Nikki,+Jessica,+Danielle

= = toc =Standing Waves on a String Lab= Period 4 5/20/11 Due 5/23/11

To find the relationships between frequency and the number of antinodes, wavelength, and tension.
 * Objective**

We hypothesize that as frequency increases, the number of antinodes will increase. Wavelength will decrease. This because there will be more waves in the same length of string. As we increase tension, frequency will decrease. We hypothesize this because increasing tension increases the velocity(based on the equation ). Because velocity is increasing and the wavelength is constant (v=wavelength x f) the frequency will increase.
 * Hypothesis**

Red braided string String vibrator Sine wave generator Masses
 * Materials**

1. Attach the string to a generator that is clamped to the table. Send the string through the pulley and attach a set mass to the end of the string. 2. Turn on the generator and make it at its maximum amplitude. 3. Move the dials until the string is vibrating at its maximum amplitude for that amount of antinodes. Record that frequency. The string vibrator should be making little to no sound. 4. Repeat step three but find the frequency for a different amount of antinodes. Repeat these steps until you get significant data to make a conclusion.
 * Procedure**


 * Sample Calculation**


 * Data**


 * Discussion Questions:**

1. Calculate the tension T that would be required to produce the n=1 standing wave for the red braided string. F=ma T=W T=mg

2. What would be the effect if the string stretched significantly as the tension increased? How would that have affected the data? The velocity would increase as string gets tighter. This is because of the equation v=SQRT(elastic property/inertial property). As the elastic property increases, so does the velocity. In order to get the same number of antinodes, the frequency would have to be higher because the velocity is higher and v=(wavelength*frequency).

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. A string that has a higher elastic property, would have a higher velocity. Therefore, the frequency would be increased as well in order to maintain the number of antinodes. Also, there would need to be less mass to keep the number of nodes constant because the string would need the lower elastic property. This works in the reverse as well. A string with a lower elastic property would need more mass to increase its elastic property so the number of nodes would remain the same.

4. What is the effect of changing the frequency on the number of nodes? Changing the frequency increases the number of nodes because the string is vibrating more frequently so it is also reflected more frequently. Therefore there are more crests and troughs so there is an increased amount of both types of interference

5. What factors affect the number of nodes in a standing wave? Frequency is the primary factor that affects the number of nodes in a standing wave but subsequently therefore all things that affect frequency of a standing wave affect the number of nodes. The properties of the string such as length, elasticity, and mass specifically affect the frequency. All objects have a natural frequency, but by changing the length, amount of tension, or mass of the spring changes the velocity. This is based on the two equations for velocity of a string: which show increased tension and length increase velocity and increased mass (which increases inertia) decreases velocity. Therefore, if velocity is increased and wavelength remains constant then the frequency will increase (based on the eqn ), which changes the number of nodes in the wave.

The data we collected and graphed illustrates the very relationships we hypothesized would be true of a standing wave in a string. We found that the number of antinodes and the frequency were in fact directly related, which is shown in our first graph because it is linear. As frequency increased the number of antinodes increased. Frequency and wavelength on the other hand have an inverse relationship as shown in the second graph by the downward curve. As frequency increases wavelength must decrease to keep the velocity constant (consider velocity of a wave is always the same in the same medium). Also, the equation of the graph also illustrates the correct relationship because it is y=mx^-1 where y is frequency the slope is velocity and x is wavelength. This equation is derived from the equation v=λf by solving for frequency and although the exponent in the equation of our graph is not exactly -1 it is extremely close. Lastly, tension had a direct square relationship with frequency based on the equations: and v=λf so that when you use substitution for velocity Therefore our graph should be a curve which it is and it shows how whatever factor frequency is increased by Tension is increased by the square root of that factor (it descreases), which is why our curve starts at the axis and curves down as it moves from left to right across the graph. So in the equation y=mx^.5 x= tension, y= frequency, and the slope is the square root of l over m divided by wavelength.
 * Conclusion**

