span style="font-family: Arial,Helvetica,sans-serif;"> With these adjustments, the data should be more accurate. The spring's mass is part of the mass that the spring must account for in its period. However, according to the graph, the Rsquared value changed very little. The graph should be equally accurate as before. However, the effect of this change can been seen in the coefficient of the equation.

<span style="font-family: Arial,Helvetica,sans-serif;"> Conclusions: <span style="font-family: Arial,Helvetica,sans-serif;"> Our hypothesis that Mass is directly related to period, but with a square root exponent. Because T=2pi*sqrt(m/k), increasing m, increases the period, but not in a linear manner. This was proven to be correct. We were trying to find the spring force constant directly and indirectly. We were able to find the spring force constant directly using hooke's law where we found that k, the constant was equal to 26.86. Indirectly we were finding the period. We found indirectly that k, the spring force constant is equal to 25.904. Very similar with low percent errors.

<span style="font-family: Arial,Helvetica,sans-serif;"> The percent error in this project was very little. The largest was 5%. Although we had little error there was still a large chance for error. There was a lot of human error with reaction time. Because we used a stopwatch to time the period of oscillation the error comes from the reaction time stopping and starting the stopwatch while also counting the number of cycles. The error is very close to 0% because we were able to plug our numbers directly into excel and make sure we got results which make sense. Our accurate results came from numerous trials.

<span style="font-family: Arial,Helvetica,sans-serif;"> We can use the information we explored in this lab and apply it to real life.

= = =<span style="font-family: Arial,Helvetica,sans-serif;">Mobile Project = <span style="font-family: Arial,Helvetica,sans-serif;"> Due: May 9, 2011

<span style="font-family: Arial,Helvetica,sans-serif;">**Hypothesis** <span style="font-family: Arial,Helvetica,sans-serif;"> When balancing masses on a wooden dowel, the concept of equilibrium should mathematically hold true. A balanced mobile requires both balanced torque and forces.

<span style="font-family: Arial,Helvetica,sans-serif;">**Materials** <span style="font-family: Arial,Helvetica,sans-serif;"> Ringstand <span style="font-family: Arial,Helvetica,sans-serif;"> Dowels <span style="font-family: Arial,Helvetica,sans-serif;"> Star Wars action figures (or objects of any theme) <span style="font-family: Arial,Helvetica,sans-serif;"> Fishwire <span style="font-family: Arial,Helvetica,sans-serif;"> Small saw <span style="font-family: Arial,Helvetica,sans-serif;"> Hot Glue <span style="font-family: Arial,Helvetica,sans-serif;"> Scale (for weighing) <span style="font-family: Arial,Helvetica,sans-serif;"> Ruler

<span style="font-family: Arial,Helvetica,sans-serif;">**Procedure** <span style="font-family: Arial,Helvetica,sans-serif;"> 1. Mass all the objects that might be hung on the mobile, including the dowels. <span style="font-family: Arial,Helvetica,sans-serif;"> 2. Working from the bottom of the mobile up, use torque to calculate the distances masses should be from each other on the dowel. Be sure to follow regulations, such as 2 pivot points not at the center of mass. <span style="font-family: Arial,Helvetica,sans-serif;"> 3. Chisel indentations into the dowels as the positions where a string will hang. <span style="font-family: Arial,Helvetica,sans-serif;"> 4. Hang the objects <span style="font-family: Arial,Helvetica,sans-serif;"> 5. Using a hot glue gun, secure the strings of the dowel in place.

<span style="font-family: Arial,Helvetica,sans-serif;"> <span style="font-family: Arial,Helvetica,sans-serif;">The bolded items represent the items used in the actual mobile. <span style="font-family: Arial,Helvetica,sans-serif;">

<span style="font-family: Arial,Helvetica,sans-serif;">**Photo of Mobile on Board:** <span style="font-family: Arial,Helvetica,sans-serif;">

<span style="font-family: Arial,Helvetica,sans-serif;">**Dowel 2** <span style="font-family: Arial,Helvetica,sans-serif;"> <span style="font-family: Arial,Helvetica,sans-serif;"> First Condition of Equilibrium <span style="font-family: Arial,Helvetica,sans-serif;">

<span style="font-family: Arial,Helvetica,sans-serif;"> Second Condition of Equilibrium <span style="font-family: Arial,Helvetica,sans-serif;">

<span style="font-family: Arial,Helvetica,sans-serif;">**Dowel 7** <span style="font-family: Arial,Helvetica,sans-serif;"> <span style="font-family: Arial,Helvetica,sans-serif;"> First Condition of Equilibrium <span style="font-family: Arial,Helvetica,sans-serif;">

