Emily,+Emily,+Navin,+Andrew

=**Lab: Transverse Standing Waves on a String**=

- frequency and harmonic number - frequency and wavelength - frequency and tension
 * Purpose**:To determine the relationships between:


 * Hypotheses**:The relationship between frequency and harmonic number will be a direct one. Because v = wavelength x ƒ, and the L = .5n x wavelength, with some algebraic manipulation, we attain an equation where frequency and harmonic number are on opposite sides of the equation, and given that every other variable is constant, the relationship between two variables on opposite sides of the equations is direct. A greater frequency means a smaller period, meaning that there will be more cycles, meaning more antinodes.

The relationship between frequency and wavelength is indirect. Because v = wavelength x ƒ, and because we are not changing the medium, we know the velocity to be constant. We hypothesize that as the frequency increases, wavelength will decrease because when frequency increases, the period shortens, meaning that more cycles happen in less time, meaning that the wavelength (the distance between two identical points on two different waves) decreases, showing an indirect relationship.

The relationship between frequency and tension is direct squared. v =sqrt(T/(m/L), and v = wavelength x ƒ. By setting these two equations equal to each other, we are able to see that the relationship between frequency and tension is directly squared or square rooted relationship, depending on which way you write the equation. This does make sense, for as you increase the frequency, the elastic quality of a string increases.


 * Materials**: Electrically-driven oscillator, Pulley and Table Clamp Assembly, Weight Holder, Different Masses, String, and Electronic Balance

1. Attach string to oscillator. 2. Set the sensor to the frequency that gives one antinode,and two nodes. 3. Change the frequency to get two antinodes and three nodes, etc. 4. The relationship that should be seen is by doubling the frequency, you double the number of antinodes. 5. Record the frequency when you clearly see a certain number of antinodes. 6. Measure the wavelength at each of these frequencies. 7. Graph results.
 * Procedure For Finding the Relationships Between Frequency and harmonic number, and Frequency and wavelength:**

1. Initially add a 1 kg mass to the string, and record the frequency, harmonic number, and wavelength at this point. 2. Add another mass ( 1 kg) to the string, and find the frequency that creates the same harmonic number and same wavelength that was measured intially. 3. Repeat until you have gathered sufficient data. 4. Graph results.
 * Procedure for Finding the Relationship Between Frequency and Tension**:

media type="file" key="NAvin's lab setup.mov" width="300" height="300"media type="file" key="Navin's lab setup2.mov" width="300" height="300" align="left"
 * Setup:**


 * Data**:




 * Sample Calculations and Error Calculations:**
 * Discussion Questions**:
 * 1. Calculate the tension T that would be required to produce the n=1 standing wave for the red braided string.**



The way our lab was set up the pulley-wheel cut off the length of the string. So in order to increase the tension in the string, we had to add mass to the end of the string that would stretch it significantly, thus increasing the tension. The pulley-wheel though would stay a consistent distance from the oscillating machine, thus ensuring that even though the string may be stretched, the wavelength of the wave would be constant. So the fact that the string was stretched was expected and completely necessary for our data collection of how frequency and tension are related.
 * 2. What would be the effect if the string stretched significantly as the tension increased? How would that have affected the data?**

Through this experiment, we found that as the hanging mass increased (as the tension in the string increased)
 * 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.**

We found that as the frequency as which the string oscillated increased, the number of nodes present increased as well. They both varied directly with one another.
 * 4. What is the effect of changing frequency on the number of nodes?**


 * 5. What factors affect the number of nodes in a standing wave?**

The number of antinodes (which is always one less than the number of nodes), and therefore the length of the string and the wavelength of the standing wave would affect the number of nodes in a standing wave.