Waves have real life applications in terms of a great deal of technology especially that used by the military, which involves the use of radio waves, sonar, etc. Specifically, these relationships are useful when needing to match a certain frequency of a radio the operator can change the wavelength in order to achieve effective communication. = = = = = = = = = = = =

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

Period 4 5/13/11 Due 5/16/11


 * Objectives**
 * To directly determine the spring constant k of a spring by measuring the elongation of the spring for specific applied forces
 * 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
 * To compare the two values of spring constant k

There is a direct relationship between the mass on a spring and its period of oscillation.
 * Hypothesis**

Springs, masses, spring clamp/rod with ruler attached, balance, timer
 * Materials**

Direct: 1. Find the equilibrium point of the hook hanging off a spring. 2. Add mass to the hook (make sure to include the mass of the hook in calculations). Measure the displacement of the hook from the equilibrium point. 3. Repeat for multiple masses. 4. Use Hooke's law to solve for k. Indirect: 1. Put mass on the hook. 2. Pull the mass down a certain displacement and release. To find the period, count ten oscillations of the spring while timing. One person should count oscillations and tell another person with the timer when it has reached ten. 3. Repeat multiple trials for each mass. 4. Add more mass and repeat. Make sure to displace the hook the same amount each time, not just pull the hook down to the same number. 5. Use variation of T equation to solve for k.
 * Procedure**

Direct:
 * Data**



Indirect:



Derivation:
 * Calculations**

Direct:

Indirect:

__**Discussion Questions:**__
 * Conclusion**

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 because our graph of the two illustrated a linear relationship where the slope was the spring force constant (y=kx), which shows the spring force was in fact constant. 2. Why is the time for more than one period measured? BecBecause the periods occur so quickly it is much easier to count a certain amount of oscillations and record the time for that and then calculate the period. 3. Discuss the agreement between the k values derived from the two graphs. Which is more accurate? The k values generated from the direct and indirect methods were relatively close in value. The k value from the direct method was 36.8 and the average k value from the indirect method was about 38.3 so the difference between the two was about 1.5. The value calculated via the direct method is more accurate because there is less human error involved. Also, as can be seen in the data the k value from the direct method is more constant for different masses than it is with the indirect method. 4. Generate the equations and the corresponding graphs for i. position with respect to time. ii. velocity with respect to time. = =  iii. acceleration with respect to time.

= =

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 (where //m// is the hanging mass and //ms// is the mass of the spring)? Redo graph #2 using, and explain these results.

Conclusion: In this lab, we were trying to find the spring force constant in two ways, directly and indirectly. We wanted to prove that they would be same either way we tried. Because we only got a percent difference of 4.148%, we can say that our lab was successful. Our spring force constant of 36.88 from the direct method is extremely similar to the 38.06 that we got indirectly. Because these numbers are so close, our hypothesis was proven correct.

The error might have occurred would have happened during both the direct and indirect scenarios. Because we were measuring the displacement in the direct method, we could have been slightly off with our measurements. In the indirect method, our error comes from a different source. The reaction time of the person timing varies each time we did this experiment. Also the first trial we did, we set the time and counted the oscillations. This proved to not be as accurate because we would be stopping mid-oscillation, which is more difficult to count than if it was half a second.

We fixed most of our errors before they really hurt the experiment. After we realized there was a more accurate way to get data for the indirect experiment, we switched methods. We then started setting the number of oscillations and counting the time. Also to make the reaction time less significant, we had the same person time each trial. We also did multiple trials of each weight just to make sure that we were getting similar results. If the reaction was affected every trial, it in turn has very little effect. To make the direct method more accurate, we could have used a measuring tool that measures more accurately. = = = = = = = = = =

=Mobile Design Project= Objective: Design and build a kinetic mobile that reflects a physical balance.



