Group3_4_ch11

Lab Group 14: Robert Kwark, Noah Pardes, Max Llewellyn, Kosuke Seki

Speed of Sound -- Resonance Tube
Members: A - Kosuke Seki, B - Noah Pardes, C - Maxx Grunfeld, D - Max Llewellyn
 * Objectives:**
 * 1) Determine several effective lengths of the closed tube at which resonance occurs for a frequency.
 * 2) Determine several effective lengths of the open tube at which resonance occurs for a frequency.
 * 3) Determine the speed of sound from the measured wavelengths and known frequency of the sound.


 * Hypothesis:**
 * 1) For the closed tube, resonance should occur at even increments along the tube.
 * 2) For the open tube, resonance should occur at even increments as well, but in more quantity than the closed tube
 * 3) The speed of sound we find here should be the same as the speed of sound in air

For this lab, we are finding the length of resonance. To do this, first get a large plastic tube, with a smaller plastic tube inside of it. Next we get a frequency generator and hook a speaker up to it that creates a constant sound when turned on. After that, we aim the speaker into the tube and turn it on. From here, we increase and decrease the length of the tube, as well as open and close the tube, looking for different resonance points. After this, we record the data and analyze it.
 * Procedure:**


 * Data and Corresponding Graphs:**
 * Sample Calculations:**



(theoretical)

( The parantheses represent ¼ wavelength for a closed tube. The fifth resonance has a harmonic number of five, and therefore 5/4 wavelengths, which is how we get this picture. )( In this case for an open tube, each parentheses represents ½ wavelength. The fifth resonance still has a harmonic number of five but now has 5/2 wavelengths, which is how we get this picture.
 * Discussion Questions:**
 * 1) **What is the meaning of the slope of the graph for the open tube? For the closed tube?** The slope of the graph for the open tube is one half of the wavelength. The slope of the graph for the closed tube is one fourth of the wavelength. The equation for the open tube is L = n*1/2*wavelength and the equation for the closed tube is L = n*1/4*wavelength
 * 2) **Why was the length of the tube always smaller than expected?** The length of the tube is always smaller than expected because the antinode shifts in the tube, so the length needed is a little shorter than it is ideally (end shift). Additionally, as the diameter of the tube increases, the length requirement decreases.
 * 3) **Suppose that the temperature had been 10 ˚C higher than the value measured for the room temperature. How much would that have changed the measured value of L?** Due to the direct relationship between temperature and velocity, as one increases the other does too. So if the velocity were to increase, the temperature would increase, and the wavelength would increase as well (velocity = frequency*wavelength).
 * 4) **Draw a figure showing the fifth resonance in a tube closed at one end. Show also how the length of the tube L5,is related to the wavelength, λ. **
 * 1) **Draw a figure showing the fifth resonance in a tube open at one end. Show also how the length of the tube L5,is related to the wavelength, λ. **
 * 1) **What does this have to do with making music?** This has a lot to do with making music, especially in relation to string instruments. By changing the length of the string, people are able to change the sound that it produces. Also by changing wavelength, the frequency changes, and music is created as a result of this.


 * Conclusion:**

Standing Waves on a String

 * Objectives:**
 * 1) What is the relationship between frequency and the tension of transverse waves traveling in a stretched string?
 * 2) What is the relationship between frequency and harmonic number?
 * 3) What is the relationship between frequency and wavelength?
 * Hypothesis:**
 * 1) Likely, it is some kind of exponential growth. This is because as tension goes up, velocity goes up. And when velocity goes up, frequency also goes up.
 * 2) More frequency will result in more harmonic numbers, because there will be an increase in nodes and antinodes.
 * 3) They are inversely proportional.


 * Procedure:**

__//Part 1//__ //Graph Interpretation://
 * Data and Corresponding Graphs:**
 * For this graph, there was the standard equation of: y = Ax B
 * The value for A on this graph can be derived as[[image:Screen_shot_2012-05-14_at_5.57.35_AM.png width="53" height="64"]]
 * The B value in this equation should have been 0.5
 * Our value for this part of the graph was quite close, at a percent error of only 2.88%

__//Part 2//__ //Graph Interpretation://
 * For the f vs n graph:
 * The equation is: y = Ax
 * A in this case = fundamental frequency (f n = **f 1 ** x n)
 * For the f vs wavelength graph:
 * The equation is: y = Ax B
 * B in this case should equal -1
 * We had a much larger percent error for this value, at 25%