<span style="font-family: Arial,Helvetica,sans-serif;"> Second Condition of Equilibrium <span style="font-family: Arial,Helvetica,sans-serif;">

<span style="font-family: Arial,Helvetica,sans-serif;">**Dowel 4** <span style="font-family: Arial,Helvetica,sans-serif;"> <span style="font-family: Arial,Helvetica,sans-serif;">

<span style="font-family: Arial,Helvetica,sans-serif;">

<span style="font-family: Arial,Helvetica,sans-serif;">**Dowel 5** <span style="font-family: Arial,Helvetica,sans-serif;"> <span style="font-family: Arial,Helvetica,sans-serif;">

<span style="font-family: Arial,Helvetica,sans-serif;">

<span style="font-family: Arial,Helvetica,sans-serif;">**Dowel 3** <span style="font-family: Arial,Helvetica,sans-serif;"> <span style="font-family: Arial,Helvetica,sans-serif;">

<span style="font-family: Arial,Helvetica,sans-serif;">

<span style="font-family: Arial,Helvetica,sans-serif;">**Dowel 6**

<span style="font-family: Arial,Helvetica,sans-serif;">

<span style="font-family: Arial,Helvetica,sans-serif;"> <span style="font-family: Arial,Helvetica,sans-serif;">

<span style="font-family: Arial,Helvetica,sans-serif;">**Dowel 1/Overall Torque and Forces Equations** <span style="font-family: Arial,Helvetica,sans-serif;"> <span style="font-family: Arial,Helvetica,sans-serif;"> <span style="font-family: Arial,Helvetica,sans-serif;">

<span style="font-family: Arial,Helvetica,sans-serif;">**Conclusion** <span style="font-family: Arial,Helvetica,sans-serif;"> Our calculations of torque and forces show that our mobile is almost at perfect equilibrium. We planned the lever arms of the masses, which allowed us to mathematically achieve equilibrium before hanging the masses. Due to the fact that the calculations were used to place the mobiles, there is little to no error. Our mobile successfully balances on all levels, and consequently, it balances overall. The net torque is 0, as well as the net forces, because the mobile is not tilted out of equilibrium.

<span style="font-family: Arial,Helvetica,sans-serif;"> Like always, there are sources of error. For example, to keep our masses in place, we chiseled the wood and applied hot glue. The chisels in the wood could have lent to changes in CM and mass, because we removed wood from the dowel when chiseling. The hot glue gun probably had an even greater effect. The glue comes out of the glue gun in clumps that can add weight on top of the weight from the hanging objects. By inconsistently changing the forces applied at different lengths of the wood, we could accidentally cause disequilibrium. The glue gun, was however necessary, because the chisel marks were insufficient in holding the hanging objects in place. Had we not used the glue, there would have been even greater error from changes in lever arm caused by moving strings. To correct these changes, students could be provided with some type of string to hang objects from, that will be able to stick to the wood, without other added substances. Doing so might require the dowels to be made of a different substance too. Another place of error would be weight of the string. individually, the string does not weigh much at all. But with all the string throughout the mobile combined, it could alter weights and change the balance of the mobile. A real world application for mobiles and knowing how to solve for static equilibrium would be the art style of mobiles. People such as Calder make mobiles for art. In order to create such art, calculations are essential. Without prior knowledge of how calculate for the static equilibrium of the mobile, their creation would be extremely difficult and overly time consuming.

=<span style="font-family: Arial,Helvetica,sans-serif;">Levers: Static Equilibrium = <span style="font-family: Arial,Helvetica,sans-serif;"> 4/15/11 <span style="font-family: Arial,Helvetica,sans-serif;"> Due: 4/18/11

<span style="font-family: Arial,Helvetica,sans-serif;">**Objective** <span style="font-family: Arial,Helvetica,sans-serif;"> We want to put the system at equilibrium using concepts of torque.

<span style="font-family: Arial,Helvetica,sans-serif;">**Hypothesis** <span style="font-family: Arial,Helvetica,sans-serif;"> The counterclockwise torque should be equal to the clockwise torque. At equilibrium, net torque is equal. Therefore, we should be able to put the system at equilibrium by manipulating masses and distances.