 * Conclusion:**

 In this lab, we investigated the relationships between frequency and three other variables: harmonic number, wavelength, and tension. Based on our data, we determined these relationships as follows, which were exactly what we hypothesized they would be:

 For the frequency vs. harmonic number graph, we attained a slope of 9.5867, which is the natural frequency, or fundamental frequency, which means the frequency when the harmonic number equals 1. This number is extremely important, as all of the rest of the harmonics build off of this number. In order to verify the accuracy of this graph, we set the slope of the graph (9.5867) equal to v/2L and solved to velocity, found to be 96.826 m/s. We then compared this value to the slope of the graph of frequency vs. wavelength, which is another approximation for the velocity of the wave moving through the string, which is 90.922 m/s. Because these two values are both experimental, we calculated percent difference between them, which we found to be 6.29%. This relatively small difference could be the result of measurement techniques, as we could not precisely create two antinodes (we may have been a few Hertz off, as well as a few centimeters off with the wavelength). Both of these problems could be the cause of this error.

 For the frequency vs. wavelength graph, we attained an inverse linear relationship, whose slope 90.922, is the velocity of the wave, in m/s. The y is the frequency, and the x is the wavelength. The exponent of the x is -1.0179, which is only a 1.79% error from the expected exponent, which is -1. We compared these two values in order to determine how close the relationship we found between wavelength and frequency came to the theoretical relationship. We suspect that this error comes from our measurements of the wavelength, for we were not precisely measuring from the crest of one wave to the crest of another (it was impossible to determine the same two points on two different waves). Also, in order to analyze the effectiveness of this experiment in greater detail, we compared this velocity with the theoretically velocity attained from using the equation v = sqrt(T/(m/L). From this equation, we found the velocity in one of our trials to be about 90.82 m/s, while the slope of our graph is 90.922 m/s, a percent error of about .11%, further confirming the inverse relationship between frequency and wavelength.

 For the relationship between tension and frequency, we found a square root relationship (ƒ = sqrt(T)). We then compared our value for the exponent of the T (0.4738) to the value we theorized (0.5) by doing percent error. We found that the error in this comparison was 5.24%. We suspect that our exponent value differs from the theoretical value because we could not be certain that by readjusting the frequency after adding more hanging mass that we were able to perfectly restore the amount of antinodes. Our measurements were not perfect.

=**LAB: WHAT IS THE RELATIONSHIP BETWEEN THE MASS ON A SPRING AND IT'S PERIOD OF OSCILLATION?**=

- To directly measure the spring constant "k" of a spring by measuring the elongation of the spring for specific applied forces and creating a graph of F vs X and finding the slope. - 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, and using the theoretical formula. - To compare the two values of spring constant "k"
 * Purpose:**


 * Hypothesis**: We hypothesize that if we use these two different methods of finding k, they will give us almost the exact same value of "k", for they both allow us to find the spring force constant, which never changes. Also, we predict, based on the theoretical equation T = 2*pi*SQRT(m/k), that as the mass hanging on a spring increases, so will it's period of oscillation.


 * Materials:** For this lab, to find "k" directly, we hung **various masses** from **a spring** and used **a ruler** to measure the distance traveled. To find "k" indirectly, we used **the same spring set-up** and **stopwatches** to manually find the each mass's individual period on the spring. Finally, we used **Andrew Miller** and an **excel program** to organize our data.

//Directly Finding "k"//
 * Procedure:**
 * 1.** Set up a spring on a stand with a ruler attached to it like so:
 * 2.** Make sure that the end of the spring, at rest without any mass on it, lines up with the zero mark on the ruler.
 * 3.** Place one known mass on at a time, and measure the distance it stretches the spring (in meters)
 * 4.** Record on excel and create a Force vs. Distance graph where force is the weight of each mass and the spring mass is negligible.
 * 5.** Find the slope of the trendline of the graph. This value will be "k".

//Indirectly Finding "k"//
 * 1.** Using the same spring set-up, (where at rest the spring's end is at zero), test one mass at a time.
 * 2.** Let the spring fall and record how long it takes the spring to go through 10 periods or oscillations.
 * 3.** Record the spring's time for one single period on excel as well as the individual mass used in each trial
 * 4.** Graph T^2 vs. Mass graph. The slope will be (4*pi^2)/k. Set this equation equal to the actual slope and solve for "k".





