Jae,+Ross,+Rory,+Richie

=Standing Wave= Jae, Ross, Rory, Richie 5/20/11

Objective: To find the relationship between 1. Frequency and the number of antinodes 2. Frequency and wavelength 3. Frequency and Tension of the string

Hypothesis: 1. We believe that the frequency and the number of antinodes will be a linear relationship. This is because we know that 2. We believe that the frequency and the wavelength will have an inverse square relationship. We get this from the equation v=(lamda)f. Because our velocity was constant, as frequency goes up, wavelength will decrease. 3. We believe that the frequency and the tension in the string will be a direct square root. We believe this because the square root of Tension is directly related to velocity and increasing tension will increase the velocity. If velocity increases, and wavelength stays constant, then the frequency will also increase.



Procedure: 1. Place a mass (500g) on the end of your string, hanging off a pulley. 2. Turn the frequency up until you have one antinode in your string and record the frequency. 3. Repeat step 2 multiple times with a different number of antinodes. 4. For objective two measure the length from one node to the next one and record this number at each frequency. 5. For objective three decide how many antinodes you want to find 6. Add more mass onto the end of the string and record the frequency when reached. 7. Repeat step 6 at least 5 times.

Materials: 1. Electrically-driven oscillator 2. Pulley & table clamp assembly 3. Weight holder & selection of masses 4. String 5. Electronic balance

Data:



Calculations: Percent Difference for slope of Frequency vs Number of antinodes.

Discussion Questions: 1. Calculate the tension T that would be required to produces the n=1 standing wave for the red braided string.

2. What would be the effect if the string stretched significantly as the tension increased? How would that have effected the data? The velocity would increase as the tension stretched the string due to the fact that v=SQRT(elastic property/inertial property). As the velocity increased the frequency would have to increase to keep the same amount of nodes. For our data the frequency would have had to have been greater to keep the same amount of antinodes (3) in the string.

3. What is the effect of the type of string on the amount of hanging mass needed to create a set of nodes? Explain this. If the string has a higher elastic property then the velocity would be greater in the string, and the frequency would have to be greater to keep the same amount of nodes. So the mass (which creates the higher tension resulting in the higher velocity) that is needed is dependent on how much tension is needed in the string to make the velocity just right.

4. What is the effect of changing the frequency on the number of nodes? As the frequency increases, so does the number of nodes.

5. What factors effect the number of nodes in a standing wave? The frequency, the tension in the string, and the type of string.

Conclusion: We were right with some of our predictions, yet not with all of them. We were right to hypothesize that the relationship between the frequency and the number of antinodes would be a linear direct relationship. We also found that the slope of our graph was 19.11. In class we discussed how the slope was actually the first harmonic, the natural frequency of our wave. When we compared the two numbers we only got % error, showing that our data collection was pretty consistent and had minimal error. Yet, we were wrong in our prediction of the relationship between frequency and the wavelength. We thought it would be an inverse linear relationship, however we found that if we rearranged the equation we used four our hypothesis, the f=v/wavelength. Our graph would come with an equation such as y=(coefficient)*x^(-1), meaning that the coefficient is the velocity that our string was moving at. We did get out hypothesis correct for our last relationship and we concluded that the coefficient using this manipulation of equations: There was not much room for error in this lab as we were using machines to generate the waves. Yet there were some things that could be improved. At some points we were getting that different frequencies were producing the same exact wave. THis was confusing as we did not know which frequency to pick. (We knew it had to be the one that created the biggest antinode, yet it was hard to tell when the string was moving so fast).

= **__Spring Force Constant__** =

Jae, Ross, Rory, Richie 5/13/11

__Objective:__
 * 1) = To directly determine the spring force constant of a spring by measuring the elongation of the spring for specific applied forces. =
 * 2) = To indirectly determine the spring force 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:__ We believe that the two spring constants (k) that we calculate will be equal to each other. We believe that this because we are using both Hook's Law (F=-Kx) and the equation for Period T=(2pi)*SQRT(m/k) to find the value of k. Being that k is the same in both of these equations and we know that each of these equations has to be true for a spring in simple harmonic motion, we can conclude that the k's in both equations will be the same.