__//Mass Length of String://__ __//Tension://__ __//Wave length (for 3 nodes w/ Tension = 9.8 N)//__ //Method 1:// __//Wave speed (for 3 nodes w/ Tension = 9.8 N)//__ //Method 1:// __//Fundamental Frequency (for 3 nodes w/ Tension = 9.8 N)//__ __//Percent Error//:__ //Comparing Theoretical exponent vs Actual exponent in Frequency vs. Tension Graph// //Comparing Theoretical exponent vs Actual exponent in Frequency vs. Wavelength Graph//
 * Sample Calculations:**


 * Discussion Questions:**
 * 1) Calculate the tension T that would be required to produce the n = 1 standing wave for the red braided string.
 * 1) What would be the effect if the string stretched significantly as the tension increased? How would that have affected the data?
 * 2) If the string stretched, it would change the mass/length ratio that we had originally. That would increase the velocity, because v=sqrt(T/(m/l)). Because our ratio was over one, a smaller ratio would result in an increase in velocity.
 * 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.
 * 4) The type of string changes the mass/length ratio, which would have a drastic effect on the tension needed to crease a set number of nodes. Based on the equation v=sqrt(T/(m/l)) and v=f(lamda), the tension would have to change significantly in order to hit a specific frequency and speed.
 * 5) What is the effect of changing frequency on the number of nodes?
 * 6) Changing frequency increased or decreased the amount of nodes. If you added the fundamental frequency, then the number of nodes would increase by one, otherwise, it just created a distorted wave.
 * 7) What factors affect the number of nodes in a standing wave?
 * 8) The number of nodes is affected by many things including tension, frequency, string length, wave speed, and string density.

We were able to answer the three objectives that were given to us at the beginning of this lab. The first objective was to find the relationship between frequency and tension of a transverse wave. We hypothesized that as tension increased, the frequency would increase exponentially. The second objective was to find the relationship between frequency and harmonic number. We hypothesized that as the harmonic number increased, the frequency would increase linearly. Finally, the last objective was to was to find the relationship between frequency and wave length. We hypothesized that as wavelength increases, frequency will decrease by a power of -1 (or that they are inversely proportional). To find out whether our predictions were correct, we attached a string to an electrically driven oscillator and used this to gather the needed information. After completing the lab, we were able to see that our hypotheses were, for the most part, correct. We say //for the most part// because all of our predictions were right except for the first one, which was only slightly off. Instead of increasing exponentially as we had predicted, it increased by a power of almost 1/2. Additionally, the data we gathered was, for the most part, precise. However, as we look at the percent error, there is one value in particular that is quite high. There were some sources of error that could have accounted for such a high value of percent error. One, as always, is human error. While twisting the nob to create different frequencies, it is often difficult for one to get the measurement that is exactly correct. Although it might look stable, the frequency could have been a few decimal places off from what it actually should have been. To fix this error, we could have solved for frequency before and simply set the machine to that number to get the exact result. Another source may have come from the pulley holding the string on the opposite end of the system. While this does allow us to adjust the tension force with relative ease, it makes the system less stable than say, pulling the string to a certain tension and then clamping it to a heavy object. This prevents any excess movement at the node, where there should theoretically be none. Overall, other than that one value, all of our results came out quite accurately and precisely
 * Conclusion:**

Mass on a Spring II
For the first part, first mass the mass holder. Attach holder onto the hook on the spring and measure the distance. Next, add mass onto the hook and find the displacement from the original distance. Plot those displacements on excel to find the spring constant. For the second part, add enough mass to get a reasonable oscillation. Measure 10 oscillations and divide that time by 10 (to get a more accurate time). Graph the two variable (mass and period) and use the equation to solve for the spring constant.
 * Objectives:**
 * 1) To directly determine the spring constant k of a spring by measuring the elongation of the spring for specific applied forces
 * 2) to indirectly determine the spring constant k from measurements of the variation of the period T of oscillation for different values of mass on the end of the spring.
 * 3) To compare the two values of spring constant k.
 * Hypothesis:**
 * 1) We believe that a larger mass is equal to a longer period. This is because of the equation "T=2pi(m/k)^1/2," which suggests that increasing mass increases the period. We also believe that we can find the spring constant of the spring by using F=kx. These two k values should be the same as each other; the k value is a constant and does not change.
 * 2) We also believe that we should get the same spring force constant from both methods.
 * Procedure:**
 * Data:**

//Part 1: Hooke's Law// F=-kx Our graph: y=25.528x The y value represents Force. Therefore, k=25.528 N/m
 * Sample Calculations:**

Part 2:

__Result analysis:__ -Take the average of the k in the first part and the k in the second part to get the average experimental value.