Large Dowel = 75 cm, CG=37.5 cm Smaller Dowels = 30.5 cm, CG= 15.25 cm


 * Data:**



Percent Error:

Sample Calculation (finding r)
 * Calculations**:

Net Torque Overall Mobile

Net Force Overall Mobile
 * We found the tension in the string holding up the entire mobile with the force sensor





Percent Error:

Our mobile has almost a net torque and force of 0, the two conditions of equilibrium. The source of error stems from the hot glue we had to add to keep strings attached to the dowels and the crayons to stick together. The hot glue, although seemingly insignificant, has enough mass to throw the mobile off balance. Because the mass of the hot glue is not in the torque and force equations, the net torque/force cannot be zero in the calculations.
 * Conclusion**

=Levers-Static Equilibrium Lab= Period 4 4/15/11 Due 4/11/11

**Purpose** What is the relationship between the torques acting on an object at equilibrium?

Meterstick, pivot, knife-edge level clamps, 3 mass hangers, set of masses, unknown mass, balance, string, masking tape
 * Materials**

The torques (counterclockwise and clockwise) acting on an object at equilibrium are equal.This is because in order for an object to be at equilibrium, the sum of the forces, or torques on either side have to equal zero. Since the object is at equilibrium, the counterclockwise and clockwise torques have to equal zero, therefore making them equal.
 * Hypothesis**


 * Procedure**
 * 1. Find the center of mass of the meterstick. **
 * 2. For trial 1, take a .020 kg mass and a .1 kg mass and place one on each side of the meter stick. The heavier mass should be closer to the fulcrum than the other mass is. Balace the meter stick. **
 * 3. For trial 2, place a new heavier mass, .2 kg, on one side of the meter stick alone. Place the other two masses on the other side of the meter stick. The heaviest mass should be closer to the fulcrum than the other two masses are. **
 * 4. For trial 3, set the fulcrum off at least 20 cm from the center of mass. Use the heavier mass on the shorter side of the meter stick and the lighter mass on the long side to balance it. **
 * 5. For trial 4, keep the fulcrum offset. Use two masses. Use the heavier mass on the shorter side of the meter stick to balance it. **
 * 6. For trial 5, we must solve for the unknown mass. Set the fulcrum back to the center of mass. Use the unknown mass and one other mass on the other side of the meter stick to balance it out. Calculate for unknown mass using knowledge that the torque equations. **
 * 7. For trial 6, measure the mass of the meter stick. Measure the angle that the cable is at. Measure the distance away from the fulcrum (the hinge) that the cable is at. Using all of this experimental data, solve for the tension in the cable. **

**Data**

Top Section = experimental Bottom Section = theoretical



**Calculations:** 1. Two different masses with the fulcrum at the center of mass of the meterstick (COM).    2. Three different masses with the fulcrum at the COM.    3. Two different masses with the fulcrum at least 20 cm off COM.

<span style="font-family: Arial,Helvetica,sans-serif; font-size: 12px; margin: 0px 0px 0px 0.5in; padding: 0px; text-indent: -0.25in;"> <span style="font-family: Arial,Helvetica,sans-serif; font-size: 12px; margin: 0px 0px 0px 0.5in; padding: 0px; text-indent: -0.25in;">

<span style="font-family: Arial,Helvetica,sans-serif; font-size: 12px; margin: 0px 0px 0px 0.5in; padding: 0px; text-indent: -0.25in;">4. One mass with the fulcrum at least 20 cm off COM. <span style="font-family: Arial,Helvetica,sans-serif; font-size: 12px; line-height: 0px; margin: 0px 0px 0px 0.5in; overflow: hidden; padding: 0px; text-indent: -0.25in;"> = =

<span style="font-family: Arial,Helvetica,sans-serif; font-size: 12px; margin: 0px 0px 0px 0.5in; padding: 0px; text-indent: -0.25in;">5. One object of unknown mass, everything else to be determined by you. <span style="font-family: Arial,Helvetica,sans-serif; font-size: 12px; margin: 0px 0px 0px 0.5in; padding: 0px; text-indent: -0.25in;"> <span style="font-family: Arial,Helvetica,sans-serif; font-size: 12px; margin: 0px 0px 0px 0.5in; padding: 0px; text-indent: -0.25in;">*we added a second known mass in order to solve for the mass of the unknown

