Chloe,+Ani,+Jill,+Elena

= = =Lab: What is the relationship between frequency and harmonic number, wavelength, and tension? = **Group Members: Ani Papazian, Jillian Laub, Chloe Murtagh, Elena Solis ** **Date Completed: 5/23/11 ** **Date Due: 5/24/11 **

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

**Purpose:** To find all of these relationships using a string vibrator/generator and comparing them using graphs.

**Hypothesis/Rationale:** -The higher the frequency, the greater the harmonic number because the more waves per second, so the interference is higher, creating more nodes (there is a direct positive linear relationship) <span style="background-color: transparent; color: #000000; font-family: serif; font-size: 16px; text-align: start; text-decoration: none; vertical-align: baseline;">-The higher the frequency, the smaller the wavelength becomes because (L being length of string, which is constant, n which is harmonic number, and λ, which is wavelength) In this equation, as frequency is increased (increasing the harmonic number), the wavelength decreases (there is a inverse linear relationship)

<span style="background-color: transparent; color: #000000; font-family: serif; font-size: 16px; text-align: start; text-decoration: none; vertical-align: baseline;">-As the mass on the string increases, increasing the tension, the frequency required to achieve the same harmonic number for each mass will have to be greater because as the tension becomes greater, the frequency increases according to the equation where T is tension and f is frequency (there is a direct positive square relationship)

Materials: electrically-driven oscillator, pulley & table clamp assembly, weight holder, masses, string, electronic balance

Procedure: Frequency vs. Harmonic Number: 1. Set up oscillator on table with string 2. Attach a 1,500 g mass to the end of string 3. Turn on oscillator, adjusting frequency until a harmonic number is reached 4. Record harmonic number and frequency in excel 5. Repeat steps 3-4 for a variety of harmonic numbers and record, make graph

Frequency vs. Wavelength: 1. Turn on oscillator, adjusting frequency until a harmonic number is reached 2. Using the equation L = (n/2)λ, find wavelength 3. Record wavelength and frequency in excel 4. Repeat steps 2-3 for multiple wavelengths, make graph

Frequency vs. Tension: 1. With mass of 1,500 g, adjust frequency until harmonic number of 3 is found 2. Record frequency and mass in excel 3. Repeat steps 1-2 with different masses (2,000g, 2,500g, 3,000g, etc.), make graph

<span style="background-color: transparent; color: #000000; font-family: serif; font-size: 16px; text-align: start; text-decoration: none; vertical-align: baseline;">Data/Graphs:

According to our graph, frequency is 19.927 times the harmonic number. If the harmonic number is n=3, the frequency would be about 59.78 Hz. This has a percent difference from our experimental data of 2.7%. The slope of our graph is about 20, which is the natural frequency of the string, or the frequency when the harmonic number of 1.



According to our graph, frequency is wavelength to the -1.0257 power times 65.758. If the wavelength is 1.1, frequency is 59.6 Hz. This has a percent difference from our experimental data of 4.4%. Since frequency=velocity/wavelength, wavelength should be to the -1 power. The approximate speed of the wave in a string is 60m/s, so the coefficient should be around 60.



According to our graph, frequency is tension to the .4506 power times 30.293. If the tension is 14.7, frequency should be 101.7 Hz. This is a 1% difference from our experimental data. The power should be about .5, because there is a square root relationship between frequency and tension, according to the equation in the rationale.



<span style="background-color: transparent; color: #000000; font-family: serif; font-size: 16px; 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. from graph: using frequency equation: using velocity equation:

2. What would be the effect if the string stretched significantly as the tension increased? How would that have affected the data?

As the tension increases and the string gets tighter, the elastic property of the string will increase, making the velocity increase. With a higher velocity, a greater frequency will be required to attain the same number of anodes (v=λf). This is consistent with our data that shows that as the tension increases, <span style="font-family: Arial,Helvetica,sans-serif;">the frequency needs to be greater to achieve the harmonic number 3.

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 amount of hanging mass needed to create number of nodes because, at a set frequency, the tension must be one certain value to achieve a number of nodes. If the string is stretchier, it will require more weight on the end to achieve that needed tension, because the string will give with some of the weight.

<span style="font-family: Arial,Helvetica,sans-serif;">4. What is the effect of changing frequency on the number of nodes?

<span style="font-family: Arial,Helvetica,sans-serif;">As the frequency increases, as does the number of nodes. This is because as the number of waves per second increases, there are many more opportunities for interferences (the cause of nodes). The relationship between frequency and harmonic number (number of nodes) is f=xn, where x is the frequency where there is only one node.

<span style="font-family: Arial,Helvetica,sans-serif;">5. What factors affect the number of nodes in a standing wave?