Yes. There is a linear relationship between the Force applied and the displacement. F=kx is a linear relationship; as x increases, so does F. Since F was on the y-axis and displacement on the x-axis, this meant that the slope was equal to the spring constant. Our results were spectacular; the points matched up to the line almost perfectly, and we had a .998 r^2 value. If you only measure the time for one oscillation, then the results would be highly variable. If you take the time for 10 oscillations and divide by 10 to get the time for one period, then there is less room for error. By measuring only one period, reflexes among other things has a much larger impact on each trial. By measuring 10 oscillations, you can get a much more accurate time b/c you are taking the average of 10 periods while not relying so much on reflex. The value of k from both our graphs were both relatively close. They were a little over 1 N/m apart, and had a percent difference of 2.52%. This is an amazing result. The k value determined from the first graph was probably more accurate because less human interaction was involved. Timing the a moving spring created more room for error, as opposed to simply measuring the distance of a spring at equilibrium at various masses. Timing a moving object always requires reflex and other human senses, which may be different at each trial and from person to person. That is why at track meets FAT (fully-automated timing) is used to determine times. The field events are still measured by people and a measuring tape, even in the Olympics, because there is a lot less room for error than for the races.
 * Discussion Questions:**
 * 1) 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 this range of forces?
 * 1) Why is the time for more than one period measured?
 * 1) Discuss the agreement between the k values derived from the two graphs. Which is more accurate?
 * 1) Generate the position with respect to time equation and the corresponding graph for
 * 2) position with respect to time.
 * 3) x=Acos(2πf*t)
 * 4) velocity with respect to time.
 * 5) v=2πf*x (this is dependent on the position/displacement... to make it dependent on the time, you must make x =the position time equation above) = 2πf*(Acos(2πf*t))
 * 6) acceleration with respect to time.
 * 7) a=(2πf)^2*x (same reasoning as #2) = (2πf)^2*(Acox2πf*t)).
 * 8) [[image:http://dev.physicslab.org/img/4f9484e0-5b2f-49eb-83c1-ef03cc1883c0.gif width="185" height="285"]] [[image:http://01.edu-cdn.com/files/static/wiley/0071621318/GENERATING_SINE_WAVES_01.GIF width="420" height="271"]]
 * I didn't understand what you meant by this question, so I did the best I could.
 * We couldn't really make an actual position time graph, b/c we did not and could not have taken the amplitude of an oscillation (would have been wildly inaccurate).


 * 1) A spring constant k = 8.75 N/m. If the spring is displaced -0.150 m from its equilibrium position, what is the force that the spring exerts?
 * 1) A massless spring has a spring constant of k = 7.85 N/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?

-Yes, our results were much better. The exponent is much closer to .5 than before; our percent error became almost 0%. The k value, when calculated, is around 26.1 N/m. This was closer to the k value of Part 1 (25.53 N/m). The reason the results improved was because previously, we did not account for some of the mass affecting the spring.
 * 1) We neglected to take into account the mass of the spring itself. Are your results any better when using the more accurate relationship m+1/3m (where //m// is the hanging mass and //ms// is the mass of the spring)? Redo graph #2 using sqrt(m+1/3m), and explain these results.[[image:Screen_shot_2012-05-06_at_5.14.48_PM.png]]

Out of all the labs this year, this lab was one of the more straightforward ones. In it, we attempted to find the spring force constant using two different methods. The reason behind the two methods was because we wanted to see which one yielded better results, as well as whether the two methods produced the same results. We hypothesized that our values would indeed be equal for each method and that the spring force constant should remain //constant//. After collecting and graphing all of the appropriate data, however, we discovered that our hypothesis was not exactly correct and there was a very small percent error. We experienced a 3.1% error, getting a K of 25.528 N/M for method one and a K of 24.274 N/M for method two. We were thus able to conclude that the spring force constant remains the same for a range of forces. We also discovered that method one was a more accurate way than method two in calculating the spring force constant. There were several sources of error for this experiment. For instance, while stretching the coils, they could have been overstretched from over use, leading to incorrect values for the spring force constant. Also, it was difficult to get excellent measurements because of the nature of the lab. With coils moving up and down, it is hard to find perfectly exact values without being slightly off. In order to change this in the future, we would use materials and devices that perhaps allowed for more precise measurements to be taken; for example, a motion sensor or a FAT system instead of using a hand timer and eyeballing the measurement. An application of this could be trampolines. A trampoline works by someone jumping on the fabric, which presses down the material, which presses the springs down, which in turn creates an upward force. Because we now know that weight affects oscillation, we will be able to reason why some people can jump higher than others.
 * Conclusion:**