Rachel,+Spencer,+Jimmy,+Sam

=Transverse Standing Waves on a String=
 * Group:** Rachel, Sam, Jimmy, Spencer
 * Due Date:** 5/24

1. There will be a linear direct relationship between frequency and harmonic number. 2. There will be an indirect power relationship between frequency and wavelength. 3. There will be a direct power fit between frequency and tension. 1. We know that the natural frequency of the string is somehow directly related to the frequency and harmonic number. 2. We know this because of the equation where frequency is equal to velocity over wavelength. 3. We know that it is a direct root because of the equation where velocity is equal to the square root of tension over mass over length. Frequency comes into play because frequency times wavelength is equal to velocity.
 * Objective:** To find the relationship between frequency and tension, frequency and wavelength, and frequency and harmonic number on a transverse wave.
 * Hypothesis:**
 * Rationale:**

1. Set up string with oscillator on one end and the pulley, table clamp, and weights hanging off the other end of the table. 2. Chose a frequency with the dial. 3. Measure each wavelength. 4. Repeat 2 and 3 for numerous trials. 5. Pick a set wavelength and harmonic number. 6. Add mass to the end of the string. 7. Change the frequency dial until you reach your set wavelength and harmonic number. 8. Repeat 6 and 7 for multiple trials.
 * Materials:** Electrically-driven oscillator, pulley, table clamp, weight holder, masses, string, electronic balance, meter stick
 * Procedure:**

media type="file" key="Movie 25.mov" width="300" height="300" = = Data Table: These were our measurements in all three of our experiments. The graphs showing the relationships between the values can be shown below.



In this graph, we put the frequency on the y-axis and the harmonic number on the x-axis. This produced a linear relationship with a strong fit (r squared = .9994). The equation was frequency = (natural frequency) (harmonic number). The natural frequency of the red string was the slope. Our value for this was 19.553 which is a good value considering the actual natural frequency is around 20.



In this graph, we put frequency on the y-axis and the wavelength on the x-axis. This produced an negative power relationship with a strong fit (r squared =.9922). The equation and how we arrived there is shown below. The coefficient for x on the graph is the velocity of the wave, 59.538m/s. Our value, using our measurements, was 54.11m/s, which is not far off from what our graph gave us. Also, the exponent for x should be -1 according the equation. We got -1.0213, which is very close to the expected value.





In this graph, we put frequency on the y-axis and tension on the x-axis. This produced a positive power relationship with a strong fit (r squared = .999). The equation and how we arrived there is shown below. The slope is everything that is being multiplied by the square root of tension. The square root of tension is being taken in this graph and that is why it has the curve it does. Since this is true, exponent of x should be .5. It is actually .4974, which is very close.




 * 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 stretch significantly as the tension increased? How would that have affected the data?

If the string stretched that means it would become longer and thus the wavelength would be longer. This would of contributed to some erroneous data because most of the calculations were based on a set length of the string.

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 tighter and thinner the strong is the more nodes can be produced. As tension increases so does frequency. The harmonic number is dependent on frequency so that means it is also dependent on tension. As more tension is applied the harmonic number changes and more or less nodes are created. The tension of the strong depends on the type of string. If a string is stiff and thick tension will not affect the harmonic number as much, but if it is forgiving and stretchy than tension will affect the harmonic number a lot. To achieve a set number of nodes it would be easier to have a stretchy string because pinpointing the right tension (or hanging mass) wouldn’t be too difficult.

4. What is the effect of changing frequency on the number of nodes?

As the frequency increases, the number of nodes increase. As the frequency decreases, the number of nodes decrease.

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

The tension force of the mass hanging on the string and the frequency.


 * Conclusion:**

In this lab, we found three relationships - frequency vs. harmonic number, frequency vs. wavelength, and frequency vs. tension. We proved equations that explained these relationships to be true in these experiments. Frequency was found to have a direct relationship with harmonic number, and the equation is frequency = (natural frequency) x (harmonic number). When the harmonic number increases, the frequency increases, and vice versa. Frequency was proved to have a negative power relationship with wavelength, the equation being shown below (first one). When wavelength increases, frequency decreases, and vice versa. Frequency was shown to have a positive power relationship with tension. The equation is also shown below (second one). When tension increases, frequency increases, and vice versa. The three relationships between the different factors were proved to be true.

=Mobile Project=
 * Group:** Spencer, Rachel, Sam, Jimmy
 * Due Date:** 5/9/11


 * Objective:** Design and build a kinetic mobile that reflects a physical balance.
 * Hypothesis:**
 * Procedure:**


 * Calculations:**




 * Conclusion:** We proved our mobile to be in equilibrium through proof of the first 2 conditions of static equilibrium. Both the sums of the torques and the forces in the y-direction are equal to 0. We had a little percent error due to the fact that we chose our masses and used calculations to find the distance of the radius. For example, on our first dowel we only had a 2.66% difference between the right and the left side. The right had a torque of 0.24 and the left had one of 0.2337. These should be equal to each other however there were other factors that could have influenced the difference in torques. The error could have been due to the slight added weight of the tape and the string. Also, since the string had a width, it is very difficult to have our hanging mass be exactly where the radius should begin.

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

media type="file" key="Movie 23.mov" width="300" height="300"

Data Charts and Graphs:

For this part of the lab, we hung different masses on the end of the spring and found the displacement of the spring. We then graphed our observations and fitted it to a linear model. F=kx was the equation we used. K was the slope of this equation, so that is how we found the spring force constant. k = 28.35 N/m

Then we hung different masses on the end of the spring and found the its period while it was oscillating. We then graphed our observations and fitted it to a linear model. The equation of the graph and how we found k is shown below.

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? Yes, it is constant. We can tell this by the linear relationship of the displacement versus the applied force. Our good r squared value of 0.9956 shows that a linear relationship fits for this model. Since it holds a linear relationship (F=kx), k, the spring force constant, will always remain the same for all forces.

2. Why is the time for more than one period measured? We measured ten periods and then divided this number by ten to get the time of one period. This is done to decrease human error. This type of error is common when using a stopwatch, but our way mitigates this error by completing ten trials.

3. Discuss the agreement between the k values derived from the two graphs. Which is more accurate? The Force vs. Displacement graph has a k value of 28.35 N/m, while the Mass vs. Period Squared graph has a k value of 26.23 N/m. This is pretty close to one another. The percent difference is 7.77% and is shown below. We trusted the Force vs. Displacement graph more than the other one. We thought there was much less error in measuring displacement than there is in using a stopwatch to find the period of a spring. Since there was less error in that graph, we think it was more accurate.

4. A spring constant k=8.75N/m. If the spring is displaced -0.150m from its equilibrium position, what is the force that the spring exerts? 5. A massless spring has a spring constant of k=7.85N/m. A mass m=.425kg 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 m+ 1/3ms (where m is the hanging mass and ms is the mass of the spring)? Redo graph #2 using (m+1/3ms)^1/2, and explain these results.

k = (4π^2(m+1/3ms))/T^2 This is derived from T = 2πsqrt(m/k), if m = m+1/3ms

By just analyzing the equation it can be determined that k will be bigger than the previous k found without the mass of the spring. This will thus be more accurate and closer to the value of k determined in the F vs. x graph.



0.6914*4π^2 = 27.295 which is really close to 28.353