# Acceleration due to gravity lab

1. Dec 2, 2011

### hotshot1kille

1. The problem statement, all variables and given/known data

We did a lab using a pendulum attached to a string about 1.12m long. We let it go from 20 degrees and counted 10 cycles and recorded how much time it took for the pendulum to complete 10 cycles. This lab is done assuming, there is no air resistance or any other force other than just gravity. We did 3 test cases.

Test case 1- time it took for the pendulum to complete 10 full cycles is 23.41 s.

Test case 2- time it took for the pendulum to complete 10 full cycles is 23.41 s.

Test case 3- time it took for the pendulum to complete 10 full cycles is 23.44 s.

Find the average time for 10 cycles.

Using this average time, calculate the period of the pendulum.

Using the period of the pendulum and its length, calculate the value of "g".

I don't know how to do average time, and if i get the average time wrong then i get number 2 wrong and then my answer for number 3 will be wrong. So i did make an attempt, can someone check if i am doing it right and if my answer seem fine for all questions. Please and Thank you.

2. Relevant equations

T= 2∏ √ L/g

Period: Δt/Nc

Average time for 10 cycles : Δt/ Nc

3. The attempt at a solution

1. Find the average time: Test case 1: 23.41s/10 =2.341 seconds per cycle.

Test case 2- 23.41s/10 =2.341 seconds per cycle.

Test case 3- 23.44/10 = 2.344 seconds per cycle.

2. Period= Δt/ Nc

Test case 1 and 2: 2.341s / 10 = 0.2341

Test case 3: 2.344/10= 0.2344

3. Find g.

4∏^2 (L)
---------- = g
T^2

4∏^2 (1.12m)
--------------- = g
0.2341^2

g= 806.81 m/s^2

2. Dec 2, 2011

### Delphi51

The average of 2.341, 2.341 and 2.344 is their sum divided by three. That is, about ten times the .2341 you used in the calc. That will give you a decent value for g, but not a great one. In my experience, it takes considerable practise to get good times and far better to count 100 swings rather than 10. Also, mistakes can be made in measuring the length - not just the length of the string, but the distance from the pivot point to the center of mass of the bob.

3. Dec 2, 2011

### hotshot1kille

So average time is : (23.41+23.41+23.44) / 3 = 23.42

period= (23.42)/10 = 2.342

so then gravity would equal to

4∏^2 (1.12m)
--------------- = g
2.342^2

g= 8.06 m/s^2

does this seem fine?

4. Dec 2, 2011

### Delphi51

Yes, the calc looks good! How about the measurement of L; did you include the distance from the end of the string to the center of the bob? If not, take an estimate and add it.

5. Dec 2, 2011

### hotshot1kille

Yea what we did in class was that they were already made. We just got up and took one, the lab says it has to be roughly within 1.2- 1.5m. I told my teacher i have about 1.1m, and he said it is fine, because like the 0.1m won't make a big difference for such a small calculation. He also mentioned that imagining that there is only gravity acting upon it do not worry about resistance, shape of the ball and how round it is.

6. Dec 2, 2011