<span style="font-family: Arial,Helvetica,sans-serif;">Frequency, string length, tension on the string

<span style="background-color: transparent; color: #000000; font-family: Arial,Helvetica,sans-serif; font-size: 16px; text-align: start; text-decoration: none; vertical-align: baseline;">Conclusion: Our hypotheses were generally supported through our lab. We hypothesized that frequency and harmonic number were directly related, the higher the frequency, the greater the harmonic number. We started with the first harmonic and found it had a frequency of 19 Hz and we worked our way to the sixth harmonic, which we found had a frequency of 118.2 Hz. Both our data and graph supported our hypothesis of a positive linear relationship between frequency and harmonic number. We also hypothesized that frequency and wavelength would have an inverse and linear relationship. Our data supports that they are inversely related because as frequency increases, the wavelength decreases. For example, we found that with a frequency of 19 Hz there was a wavelength of 3.3 m and for a higher frequency of 118.2 Hz there was a wavelength of .55 m (which is smaller than the wavelength for a lower frequency). We predicted a linear relationship but our graph showed a negative exponential relationship but it still showed a negative relationship, which supports our hypothesis. Finally, we hypothesized that frequency and tension have a direct positive square relationship. We changed the hanging mass, which changes the tension of the string. We add more mass, increasing the tension and increasing the frequency. When we made the tension greater we had to increase the frequency to make sure there were 3 nodes (our constant for these trials). As seen on the graph above, frequency and tension do have a positive square relationship thus verifying our hypothesis. Overall, our results supported our hypotheses of the relationships between frequency and harmonic number, wavelength, and tension. = = =Lab: What is the relationship between the mass on a spring and its period of oscillation?=
 * Group Members: Ani Papazian, Jillian Laub, Chloe Murtagh, Elena Solis**
 * Date Completed: 5/16/11**
 * Date Due: 5/17/11**


 * Purpose:**

To identify the relationship between the mass on a spring and its period of oscillation, which will result in our finding of k.

After plugging in our period and mass values into, and graphing T^2 vs. mass, we will get a very similar k value. Since both methods are supported as true, theoretically, they should both produce corresponding k vales.
 * Hypothesis/Rationale:**

**Materials:** <span style="font-family: Tahoma,Geneva,sans-serif;">spring stand, spring, ruler, assortment of masses, timer, balance, clamps and rods

**Procedure:** 1. Set up the spring stand/ruler with the spring attached. 2. Hang a certain mass from the bottom of the spring, record the mass. 3. Pull the spring downward a certain amount, recording the x (distance) value, and release. 4. Begin timer when spring is at maximum height and continue for 10 oscillations. 5. Repeat with various masses. 6. Plug data into excel and use calculations in excel to find force and k value.


 * Data:**

**Excel Data Tables:**

**Graphs:** T^2 vs. Mass:

Force vs. x:

**Sample Calculations:**

Find k using graph of T^2 vs. mass, using force and Hooke's law, and the percent difference:



1. Does 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, our data indicates that the spring force constant remains the same for different forces. We tested different masses, which according to Hooke's law equals the force. Therefore we had different forces but found that for each trial k was equal to about 4 N/m. The values ranged from 4.022 N/m to 4.101 N/m and they averaged 4.063 N/m. Because the values were so close our data indicates that k is constant for all forces.
 * Discussion Questions:**

2. Why is the time for more than one period measured? - If the period is measured over a larger period of time, we are able to time it more exact. This way, our period is a more exact number and there is less room for human error.

3. Discuss the agreement between the k values derived from the two graphs. Which is more accurate? For the graph T^2(s^2) vs Mass the k value was found to be 4.07N/m. This value was found be manipulating the equation for the period of a spring(as seen above) and then using the slope of the line to solve for k. For the graph Fs vs x we found k to equal 3.92. This graph was derived from Hooke's law F=kx and the slope of the line is the k value. The first graph is more accurate because it is closer to our theoretical value for k which was 4.07. This graph accounts for the period of the function which is more useful than the other graph because the spring was moving during the experiment and Hooke's law applies to systems at equilibrium. Therefore graph T^2(s^2) vs Mass is more accurate.

4. equations and graphs for: -position: - velocity: - acceleration:

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? 6. 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?



7. We neglected to take into account the mass of the spring itself. Are your results any better when using the more accurate relationship? Explain these results. We originally found the value for k to be 3.92 on the graph of Fs vs. mass. When we incorporated the value of the spring mass, the k value on the graph stayed the same. Despite adjusting the mass, the value of the spring constant stayed the same.


 * Conclusion:**

Our hypothesis was definitely supported by our conclusion. We hypothesized that our k value from each of the methods we used to solve for it would be almost the same. We solved for k once using the period, and also using Hooke's law. After comparing both results, they were only 3.63% different.

Because we did not have expected/theoretical results, we had to compare our results to the others, and find percent difference. Our results were very similar from one way to solve for k to the other. When using the period to solve for k, the average k value was 4.063. When using Hooke's Law to solve for k, the k value was 3.92. The difference between these two values was 3.63%. The error most likely came from the human error in using a stopwatch to time the periods. Because reaction time would delay us from stopping the watch exactly at the top of the amplitude, results would be a little skewed. Also, the imprecision of the ruler measuring the 'x' value could have also contributed to the error. Finding some more automated way to measure the exact time of the period would greatly increase the precision. Using a more precise ruler to measure the 'x' value would also improve the results.