Anthony,+Aaron,+Justin,+Nicole

=Lab: Transverse Standing Waves on a String = **Group: Justin T, Anthony I, Aaron C, Nicole M, Sammy W ** **Date Completed: 5/23/11 ** **Date Due: 5/24/11 ** **Class: Period 2 **

**Objective: ** What is the relationship between frequency and the tension of transverse waves traveling in a stretched string? What is the relationship between frequency and harmonic number? What is the relationship between frequency and wavelength?

**Hypothesis: Sammy **

**Materials: ** Electrically-driven oscillator, pulley and table clamp assembly, weight holder and selection of masses, string, electronic balance

<span style="background-color: transparent; color: #000000; font-family: Arial,Helvetica,sans-serif; font-size: 90%; margin: 0px; padding: 0px; text-align: start; text-decoration: none; vertical-align: baseline;">**<span style="background-color: transparent; color: #000000; text-align: start; text-decoration: none; vertical-align: baseline;">Procedure: Sammy **

<span style="background-color: transparent; color: #000000; font-family: Arial,Helvetica,sans-serif; font-size: 90%; text-align: start; text-decoration: none; vertical-align: baseline;">**<span style="background-color: transparent; color: #000000; text-align: start; text-decoration: none; vertical-align: baseline;">Data: **



<span style="background-color: transparent; color: #000000; font-family: serif; font-size: 16px; text-align: start; text-decoration: none; vertical-align: baseline;">From our graph we found that frequency is equal to the harmonic number times 19.397. 19.397 is the natural frequency of the string which should be around 20. The frequency would be 58.191 Hz, if the harmonic number is 3. This would give a percent difference of 0.359% from our experimental data.

<span style="background-color: transparent; color: #000000; font-family: Arial,Helvetica,sans-serif; font-size: 90%; text-align: start; text-decoration: none; vertical-align: baseline;">**<span style="background-color: transparent; color: #000000; line-height: 0px; overflow: hidden; text-align: start; text-decoration: none; vertical-align: baseline;"> ﻿ ** <span style="background-color: transparent; color: #000000; font-family: serif; font-size: 16px; text-align: start; text-decoration: none; vertical-align: baseline;">From our graph we found that frequency is equal to 62.509 times the wavelength to the power of -.9946. Because of the equation, frequency=velocity/wavelength, wavelength is taken to the power of -1. The coefficient represents the speed of the wave in the string, which should be about 60. If the wavelength is 1.073, then the frequency is 58.278 Hz. This would give a percent difference of 0.209% from our experimental data.

<span style="background-color: transparent; color: #000000; font-family: serif; font-size: 16px; text-align: start; text-decoration: none; vertical-align: baseline;">From our graph we found that frequency is equal to 26.728 times tension to the .4962 power. From the equation, we found there is a square root relationship between tension and frequency. Therefore the power should be around .5. If the tension is 14.7, then the frequency should be 101.435 Hz. This would give a percent difference of 1.22% from our experimental data.

<span style="background-color: transparent; color: #000000; font-family: Arial,Helvetica,sans-serif; font-size: 90%; text-align: start; text-decoration: none; vertical-align: baseline;">**<span style="background-color: transparent; color: #000000; text-align: start; text-decoration: none; vertical-align: baseline;">Percent Error: **

<span style="background-color: transparent; color: #000000; font-family: Arial,Helvetica,sans-serif; font-size: 90%; text-align: start; text-decoration: none; vertical-align: baseline;">**<span style="background-color: transparent; color: #000000; text-align: start; text-decoration: none; vertical-align: baseline;">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? If the string stretched significantly as the tension increases, then the velocity would increase also. We know this from the equation. If the tension and velocity both increased, then we would need a higher frequency to get the same number of antinodes.

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.
 * Sam**

4. What is the effect of changing frequency on the number of nodes? As the frequency increases the number of harmonics increases, which in turn increases the number of nodes. The vice versa happens when the frequency decreases.

5. What factors affect the number of nodes in a standing wave? The frequency and amount of tension in the string affect the number of nodes in a standing wave.

<span style="background-color: transparent; color: #000000; font-family: Arial,Helvetica,sans-serif; font-size: 90%; text-align: start; text-decoration: none; vertical-align: baseline;">**<span style="background-color: transparent; color: #000000; text-align: start; text-decoration: none; vertical-align: baseline;">Conclusion: **

The purpose of this lab was to determine the relationships between frequency, tension of the transverse waves, harmonic number, and wavelength. After we graphed the data of the different experiments, we were able to determine whether or not a relationship existed and even what the relationship is.

For the frequency vs. harmonic number graph, we found the slope to be 19.39, which is the natural or fundamental frequency. This is the frequency when the harmonic number is one. This is the foundation of the future harmonic numbers and harmonic motion in general.

When testing frequency vs. wavelength, we found a relationship with a slope of 62.51, which the velocity of the wave in m/s, and an exponent of -.995 when it should have been negative one because frequency and wavelength have an inverse relationship. As a result, we had only a .5% error. Even though this error is very small, we could still have improved our experiment to limit the error. One thing we could have done was more accurately measure the wavelengths and make sure that we frequencies we chose produce an exact harmonic number that is not too small or too large.