<span style="font-family: Arial,Helvetica,sans-serif;">**Materials** <span style="font-family: Arial,Helvetica,sans-serif;"> Meterstick <span style="font-family: Arial,Helvetica,sans-serif;"> Fulcrum <span style="font-family: Arial,Helvetica,sans-serif;">Hanger, clamps <span style="font-family: Arial,Helvetica,sans-serif;">Various masses

<span style="font-family: Arial,Helvetica,sans-serif;"> **Procedure** <span style="font-family: Arial,Helvetica,sans-serif;">1. Weigh the clamps and hangers. <span style="font-family: Arial,Helvetica,sans-serif;">2. Place the fulcrum at the center of mass of the meterstick. <span style="font-family: Arial,Helvetica,sans-serif;">3. While simultaneously recording data in an excel sheet, perform the following trials. <span style="font-family: Arial,Helvetica,sans-serif; font-size: 12px;"><span style="font-size-adjust: none; font-stretch: normal; font-style: normal; font-variant: normal; font-weight: normal; line-height: normal;">- Two different masses with the fulcrum at the center of mass of the meterstick (COM) <span style="font-family: Arial,Helvetica,sans-serif; font-size: 12px;">- Three different masses with the fulcrum at the COM. <span style="font-family: Arial,Helvetica,sans-serif; font-size: 12px;">- Two different masses with the fulcrum at least 20 cm off COM. <span style="font-family: Arial,Helvetica,sans-serif; font-size: 12px;">- One mass with the fulcrum at least 20 cm off COM. <span style="font-family: Arial,Helvetica,sans-serif; font-size: 12px;"><span style="font-size-adjust: none; font-stretch: normal; font-style: normal; font-variant: normal; font-weight: normal; line-height: normal;">- One object of unknown mass, everything else to be determined by you. <span style="font-family: Arial,Helvetica,sans-serif; font-size: 12px;">- A single mass supported by a tension at an upward angle and a wall. (Teacher will set this up for you.) <span style="font-family: Arial,Helvetica,sans-serif;"> 4. Mathematically, solve for what the lever arms of the masses should be.

<span style="font-family: Arial,Helvetica,sans-serif;">**Photo of Set-Up:** <span style="font-family: Arial,Helvetica,sans-serif;">

<span style="font-family: Arial,Helvetica,sans-serif;">**Data** <span style="font-family: Arial,Helvetica,sans-serif;"> <span style="font-family: Arial,Helvetica,sans-serif;"> <span style="font-family: Arial,Helvetica,sans-serif;"> The data table is in 3 significant figures.

<span style="font-family: Arial,Helvetica,sans-serif;"> Due to varying scenarios and trials, the data table was not a consistent format throughout. We only performed one trial of each scenario.

<span style="font-family: Arial,Helvetica,sans-serif;"> Trial 1- It is as follows on the table. <span style="font-family: Arial,Helvetica,sans-serif;"> Trial 2- In order to find the percent difference, we compared (“Mass 1 Torque” plus “Mass 3 Torque”) to “Mass 2 Torque”. <span style="font-family: Arial,Helvetica,sans-serif;"> Trial 3- Mass 3 is the mass of the meter stick. We found the percent difference similarly to Trial 2. <span style="font-family: Arial,Helvetica,sans-serif;"> Trial 4- Mass 1 is the mass of the meter stick. <span style="font-family: Arial,Helvetica,sans-serif;"> Trial 5- Mass 1 was a known mass that we put on the meter stick. We then solved for the unknown mass. This was our experimental value. We then weighed the mass to obtain our theoretical value. We then found the percent error between the two values. <span style="font-family: Arial,Helvetica,sans-serif;"> Trial 6- Instead of finding a percent difference or error, we tried to find the forces of Tension, Wall (x-component) and Wall (y-component).

<span style="font-family: Arial,Helvetica,sans-serif;">**Calculations** <span style="font-family: Arial,Helvetica,sans-serif;"> Trial 1 <span style="font-family: Arial,Helvetica,sans-serif;"> <span style="font-family: Arial,Helvetica,sans-serif;"> <span style="font-family: Arial,Helvetica,sans-serif;"> Trial 2 <span style="font-family: Arial,Helvetica,sans-serif;"> <span style="font-family: Arial,Helvetica,sans-serif;">

<span style="font-family: Arial,Helvetica,sans-serif;"> Trial 3 <span style="font-family: Arial,Helvetica,sans-serif;"> <span style="font-family: Arial,Helvetica,sans-serif;">

<span style="font-family: Arial,Helvetica,sans-serif;"> Trial 4 <span style="font-family: Arial,Helvetica,sans-serif;"> <span style="font-family: Arial,Helvetica,sans-serif;">

<span style="font-family: Arial,Helvetica,sans-serif;"> Trial 5 <span style="font-family: Arial,Helvetica,sans-serif;"> <span style="font-family: Arial,Helvetica,sans-serif;">

<span style="font-family: Arial,Helvetica,sans-serif;"> Trial 6 <span style="font-family: Arial,Helvetica,sans-serif;"> <span style="font-family: Arial,Helvetica,sans-serif;"> We have also provided the calculations that we performed in order to find the Tension and Forces of the Wall (x and y-components). <span style="font-family: Arial,Helvetica,sans-serif;">

<span style="font-family: Arial,Helvetica,sans-serif;">

<span style="font-family: Arial,Helvetica,sans-serif;">**Error** <span style="font-family: Arial,Helvetica,sans-serif;"> We have provided an example of our percent difference calculation for trial 1 as well as our percent error calculation for trial 5.