<span style="font-family: Arial,Helvetica,sans-serif; margin: 0px 0px 0px 0.5in; padding: 0px; text-indent: -0.25in;"> 6. A single mass supported by a tension at an upward angle and a wall. (Teacher will set this up for you.) Percent Error (trials 1-5 in order)

<span style="font-family: Arial,Helvetica,sans-serif;">**Analysis Questions:** <span style="font-family: Arial,Helvetica,sans-serif;">1. Does it get easier or harder to rotate a stick as a mass gets farther from the pivot point? <span style="font-family: Arial,Helvetica,sans-serif;">2. Does the weight of the mass increase as you move the mass away from the pivot point (your index finger) <span style="font-family: Arial,Helvetica,sans-serif;">3. Why is more mass required to balance the meterstick as you move another mass farther from the pivot? <span style="font-family: Arial,Helvetica,sans-serif;">4. Why must the mass of the hangers and clamps be taken into account in this experiment? <span style="font-family: Arial,Helvetica,sans-serif;">5. If you are playing seesaw with your younger sibling (who weighs much less than you), what can you do to balance the seesaw? Mention at least two things. <span style="font-family: Arial,Helvetica,sans-serif;">6. What kept the meterstick in equilibrium in the fourth trial? In other words, what counterbalanced the known mass?
 * <span style="font-family: Arial,Helvetica,sans-serif;">It gets harder.
 * <span style="font-family: Arial,Helvetica,sans-serif;">No. The weight remains the same because the formula is still w = mg.
 * As one mass moves away from the pivot point, the lever arm is increasing which throws the meter stick off balance. To get the torques equal again and bring the meter stick back to equilibrium, more force needs to be added to the other side. If something increases on one side of the equation (Toa clockwise = Toa counterclockwise), something must increase on the other side of the equation as well.
 * The masses of the hangers and clamps are part of the mass hanging off of the ruler. They affect the force (mg x 9.8) that is being applied to the meter stick. We need an accurate force to solve for torque, so those masses must be included.
 * <span style="font-family: Arial,Helvetica,sans-serif;">You should sit close to the center of the seesaw and your sibling should sit further out.
 * Because the fulcrum was not at the center of mass of the meter stick, there was more mass of the stick on one side. The known mass was put on the shorter side to balance out most of the meter stick being on the other side of the pivot.

<span style="font-family: Arial,Helvetica,sans-serif;">**Conclusion:** <span style="font-family: Arial,Helvetica,sans-serif;">Based on our data, our hypothesis was correct. Because the highest percent difference we had between the two torques was 2.22%, our experiment was extremely accurate. We proved that the torque on one side is very close to being equal to the torque on the opposing side. In our first trial, the torque of one is 1.37 and the torque of two is 1.34 showing that they are almost equal and therefore our hypothesis was proven correct.

<span style="font-family: Arial,Helvetica,sans-serif;">Although our results were fairly accurate, there are still sources of error in this lab. We never measured that the meter stick was actually at equlibrium. Instead we just guessed that it was close the point we needed. Also, we assumed that the center of mass for the meter stick and the masses was at the geometric center of the object. Although this is close to being true, there could have been deformities in the objects that would make this untrue. However, these sources of error barely effected the results of our lab.

<span style="font-family: Arial,Helvetica,sans-serif;">In order to make this lab even more accurate, we could have measured the distance from the meter stick to the table at the point of equilibrium. This way, we would know exactly when the meter stick was balanced and would not be guessing that we were "close enough" to equilibrium. Also, we could have solved for the center of mass of the objects before starting just to prove that it really is in the geometric center. This lab is useful when figuring out how to balance two objects from a beam. For example, when constructing a mobile you would have to use similar methods to ones we used during this lab. Also, if you become an engineer and have to design a light fixture or a work of art, this technique of balancing objects and making their torques equal would prove to be extremely useful.