For the test in the relationship between tension and frequency, we theorized that the exponent of x should be .5 because f=sqrt(T). Our .496 exponent yields as .8% error. This error can be attributed to incorrect measure of the tension of the string or the problems explained in the paragraph above.

= = =Lab: Period of Mass on a Spring=
 * Group: Justin T, Anthony I, Aaron C, Nicole M, Sammy W**
 * Date Completed: 5/16/11**
 * Date Due: 5/17/11**
 * Class: Period 2**

What is the relationship between the mass on a spring and its period of oscillation?
 * Objective:**

The relationship between the mass and the period can be characterized by the equation: Since k is the spring force constant, it will stay the same. Therefore as mass goes up, so will the period.
 * Hypothesis:**

Springs, tape, clamps and rods, masses, balance, timers, meter stick
 * Materials:**



**Procedure:** 1. Set up the spring and the ruler. 2. Add masses to the spring. 3. Stretch the spring to a point on the ruler and record the distance that the spring was stretched. 4. Start the timer and after counting 10 periods, stop the timer. 5. Divide the time by 10 to find the length of one period. 6. Repeat as necessary.


 * Data:**






 * Sample Calculations:**
 * Using Hooke's Law to find the Spring Force Constant of the Spring**
 * Using Simple Harmonic Motion to find the Spring Force Constant of the Spring**
 * Using Period Squared vs. Mass Graph**
 * Percent Difference**


 * Discussion Questions:**
 * 1.** Even though the spring force constant should hold true for any forces, when comparing the two graphs it does not show that the spring force constant is constant.


 * 2.** The reason that we measure more than one period is so that we reduce the amount of error in our results. If we only measured one period, human reaction time greatly affects our results because we only measure one period. If we were to measure 10 periods, the effect of human reaction time would decrease because there is more data and the added reaction time is divided by 10.


 * 3.** The k values derived from the two graphs were extremely close in value. The first k value is 22.44 and the second one is 22.17 with a difference of about .27. The first k value is more accurate because it has less human error compared to the second one.

F = (8.75 N/m) * (-0.150 m) F= -1.3125 N (Sammy on Justin's username)
 * 4.** F= k*x

Despite including the mass of the spring, the slope stayed relatively the same. The spring force constant is the same because we did not take the mass in account in any trial. With the mass increasing in a constant rate, the slope did not change.
 * 5.**
 * 6.**
 * 7.**

My group's hypothesis that there is a relationship between the mass of the hanging object and the period of oscillation, specifically that if the mass increase, the period of oscillation will increase and vice versa. This relationship is shown in the data table when larger masses correspond to larger periods of oscillation. Also, the positive slope of the graphs show that as the weight or force increases, the period of oscillation also increases.
 * Conclusion:**

In this lab, our group thought it would be more appropriate to calculate percent difference because since we didn't know the actual spring force constant of the spring, there was no way to find the percent error. We found the percent difference between the spring force constants using the two different ways to calculate it. After we calculated the spring force constant of the spring using the equation and then calculated it by using the slope of the Force vs. Distance, we then found the percent difference between these two figures. We found that the percent difference was .44% using the equation posted above and 2.37% using Hooke's Law. If we look back to our procedure, there a number of times where error could have occurred. First, the timer's reaction time definitely played a factor in the percent error. Even though we measure 10 periods and divided the time by 10 in an effort to reduce the effect of the reaction time, error still exists from this part of the laboratory. We could try to reduce the effect of the error by counting 50 or 100 periods instead of 10. Secondly, even though it probably had a small effect, our group does not know if the spring was stretched out during the experiment, which would have changed and ultimately decreased the spring force constant of the spring. One way we could have avoided this would be to use small hanging masses that would not ruin the spring and change the spring force constant.

1. It gets harder to rotate a stick as a mass gets farther from the pivot point. Since torque is comprised of the force of gravity on the mass and the distance of the mass from the pivot point, as the distance from the pivot point, the torque, or energy needed to turn the stick, increases.
 * __Discussion Questions:__**

2. No, the weight of the mass does not increase as you move it away from the pivot point. Weight is mass times acceleration of gravity. Since those are not changing, the weight does not change. On the other hand, the torque is increasing.

3. When you move a mass father from the pivot, it becomes easier for the mass to unbalance the meter stick which then requires more mass on the other side to balance the torques and the meter stick.

4. The mass of the hangers and clamps are part of the mass because they are connected and have an impact on the equilibrium since they are pretty heavy and add to the force on the meter stick.

5. One way to balance the seesaw with a younger sibling is for you to move closer to the center of the seesaw. Another way to balance the seesaw is for your younger sibling to move farther back.

6. Since the fulcrum was not in the meterstick's COM, the ruler was considered one of the torques that prevented it from staying in equilibrium. By adding a mass on the shorter side of the meterstick this counterbalanced the torque created by the torque not being in the meterstick's COM.