Lab 3: Non-Constant Acceleration Problem 3/4/15

Physics Lab #3: Started on 3/4/15
Non-Constant acceleration problem/Activity problem
Annemarie Branks
Professor Wolf

Objective: Find how far the elephant travels before coming to rest, analytically and numerically. Be able to explain how you can find the answer numerically.

Procedure:
     1.  After dividing the net force by the changing mass equation + the mass of the elephant to find acceleration, we would want to used calculus to to find the position equation. This method would result in long, extensive calculus that is prone to human error. We will determine how far the elephant travels before coming to rest by setting up an Excel spreadsheet to assist us in determining the position.
     2.  In the first column we list time in whatever increments we choose. The smaller the increments, the more accurate our answer will be, but the increments only need to be so small before the changes begin to be insignificant. The next column we enter in our acceleration equation which ended up being -400/(325-t) m/s^2.
     3.  Make a column showing the average acceleration of each time interval. We need to determine the average acceleration because this will allow us to determine the change velocity. Average acceleration multiplied by the chosen Δt will give the change in velocity between two specific times.This can also be thought of as looking at the acceleration curve, drawing the curve for each time increment as a straight line, and finding the area under the curve. Δv = Δt(a1+a2)/2 
     4.  Now that we have found the change in velocity, we can find the specific velocity for each point in time. Since the elephant is starting out with 25 m/s the following time can be determined by adding the change in velocity between 0 seconds and the next time with which you chose to put your interval at. Use the Excel functions to continuously add the Δv to the previously determined velocity until the velocity reaches zero.
     5.  Using the same method we did to determine velocity, we are going to find position. Make a column listing the average velocities for each time interval and in the next column list Δx by taking the average velocities and multiplying them by Δt. This is also like looking at the velocity curve and calculating the area under the curve to find position. Δx = Δt(v1+v2)/2
     6. Similar to how we found velocity, we will find position by adding the change in position to the the position at the beginning of the interval to find the position at the end of the interval. Use Excel functions to repeat this process for each time interval until you reach the point where velocity became zero. At the time is where you will find the elephant's position when it stopped moving.

Results:
We decided that a Δt of 0.01 seconds was small enough. The velocity reached zero at t=19.69 seconds where its position was 248.698 meters. This value matched with the answer we calculated using integration which was 248.7 m. Therefore, we can conclude that this method is a great alternative to integrating difficult problems. This method also showed us that the smaller Δt is, the more accurate our answers are. Calculus would assume that Δt is infinitely small which is the most accurate but the way we did it on Excel is as good as we would want it to be.

This is our spread sheet where the cell for the velocity is closest to zero. All the way to the right in column H shows the position of the elephant at this time.

Lab 2: Free Fall 2/25/15

PHYSICS LAB #2: Started on 2/25/15
Free Fall
Annemarie Branks
Professor Wolf

Objective: Prove that gravity will act up on a free falling object, with no external forces, at a downward acceleration of 9.81m/s^2.

Apparatus: An electromagnet is holding up a large piece of metal with has a metal ring around it that will come in contact with a wire. Once the magnet its turned off and the piece is allowed to drop freely, a spark will produce every 1/60th of a second and make a mark on the strip of paper parallel to it.

Description of Procedure and Results:
     1. After the apparatus has made its marks on the strip of paper, tape down the strip on a flat surface to accurately measure the distance from your zero mark to each following mark. Record these distances. Note: The measurements should start from the end of the strip with marks that are closest together.

     2. In attempt to find equations that represent the falling object's velocity and position (so that we can later derive a numerical value for acceleration) we must determine its distance with its corresponding time. We will also determine the mid-interval speed with its mid-interval time and produce a second equation. We are going to assume that the apparatus worked flawlessly and made a mark for every 1/60th of a second. From this information we can create a create a curved line and generate an equation.

