# Lap report, paraphrasing and inclusion help

**1. **Did the period of the pendulum swing depend on the mass of the bob? On the length of the string? On the displacement? How does your data relate to the pendulum equation? You need not include your data tables and graphs, but you should explain your answer based on the information in your data tables and graphs.

Answer: Observing the data from the lab and the tables, I can say that as the mass of the bob increased, so did the period of the pendulums swing. The graph for part 1 of this lab shows that as the mass of the bob increased from 50g to 400g, the average of the period also increased very gradually with each weight that was added despite the angle of the angle of the string being the same and also the length of the string being the same. However the average of the period that increased with each mass was very little and can be negotiated as the mass of the bob did not really affect the swing of the pendulum much and any data that shows that the mass does effect the period of pendulum can be explained through either human or experimental errors.

In the data from the 3^{rd} part of the lab and graph we can see that the length of the string also caused a difference in the average period of the pendulum swing. In the 3^{rd} part I test to see if the same mass and same angle but changing the length of the string would make a difference in the period of the pendulum swing. The results confirm that the length of the string also affects the period of pendulum. The length of string differed fro 0.10m to 0.40m. In each case where a length was added to the string, it tool longer for each path of periodic motion, which further resulted in longer length of string more period of pendulum. The longer the string will result in longer period of the swing as the length the period has to travel increases.

In the second part of the lab we tested if the displacement affects the period of the pendulum swing. I tested the different angular displacements along with the period of the pendulum. Looking at the table I can say that the displacement does affect the period of the pendulum. The angular displacements tested were 45°-15° and what can be observed is that as the angular displacement increased the period of the pendulum also increased.

And so the greater the displacement, the greater the period of pendulum, as the higher the arc at which the bob was released, the more time it will take for the bob to travel the distance it has to travel .

(6 points)

Score |

**2. **Calculate the slope of the line in your graph of the square of the period of the pendulum vs. length of the string [slope = (*y*_{2 }– *y*_{1})/(*x*_{2 }– *x*_{1})]. Galileo figured out the equation that describes the behavior of a pendulum. If you square both sides of the equation, you will find that the slope of the line is related to the acceleration due to gravity (*g*). Specifically, slope = 4p^{2}/*g*. Use your data to calculate *g*. How does it compare with the accepted value of 9.807 m/s^{2}?

Calculate the percent error and show your work.

Answer: Firstly let the 2 points on the graph be (0.1,0.9) and (0.4,1.7)

Slope = (*y*_{2 }– *y*_{1})/(*x*_{2 }– *x*_{1})]

Therefor (1.7-0.9) / (0.4-0.1) = 2.67 is the slope of the line of my graph

To be able to get an answer we need to alter the formula slope = 4p^{2}/g to g = 4π² / slope to be able to substitute it with values I already have.

g=4π² / slope g=4π² / 2.67 = 14.79 m/s^{2}

Percent error = (experimental value – theoretical value / theoretical value)*100

(14.79-9.807/9.807)*100 =50.81% error

The value I received for gravity after calculation is higher than the acceptable value of gravity, which is 9.807 m/s^{2, which} resulted in a percentage error of 50.81%.

(6 points)

Score |

**3. **Did the period of the spring swing depend on the length of the displacement? On the mass? Explain your answer using your data. How do your findings relate to the spring equation?

Answer:

T=2p √m/k

F=1/T

Observing the data collecting during the lab and looking at the graph, I can say that as the length of the displacement was increased from 0.05m-0.20m, the average period of the spring also increased but the increase was not significant What we need to remember is that a 200 g mass was hung on the other end of the spring. Also looking at my data and lab, I see that the affects of the length of displacement were rather gradual on the average period. The change, which was seen, can be negotiated as the difference and increase was very minor.

In the last part of the lab, difference masses were attached to the spring to observe how the mass would affect the period of the spring swing. According to my observations, mass plays a role in the period of the spring, as the mass increased from 0.10kg-0.40kg, the period average of the spring also increased.

My observation and testing confirms that the period of the spring depends on the mass on the spring and also that the period does not necessarily depend on the amplitude of motion, however human errors or experimental errors could create a false image of showing a correlation between the amplitude and the period of spring.

When comparing to the spring formula, we can see that amplitude does not have any significant affects on the period of spring, rather the 2 factors that the period of spring is dependent upon is the K which is the spring contact and m which is the mass on the spring. The frequency is also important as when conducting the lab we calculated the time for 10 swings.

(6 points)

Score |

**4. **Calculate the slope of the line in your graph of the square of the period of the spring vs. mass

[slope = (*y*_{2} – *y*_{1})/(*x*_{2} – *x*_{1})]. Robert Hooke figured out the equation that describes the periodic motion of the spring. If you square both sides of the equation, you will find that the slope of the line is related to the spring constant (*k*). Specifically, slope = 4p^{2}/*k*. Use your data to calculate the *k* of your spring.

Answer: Let the 2 points on the graph be (0.10,0.09) and (0.40,0.24)

Slope = (*y*_{2} – *y*_{1})/(*x*_{2} – *x*_{1})].

Therefor (0.25-0.09)/(0.40-0.10) = 0.53

To be able to get an answer we need to alter the formula slope = 4p^{2}/k to k = 4π² / slope to be able to substitute it with values I already have.

k=4π² / slope slope =0.53

The k of the spring is equal to 74.49 N/m