This graph demonstrates Hooke's Law (F = kx ). (The negative sign is irrelevant because its simply connotes the downward direction of the mass on the spring). Thus since F is y and the distance traveled is x, "k" is the slope of the graph or 3.7504 N/m. This graph demonstrates the harmonic motion equation T^2 = ((4pi^2)/k)*m. Since T^2 is the y and m is the x, then (4pi^2)/k is the slope. To find "k", using this graph, set (4pi^2)/k equal to the slope, 10.611, and solve for "k", which is what we did.









Yes, because the difference between our highest and lowest values of k are just a couple of tenths of newton/meters off. For all of our trials, the resulting k values were very similar.
 * 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?**

We measure the time in sets of ten because it is too difficult to measure just one period of some of the springs move too fast for accurate human measurement. By measuring the time for 10 periods, we can divide by 10 and get an average. This prevents unnecessary human error.
 * 2. Why is the time for more than one period measured?**

Both graphs are very similar in that they both show a spring force constant that is very similar. Graph. However we think that the force vs. distance graph is more accurate because it doesn't involve calculating the period. Calculating for the period leaves more room for human error, than just simply reading a ruler.
 * 3. Discuss the agreement between the k values derived from the two graphs. Which is more accurate?**


 * 4. Generate the equations and the corresponding graphs for... TRIAL 1**
 * **position with respect to time**
 * x(t) = Acos(wt) = .17cos((2π/.8455)t = x(t) = .17cos(7.431t)
 * **velocity with respect to time**
 * v(t) = -Awsin(wt) = -.17(7.431)sin(7.431t) = -1.263sin(7.431t)




 * **acceleration with respect to time**
 * a(t) = -Aw^2cos(wt) = -1.263(7.431) = -9.388cos(7.431t)



This problem can be solved using Hooke’s Law, which is the equation we used to find "k" directly, as long as we assume the spring is at equilibrium: F = - k * x F = - (8.75 N/m) * (- .150 m ) F = 1.3125 N
 * 5. 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?**

This problem can be solved using the equation we used to find "k" indirectly: T = 2*pi*SQRT(m/k) T = 2*3.14159*SQRT(.425/7.85) T = 1.462 seconds
 * 6. A massless spring has a spring constant 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?**


 * 6. 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.**



Despite adding the mass of the spring, the slope of this graph remained the same, meaning the spring force constant stayed the same as well. Graphing the square root of m + 1/3m by the period would yield the same result as the graph above, for the graph above is period squared = total mass. Taking the square root of both sides would yield the relationship period = square root of masses.


 * Conclusion:**

In this laboratory investigation, we investigated the spring force constant in two separate ways in order to determine if this constant was really a constant. We attacked our curiosity in two ways. The first way was using Hooke's Law, which states that the restoring force on a spring = the constant times the change in distance. We applied different weights to the spring and measured the distance that the spring fell. We waited until the spring was at equilibrium, for that is the only time when the restoring force is equal to mg. For example, we hung a 67 g mass and allowed the spring to return to equilibrium, and measured the distance that the spring had moved to be 17 cm. We graphed mg by distance in order to achieve a linear graph, whose slope was the spring force constant. For this value, we got 3.7504 N/m.

Putting that result aside, we solved for the spring force constant in another way. Using simple harmonic motion, we calculated the period of one cycle, the time it took for the spring to return to its original position after we applied the mass to it. Because of human error, we timed ten cycles, then divided by 10 in order to get the time for one cycle, which is the period. For example, for the first trial, we timed ten cycles to be 8.455 seconds, so one period would be 0.8455 seconds. Using this information, and using the equation that we have for the period of a spring, we graphed period squared by mass hanging. We determined the slope of the graph to be 4π^2/k, so we set the slope equal to this, and solved for k and got 3.7205 N/m.

After experimenting with the value of k in two autonomous ways, we compared these two values to see how different they were. The percent difference between k = 3.7504 N/m and k = 3.7205 N/m was 0.800%, a very minute difference, which confirms our hypothesis that the value of k is a constant, and does not change, no matter the method by which it is attained. We attribute this difference to human error in terms of our ability to accurately time, for a human's reaction time isn't perfect; however, because it is extremely close to 0, this error is negligible, yet unavoidable.