     3. The time that has passed for each mark will increase from the previous time by 1/60th of a second. In Excel we have generated a curved line and an equation to represent the relationship between distance and time. These were our results:

 
 4.  To calculate the increasing the Mid-Interval Speed we need to find its corresponding Mid-Interval Time which is simply 1/120th of a second in addition to the smaller time boundary of the interval.We determine the change in position by subtracting a distance from the following distance measured. Now Mid-Interval Speed can be determined by taking the change in position and dividing it by 1/60th of a second. With Mid-Interval Time versus Mid-Interval Speed, we can create a linear graph and equation to represent the velocity of the free falling object. These were our results:

     

5.  To see how accurate this method of determining the acceleration of gravity is were going to derive the two equations.

Distance vs. Time

y = 479.44x^2+102.64x+0.2246

y' = 958.88x+102.64

y'' = 958.88 cm/s^2

The second derivative tells us the acceleration of the equation which is 958.88cm/s^2 or 9.59m/s^2.

Mid-Interval Speed vs. Time

y = 941.47x+105.92

y' = 941.47 cm/s^2

Since this is a Speed versus Time graph we only need the first derivative to determine the acceleration which is 941.47cm/s^2 or 9.41m/s^2.

Questions/Analysis:

     1.  Show that, for constant acceleration, the velocity in the middle of a time interval is the same as the average velocity for that time interval.

Since the velocity graph is supposed to be a straight line, the velocity in the middle of a time interval will be the average velocity for the interval. For example, if we chose the interval of t=0 to t=0.2 the average velocity should be the the velocity at t=0.1.

v(0) = 941.47(0)+105.92 = 105.92 cm/s

v(0.2) = 941.47(0.2)+105.92 = 294.214 cm/s

Average velocity for the time interval [0, 0.2]: (294.214+105.92)/2 = 200.067 cm/s

Is this velocity the same as the velocity in between times 0 and 0.2 seconds?

v(0.1) = 941.47(0.1)+105.92 = 200.067 cm/s

Therefore, the velocity in the middle of a time interval is the same as the average velocity for that time interval.

     2.  Describe how you can get the acceleration due to gravity from your velocity/time graph. Compare your results with the accepted value.

The derivative of velocity is equal to the acceleration.

y = 941.47x+105.92

y' = 941.47 cm/s^2

The accepted value for gravity is 9.81 m/s^2. In comparison, 9.41m/s^2 is not very accurate.

     3.  Describe how you can get the acceleration due to gravity from your position/time graph. Compare your results with the accepted value.

The second derivative of position is equal to the acceleration.

y = 479.44x^2+102.64x+0.2246

y' = 958.88x+102.64

y'' = 958.88 cm/s^2

Though the position equation produced a much more accurate acceleration due to gravity that the velocity equation, 9.59 m/s^ is still not very accurate when compared to the accepted value 9.81 m/s^2.

Assumptions made:

     1. When the apparatus dropped the metal bar we assumed that it made no friction with the wire or the tape. Since the apparatus wasn't completely silent, there was friction which is why the Mid-interval speed vs. Time graph was not a perfect line. The friction would also explain why our determined acceleration of gravity was so much lower than the accepted value.

     2.  We also assumed that the apparatus worked perfectly in making marks exactly 1/60th of a second apart. If the assumption we made was wrong, this would also explain why the Mid-interval speed vs. Time graph was not a perfect line.

Uncertainty:

When measuring the marks on the tape we used a meter stick with centimeter and millimeter marks. We would then be uncertain of the hundredth place in our measurements when measuring in centimeters.

measurement (cm) = best estimate ± 0.01cm

Differences between expected and experimental values:

     Absolute Difference = Experimental Value - Accepted Value

9.59m/s^2 - 9.81m/s^2 = -0.22

     Relative Difference = (Experimental Value - Accepted Value)/Accepted Value * 100%

(9.59m/s^2 - 9.81m/s^2)/9.81m/s^2 * 100% = -2.24%

Errors and Uncertainty:

Using the all the lab groups results for acceleration we can determine the standard deviation from the mean. BY doing this we can statistically say how confident we are a range of values contains the actual value for acceleration due to gravity.