__Procedure:__ **Hook's Law** 1. Measure where the spring falls when there is no mass hanging on the end. This is equilibrium. 2. Add masses in increments of 5 and measure how far the spring elongates from the point of equilibrium. 3. Graph the masses and distances and click show equation. The slope should be the the spring constant.

**Period** 1. Hang masses starting at 10 g on the hanging mass off the spring. 2. Pull it down and let it go. Time how long it takes the spring to oscillate 10 times. 3. Divide the number of oscillations by the time. 4. Divide one by the number you got from step 3 to get your period. 5. Use all of this information to solve for your spring constant.

__Materials:__ 1. Springs 2. Tape 3. Clamps and rods 4. Masses 5. Balance 6. Timers 7. Meter stick

__Data:__ Spring Force Constant (Directly) Spring Force Constant (Indirectly)

__Calculations:__ Using the data from the first table and Hook's Law we made a graph where the slope was k. In our graph we got 1.5801 as k.

Finding k indirectly Finding k indirectly from graph

Percent difference __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 does. Both methods produced close to the same number for the spring constant which means that it is constant for this range of forces.

2. Why is the time for more than one period measured? We measure the time for more than one period (we do ten) becuase human reaction time is not good enough to accurately stop the stopwatch at the real time of one period. We do ten and then divide the time by ten to get an average for 1 period.

3. Discuss the agreement between the k values derived from the two graphs. Which is more accurate? We found that all our k values were the same until the 100ths place, 1.5 something. In our first method, we found that it was 1.53 and again we did it to find that it was 1.54, and again 1.58. In our second method, each trial was a different number that started with 1.5. We believe that the first method was the most accurate as it could be carried out further than the 100ths place and the second was based on human reaction time.

4. Generate the equations and the corresponding graphs for

1. position with respect to time. y=Acos( ωt) (Trial 1)

2. velocity with respect to time. y=-A ωsin(ωt) (Trial 1) l 3. acceleration with respect to time. y=-A( ω^2)cos(ωt) (Trial 1) 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 a hanging mass and m(s) is the mass of the spring)? Redo graph # 2 using, and explain these results. This answer is closer to our measured spring force constant (1.5801) than the original result (1.522) was. This result is closer since it takes into account how the whole system acts, not just the hanging mass. Since our spring had a light mass in relation to the total system, its effect was minimal.

= = Conclusion: The goal of this lab was to directly and indirectly find the spring force constant of a spring. From prior knowledge of Hooks Law and the euqation for Period we hypothesized that out k values would indeed be constant. Our hypothesis was correct because we found that the spring force constant was always accurate to the tens place, giving us 1.5 and another number that varied. We know that when we found the spring force constant directly we were as close as we could get. Our R value was .9994 meaning that our error was either very tiny or very consistent. When we were looking for k indirectly we found that our high R value meant we had very small error. The results we obtained were very consistent and the most extreme case had only a percent difference of 2.75%. However, there were several sources of error in this lab. The first was the human reaction time while finding the periods. We counted off ten periods instead of only one and had the same person use the same stopwatch each time to minimize this error as much as possible. This reduced the error significantly and made sure that the error that did occur remained consistent. We also found a lot of error in our first trial of indirect. We were only using 10 grams as the mass and the results were far from what they should have been. We did the trial multiple times, and it never improved significantly so we concluded that 10 g was just too little of a mass. We then did other masses and found that they worked. The too little mass was also a source of our error.

=__Mobile Project__= Jae, Ross, Rory, Richie

__**Hypothesis:**__ When we hang our masses on the dowels, they will be at a state of static equilibrium and have no net force nor torque. We believe this due to the fact that we know what Newton's 2nd Law of motion is: That to be at equilibrium the net torque of an object and the net force of an object must be equal to zero.

__**Materials:**__ Wooden rods String Gorilla Glue Ringstand Scale Music-related materials

__**Procedure:**__ 1. Measure all materials' masses that will be in mobile and find the lengths of the dowels. 2. Starting with the last level, hang materials and move their positions until dowel is no longer moving and perfectly straight. 3. Repeat step 2 placing hanging dowels from other dowels until mobile is complete. 4. Place Gorilla Glue over the strings spot that is tied to the dowel in order to secure its hold.

__**Data****:**__ Net Torque Net Force

Because the picture at the top of the page is a little hard to see because the mobile was moving, we made a diagram with all the masses and distances that is much more clear.