<span style="font-family: Arial,Helvetica,sans-serif;"> <span style="font-family: Arial,Helvetica,sans-serif;">

<span style="font-family: Arial,Helvetica,sans-serif;">**Analysis Questions** <span style="font-family: Arial,Helvetica,sans-serif; font-size: 12px; margin-left: 0.5in; text-indent: -0.25in;">**1. Does it get easier or harder to rotate a stick as a mass gets farther from the pivot point?** It gets easier to rotate the stick as the mass moves farther from the pivot point. The torque is equal to the lever arm times the force so as the lever arm increases the force does not need to be as large. <span style="font-family: Arial,Helvetica,sans-serif; font-size: 12px; margin-left: 0.5in; text-indent: -0.25in;">**2. Does the weight of the mass increase as you move the mass away from the pivot point (your index finger)?** No because weight is just the mass times the acceleration of gravity. The mass and gravity are not changing as the mass is moved. <span style="font-family: Arial,Helvetica,sans-serif; font-size: 12px; margin-left: 0.5in; text-indent: -0.25in;">**3. Why is more mass required to balance the meterstick as you move another mass farther from the pivot?** As the mass is moved away from the pivot point it is able to rotate the stick easier. To counteract this you need to add more mass to the other side. <span style="font-family: Arial,Helvetica,sans-serif; font-size: 12px; margin-left: 0.5in; text-indent: -0.25in;">**4. Why must the mass of the hangers and clamps be taken into account in this experiment?** The hangers and clamps act as weights on the meter stick and are large enough to have an impact on equilibrium of the stick. <span style="font-family: Arial,Helvetica,sans-serif; font-size: 12px; margin-left: 0.5in; text-indent: -0.25in;">** 5. ** **If you are playing seesaw with your younger sibling (who weighs much less than you), what can you do to balance the seesaw? Mention at least two things**.To balance the seesaw you can have your sibling sit on the seesaw with their friend so that together they might weigh close to your weight. Another way to balance the seesaw would be to try to sit closer to the middle of the seesaw and have your sibling sit as far away from the center as possible. <span style="color: black; font-family: Arial,Helvetica,sans-serif; font-size: 12px; margin-left: 0.5in; text-indent: -0.25in;">**6. What kept the meterstick in equilibrium in the fourth trial? In other words, what counterbalanced the known mass?**In the fourth trial we used one mass and the fulcrum had to be at least 20 cm off of the COM. By moving the fulcrum towards the mass we were able to achieve equilibrium. This is because as the lever arm was decreased the mass had more difficulty turning the meter stick so the meter stick was able to stay level.

<span style="font-family: Arial,Helvetica,sans-serif;">**Conclusion**

<span style="font-family: Arial,Helvetica,sans-serif;"> In this lab, we successfully proved our hypothesis to be correct. Our data shows little error. For example, Trial 1 has only 0.621% error. We were able to make the torques almost completely equal. The percent error equation above demonstrates that our percent error is very minor, and the concept of torque can be applied in actual lab applications. In this lab, there are many sources of error. Firstly, the measure of lever arm (distance) is highly dependent on accurate center of mass assumptions. Our center of masses for the added weights are only assumptions. It is possible that the weight is not concentrated as we presume it to be. Furthermore, the clamps add weight. their oblong shape might also lead to inaccurate assumption of center of mass. To improve these sources of error, the we might use smaller masses of similar weight, because the weight would be more concentrated. A hanger with less gaps in its design could also allow us to make a better estimate of the center of mass. Also, our only way of determining when the meterstick is balanced, is with our eyes. Our original position of "balance" prior to beginning the trials could have been slightly off. Each trial is also subject to the same source of error. Finally, we might also use more different masses. Each of the masses we use are different from each other by at least intervals of 20g.

<span style="font-family: Arial,Helvetica,sans-serif;"> It is important for us to understand how torque affects balance. In the past, we learned to use forces in similar applications, but torque is sometimes a better method when there are too many variables. Motion, even bending of our limbs, is an example of torque, and corresponds to the concept of torque's dependence on lever arm and radius. Beams and balances also frequently appear in real life. Their functionality depends on the balance of clockwise and counterclockwise forces; therefore, torque is used and necessary in real life. Even simple devices such as see-saws function through a basis in torque.