The equation for standard deviation is: (∑(x-)^2 /N)^0.5

We can create an empirical curve showing our confidence in our answer. For example "We are 68% confident that the actual acceleration of gravity is within one standard deviation of the mean. These values range from 935.91cm/s^2 to 976.16cm/s^2."

     1.  What pattern (if any) is there in the values of our values of g?

In the velocity graph, the points seem to be making a lesser than the average slope but then after three points it jumps up. I'd describe the pattern as slanted stairs.

     2.  How does our average value compare to the accepted value of g?

The average value differs from the accepted value by -24.96667cm/s^2

     3.  What pattern (if any) is there in the class' values of g?

There doesn't seem to be any pattern.

     4. What might account with an difference between the average value of your measurements and those of the class? Which of these systematic errors? Which are random errors?

     My partner and I made a systematic error when we were graphing our points. We repeated the last point twice which reduced our acceleration to 926.2cm/s^2 rather than 941.47cm/s^2. I hadn't realized our mistake until after we had shared our value with the class. I kept the value at 926.2cm/s^2 just so our standard deviation was the same as the rest of the class's.

     In each trial, the apparatus dropped the metal bar differently which would explain why everyone had such different results.This would have changed the distribution of x, therefore this is a random error.

     5. Write a paragraph summarizing the point of this part of the lab. What were the key ideas? What were you supposed to get out of it?

The purpose of collecting the class' data was meant to show that we could calculate an accurate acceleration of gravity; however, due to assumptions about the apparatus we ended up just proving that this was not the best method for finding the acceleration of gravity. The assumptions we made lead to most of the class's accelerations to be smaller than the accepted value.. By finding the mean of the class's accelerations and the standard deviation of the mean we can statistically say how confident we are in our range of values. Being able to put a statistic to our findings is what scientists should do when conducting their experiments and sharing them with the world. Without a statistic of certainty, you could being giving out false information.

Lab 1: Mass vs Period for an Inertial Balance 2/23/15

PHYSICS LAB #1: Started on 2/23/15
Mass vs. Period
Annemarie Branks
Professor Wolf

Objective: This lab is intended to show that, for an oscillating system, the period is directly proportional to the mass. This means that as the mass increases so will the period due to the object's inertia. We will write an equation representing a mass and a period's directly proportional relationship. We can then use this equation to find unknown masses.

Set up:

1. As shown in the photo below, use a C-clamp to secure the inertial balance to the table. This should allow the inertial balance to oscillate horizontally.
   
2. Set up the photogate so the inertial balance oscillates through it. A long, thin piece of tape should be placed on the end of the inertial balance since the balance may be too large. The photogate emits light and will able to detect when that light is obstructed. Once the photogate is hooked up to a computer, it will tell us how often that light is being obstructed by the tape when we make the inertial balance oscillate.
   

Procedure:

1. With no extra mass on the tray, pull back the inertial balance. Take notice of how much force you are using to pull back the balance, for you must use the same force for each trial in order to obtain accurate results. Record the period of the balance.
2. For each additional trial, add an extra 100 grams to the balance and record the period. In this experiment we repeated the trail until 800 grams. Multiple trials will allow us to create a more accurate curve. Always use the same amount of force when pulling back the inertial balance.

The image above shows the 100 gram discs taped to the inertial balance. The tape that is keeping the discs together may affect your results but not enough to be significant.

     3.  Input your data to create a Mass vs. Period graph. Notice that the zero mass, like in the graph below, has a period. Theoretically, if an oscillating system has no mass then there is no period, but obviously after letting go of the balance, that your system does oscillate and the photogate is picking up a period. This is due to the inertial balance having a mass of its own which will need to be accounted for when writing an equation.