__**Calculations:**__ Net Torque overall (Rod 1) Counter Clockwise: 0.279 Nm Clockwise: 0.285 Nm Percent difference: (Rod 4) Counter Clockwise: 0.050 Nm Clockwise: 0.046 Nm Percent difference: (Rod 5) Counter Clockwise: 0.176 Nm Clockwise: 0.189 Nm  Percent difference

(Rod 6) Counter Clockwise: 0.027 Nm Clockwise: 0.027 Nm  Percent Difference

(Rod 7) Counter Clockwise: 0.028 Nm Clockwise: 0.029 Nm  Percent difference

(Rod 2) Counter Clockwise: 0.012 Nm Clockwise: 0.011 Nm  Percent difference

(Rod 3) Counter Clockwise: 0.003 Nm Clockwise: 0.004 Nm Percent difference

Net Force overall (Rod 1) Rod 2 Rod 3 Rod 4 Rod 5 Rod 6 Rod 7 __**Conclusion:**__

Newtons 2nd Law states that to be in Equilibrium the net force of an object must equal zero. Another part of this law is that the net torque of the system must equal zero, which means that your counterclockwise torque and your clockwise torque need to be equal. This is what we tried to achieve in our mobile. Upon finishing our experiment we have concluded that the clockwise torques and counter clockwise torques are close to equal, hence the percent differences we calculated. This causes the mobile to be at a state of almost perfect equilibrium. We also found that the Net torques were almost equal to zero, not one was more than .35 N. Overall, the net force of the top rod was only .028N. Prior to creating the mobile we had written out the torque equations in order to understand where the object should be placed in order to achieve equilibrium. Although these positions should have been perfect they were not which caused our group to have to alter the positions in order for the mobile to be at equilibrium. This showed that there was some error. Our overall torque's for the top rod were 2.23% different from each other. One example of the error comes from the glue that was applied in order to keep the string in place. We attempted to use the least possible amount of glue because we knew that it would add weight to the mobile that we did not calculate. This weight would then trow off the center of mass of the dowel and cause our mobile to be askew. Another source of error was the string that was used to attach the objects to the mobile. We had assumed that they were weightless, but they do in fact have a form of mass which could have also led to our mobile being askew.

= = =__Levers-Static Equilibrium__= Jae, Ross, Rory, Richie 4/15/11 Due: 4/18/11


 * __Purpose:__** What is the relationship between the torques acting on an object at equilibrium.


 * __Hypothesis:__** We believe that the clockwise (cw) and counter-clockwise (ccw) torques will be equal to each other if they are acting on an object at equilibrium. Our reasoning behind this is the second condition of equilibrium which states that Net Torque is equal to 0. If we make ccw positive and cw negative then we have Torque of ccw= Toque of cw.


 * __Materials:__**
 * 1) Meter stick
 * 2) Pivot
 * 3) Knife-edge level clamps
 * 4) 3 mass hangers
 * 5) Set of masses
 * 6) Unknown mass
 * 7) Balance
 * 8) String
 * 9) Masking Tape


 * __Procedure:__**

Trial 1
 * 1) Attach the knife-edge level clamp onto the middle of the meter stick.
 * 2) Place the meter stick onto the balance and adjust the position of the clam until the meter stick is perfectly balanced.
 * 3) Attach two mass hangers on opposite sides of the ruler and adjust the position of them until the meter stick is perfectly balanced.

Trial 2
 * 1) Attach the knife-edge level clamp onto the middle of the meter stick, at the COM (50.15cm).
 * 2) Place the meter stick onto the balance and adjust the position of the clam until the meter stick is perfectly balanced.
 * 3) Attach three mass hangers on the meter stick; one clamp attached on the left and two clamps attached on the right.
 * 4) Adjust the positions of the masses until the meter stick is perfectly balanced.

Trial 3
 * 1) Attach the knife-edge level clamp onto the middle of the meter stick, at the COM (50.15cm).
 * 2) Place the meter stick onto the balance and adjust the position of the clam until the meter stick is perfectly balanced.
 * 3) Once it is balanced move the pivot point 20 cm away from the center of mass.
 * 4) Attach two mass hangers on opposite sides of the meter stick.
 * 5) Adjust the position of the masses until the meter stick is perfectly balanced.