     4.  Since we do not know the mass of the inertial balance's tray, we are going to use the User Parameter function on LoggerPro to approximate what it may be. The goal is to find masses that have a correlation as close to 1.000 as possible so we can achieve a straight line. Multiple values for the mass of the tray may have that correlation so we are going to record the largest and the smallest mass with a correlation that is the closest to 1.000. In our experiment, we ranged the mass of the tray to be from 0.280 kg to 0.360 kg with a correlation of 0.9996.
     5.  A typical linear equation looks as such: y = mx+b

     We know our y-values will be our period (T), our slope will be some constant (A), and the x-values with be the additional masses including the mass of the tray "(m+Mtray) that need to be set to a power (n) as indicated by the curvature of the line. The "b" will become zero since like we have started before, a zero mass will have zero period.
     Now we are able to set up an equation that is related to our data: T = A(m+Mtray)^n

     6. To figure out what the values of A and n are we must log both sides of the equation above. This is our new equation: lnT = nln(m+Mtray) + lnA
     7.  Change the graph from Period vs Mass to ln T vs ln (m+Mtray) so that we may find the values that correspond with our new equation. The n will be our new slope, ln(m+Mtray) will be our new x, and lnA will be our new y-intercept.

     8. Adjust the parameter Mtray to achieve a straight line. The goal is to find a range of values that give a correlation coefficient of 0.9999. Record the maximum and minimum values of Mtray that produce a correlation as close to 1.000 as possible. For my group we found a minimum mass of 0.280 kg and a maximum mass of 0.360 kg.

     9. Record the slope and y-intersection for the maximum and minimum Mtray values. For our experiment, specifically, we found that a minimum mass of 0.280 kg resulted in a slope of 0.6465 and a y-intercept of -0.4336 seconds. The maximum mass of 0.360 kg resulted in a slope of 0.7424 and a y-intercept of -0.4852 seconds. The range from 0.280 kg and 0.360 kg had a correlation of 0.9996.

  10. With our new information we can complete the equation that will describe the relationship between mass and period; however, we will have two equations since we have a maximum and minimum mass for the tray of the balance.By plugging in we first has the equations

• minimum: lnT = 0.6465ln(m+Mtray)-0.4336
• maximum: lnT = 0.7424ln(m+Mtray)-0.4852

One we wrote those equations we were able to e both sides and write the equations in terms of T. They appear as follows:

• minimum: T = 0.648(m+0.280)^0.6465
• maximum: T = 0.616(m+0.360)^0.7424

     11. To see how well the new equations describe the mass and period relationship we are going to do tests of unknown masses.We will then compare their theoretical masses with their actual masses. We repeated the experiment using a white box and a water bottle. The white box had a period of 0.3919 seconds and the water bottle had a period of 0.6652 seconds. With our new equations we found that the white box has an approximate range for its mass which is 0.179 kg - 0.184 kg. The water bottle had a range of 0.761 kg - 0.749 kg.

     12.  Weigh the objects on a scale for find their actual masses and determine how close they are to their theoretical ranges. Determine if your equation accurately determines the mass of objects.

Specifically, for our experiment, the white box is 0.172 kg which is close to the minimum value but not within range. The water bottle is 0.601 which seems to be significantly under the minimum value of its predicted range.

Summary:

     Finding the periods of eight known masses and the unknown mass of the tray allowed us to plot points on a graph and create two equations that account for the lowest and highest possible values for the mass of the tray. These equations are meant to describe the relationship between period and mass. To determine the accuracy of our equations we ran two more experiments of unknown masses. We then compared the objects' theoretical masses and actual masses. It was found that though this method gave close results, the results are not very accurate, yet the experiments did visually prove that there is a relationship between mass and period.

Error: 

• Each trial required a person to pull back the inertial balance with a certain amount of force every time. If someone had not pulled the inertial balance enough, then the period would have been too small to fit in a trend amongst the other periods. I am sure my lab partner did his best to pull back the inertial balance with the same force for every trial, but it is most likely he did not use the same exact force every time. This could have affected our results.
• Collecting data for the period of the water bottle, we laid the water bottle down length wise. This allowed more room for the water to be moving back and forth, but the water would not have been moving in sync with the inertial balance. The force of the water in the bottle would have affected the period thus affecting our predicted value for its mass.