Trial 4
 * 1) Attach the knife-edge level clamp onto the middle of the meter stick, at the COM (50.15cm).
 * 2) Place the meter stick onto the balance and adjust the position of the clam until the meter stick is perfectly balanced.
 * 3) Once it is balanced move the pivot point 20 cm away from the center of mass.
 * 4) Attach one mass hanger on the right side of the meter stick.
 * 5) Adjust the position of the mass until the meter stick is perfectly balanced.

Trial 5
 * 1) Attach the knife-edge level clamp onto the middle of the meter stick, at the COM (50.15cm).
 * 2) Attach one mass hanger with the unknown mass onto one side of the meter stick and record what distance it is away from the fulcrum.
 * 3) Attach another mass hanger with a known mass onto the other side of the meter stick.
 * 4) Move the known mass hanger on the meter stick until the meter stick is balanced
 * 5) Record where the known mass is hanging from the fulcrum and solve for the unknown mass.

Trial 6
 * 1) Measure the angle of the tension on the meter stick
 * 2) Use the second condition of equilibrium, the mass of the ruler, the hanging mass, and the angle to solve for the tension in the string
 * 3) Use the force of the string and the Net forces of the system to solve for the forces of the wall on the meter stick.

__**Data:**__ (Torques divided by 9.8 since g's cancel out) Trial 6

__**Diagrams**__:

__**Calculations:**__ Trial 6 Trials 1-4 \ Trial 1 Trial 2 Trial 3 Trial 4 Trial 5

Percent Differences

Trial 5-percent error

__**Analysis Questions:**__ 1. Does it get easier or harder to rotate a stick as a mass gets father from the pivot point? It does not get easier to rotate a stick as a mass gets farther from the pivot point because then the center of mass is off. When you need to find a new center of mass, then it is harder to rotate the stick.

2. Does the weight of the mass increase as you move the mass away from the pivot point (your index finger)? The weight of the mass does not increase as you move it farther away from the pivot point, although it may seem heavier. What is actually happening is that the torque of the mass is getting larger as it moves farther away from your finger and the pull of the mass to one side of the other is what is increasing.

3. Why is more mass required to balance the meter stick as you move another mass farther from the pivot? More mass is required to balance the stick becaue when you move mass away from the pivot point, then the center of mass is thrown off, and the torque on each side of the stick is no longer equal. You need to balance both torques out on both sides of your pivot point, for the stick to maintain equilibrium or balance.

4. Why must the mass of the hangers and clamps be taken into account in this experiment? The mass of the hangers and the clamps need to be taken into account because they are part of the mass being put on the stick. They hangers and clamps are not part of the meter stick to begin with so they are not calculated into the center of mass on the stick. Meaning, that the masses or the hangers and clamps will add to the torque and off-balance-ness of the stick when they are added.

5. If you are playing seesaw with your younger sibling (who weighs much less than you), what can you do the balance the seesaw? Mention at least two things. You could either sit closer to the middle of the seesaw to balance out your torque with your younger sibling, or you could have your sister bring something on the seesaw to make her weight equal to your own.

6. What kept the meter stick in equilibrium in the fourth trial? In other words, what counterbalanced the known mass? What kept the meter stick in equilibrium was the center of mass of the meter stick pulling (in our case) the clockwise position down, while the known mass was pulling in the counter clockwise position. The product of the distance from the fulcrum and mass of the known object were equal to the product of the center of mass and the distance away from the fulcrum (20 cm).

__**Conclusion:**__

As our results show, we were very accurate with our hypothesis. We proved that when an object was at equilibrium by being perfectly balanced, the torques would be equal to each other, thereby negating each other. All of our observations to the reaching of equilibrium were supported by the calculated torques.

There were multiple sources of error in this lab. For one, we determined the levels of equilibrium and the center mass by eyeing it, which was not the best method. Also, we assumed that the center of mass for the weights was in the geometric center, and we did not take into account the possibility that chips in the materials and the oblong shapes of the clamps would throw off the center of mass.

This lab is important as understanding torque at equilibrium can be applicable outside the lab. For example, the fifth trial of our lab helps explains why balances work, as the torque equation will tell the mass as the lever arm is increased or decreased. Also, the principles used in the lab are also used in some exercise equipment, as individuals have to transfer their weight and position to balance themselves. Nevertheless, understanding torque and equilibrium are an integral aspect of physics.