1. I'm studying newton's Laws at the moment, and currently reading up on period of a pendulum:
T = 2π√(l/g)
If you can't read it. it's Period = 2 * PIE * Sqrt(l/g)
Where l is the length of the pendulum and g is the acceleration due to gravity.
I'm a bit confused when I finished an experiment on this topic. When I took my data and input it into excel as Period versus Length in meters and let excel put in a linear trend line and find the slope. I also did this again by rearranging the equation to give me T^2 .
T^2 = (4π^2 l) / g
Which reads, Period squared = ( 4 * PIE ^2 * l ) / g
Where l is the length of the pendulum and g is the acceleration due to gravity.
Again, I plotted it on excel and it gave me a linear trend line and a slope .
The slope reading is 3.6523 from my data. What I want to figure out is if this rearranged equation giving me that slope agree with the theory (g = 9.81 m/s^2)
How would I actually show this?

I would be very thankful if someone can explain it and answer it as well. Thanks in advance.  2.

3. Excel will draw a trend line making the best fit to a straight line, even if the data is not linear, which in the first case (plotting T versus l) it shouldn't be. So that graph is useless.

In the second case you have found that the slope (T^2/l) is 3.6523.

Now just plug that into your equation and solve for g.  4. Originally Posted by Harold14370
Excel will draw a trend line making the best fit to a straight line, even if the data is not linear, which in the first case (plotting T versus l) it shouldn't be. So that graph is useless.

In the second case you have found that the slope (T^2/l) is 3.6523.

Now just plug that into your equation and solve for g.
Slope represents what? I'm guessing it's representing T^2. To find g

The equation becomes

g = (4π^2 l ) / T^2
Where T^2 is my slope , 3.6523

What should I use for the length? Average or any length on the line of best fit or any length in my table of values?
If i use the length from a random length in my table of values, the answer will fluctuate.  5. No, like Harold said, slope is (T^2/l), not T^2.  6. How do i incorporate T^2/l into the equation then.

I arrived at the equation :

1/g = T^2 / (4pie^2 l)  7. Originally Posted by Lightz
How do i incorporate T^2/l into the equation then.

I arrived at the equation :

1/g = T^2 / (4pie^2 l)
Do you see T^2/l in that equation? Just replace it with the number you found experimentally. Then solve for g.  8. Originally Posted by Harold14370 Originally Posted by Lightz
How do i incorporate T^2/l into the equation then.

I arrived at the equation :

1/g = T^2 / (4pie^2 l)
Do you see T^2/l in that equation? Just replace it with the number you found experimentally. Then solve for g.
1/g = T^2 / (4pie^2 l)

Okay, i'm going to attempt it and can you guys review it for me so I'm accurate.

1/g = T^2 / (4pie^2 l)
(4pie^2) / g = T^2/l <- I got this by multiplying both sides by 4pie^2.
g = 3.6523 / 4pie^2

This gives an answer of 0.0925
Either I did it wrong or somethign terrible happened to the experiment / data table / graph /excel to generate slope.  9. It looks upside-down to me.   10. Originally Posted by Lightz
(4pie^2) / g = T^2/l <- I got this by multiplying both sides by 4pie^2.
g = 3.6523 / 4pie^2
Try again.  11. 1/g = T^2 / (4pie^2 l)   12. Is the l supposed to be in the exponent? It doesn't look like it given the original equation.

And, Lightz, a minor nitpick: You can eat pie, but you can't eat pi.  13. Is the l supposed to be in the exponent? It doesn't look like it given the original equation.
True. Fixed.  14. Originally Posted by Harold14370 Originally Posted by Lightz
(4pie^2) / g = T^2/l <- I got this by multiplying both sides by 4pie^2.
g = 3.6523 / 4pie^2
Try again.
Hm, I didn't do that correctly? hm.  15. Originally Posted by Lightz Originally Posted by Harold14370 Originally Posted by Lightz
(4pie^2) / g = T^2/l <- I got this by multiplying both sides by 4pie^2.
g = 3.6523 / 4pie^2
Try again.
Hm, I didn't do that correctly? hm.
No. You are making a simple, basic algebra error. As himnextdoor said, it's upside down.  16. Originally Posted by Harold14370 Originally Posted by Lightz Originally Posted by Harold14370 Originally Posted by Lightz
(4pie^2) / g = T^2/l <- I got this by multiplying both sides by 4pie^2.
g = 3.6523 / 4pie^2
Try again.
Hm, I didn't do that correctly? hm.
No. You are making a simple, basic algebra error. As himnextdoor said, it's upside down.
(4pie^2) / g = T^2/l <- I got this by multiplying both sides by 4pie^2.
g = 3.6523 / 4pie^2

So my error is in this step?

T^2= (4 * pi^2 * l ) / g
isolate for g :
g = (4 * pi^2 * l ) / T^2
Reciprocal of both sides:
1/ g = T^2 / (4 * pi^2 * l )
I'm stuck on this part..  17. (4pie^2) / g = T^2/l <- I got this by multiplying both sides by 4pie^2.
g = 3.6523 / 4pie^2

You will get   18. Originally Posted by Lightz
(4pie^2) / g = T^2/l <- I got this by multiplying both sides by 4pie^2.
g = 3.6523 / 4pie^2

(4pie^2) / g = T^2/l <- I got this by multiplying both sides by 4pie^2.
g = 3.6523 / 4pie^2

So my error is in this step?

T^2= (4 * pi^2 * l ) / g
isolate for g :
g = (4 * pi^2 * l ) / T^2
Reciprocal of both sides:
1/ g = T^2 / (4 * pi^2 * l )
I'm stuck on this part..
Is that a 'g' in the denominator?  19. Originally Posted by himnextdoor Originally Posted by Lightz
(4pie^2) / g = T^2/l <- I got this by multiplying both sides by 4pie^2.
g = 3.6523 / 4pie^2

(4pie^2) / g = T^2/l <- I got this by multiplying both sides by 4pie^2.
g = 3.6523 / 4pie^2

So my error is in this step?

T^2= (4 * pi^2 * l ) / g
isolate for g :
g = (4 * pi^2 * l ) / T^2
Reciprocal of both sides:
1/ g = T^2 / (4 * pi^2 * l )
I'm stuck on this part..
Is that a 'g' in the denominator?
yah I flipped both sides. so i can achieve T^2/l on one side.
What am i doing wrong. I'm ashamed  20. LOL. You're calling it 'g' when you should be calling it '1/g'.

Two more steps; you can do it.  21. {1\over g} = {3.6523 \over {4 \pi^2}}

I got this.... But people say it's wrong

From this :

1/ g = T^2 / (4 * pi^2 * l )

I then multiplied both sides by 4 * pi^2 * l to get rid of the denominator on the right side leaving l on the bottom. leaving T^2/ l on the right
On the left : 4 * pi^2 * l / g
I then replace T^2/l with my slope. so it becomes :

4 * pi^2 / g = 3.6523

Isolating g, it turns out to be
4 * pi^2 / 3.6523 = g
I think i got it right this time? g = 10.8
That's a significant different. Am i doing it wrong ?  22. Yay!

Maybe the 'l' is inaccurate??  23. I'm guessing these results were gathered from a high-school (or similar) class experiment? In that case, 10.8 and 9.8 are close enough to be within the experimental error. If you rerun the experiment a number of times, or a run it more carefully, or with better equipment, you should get closer to the expected result.  24. Originally Posted by MagiMaster
I'm guessing these results were gathered from a high-school (or similar) class experiment? In that case, 10.8 and 9.8 are close enough to be within the experimental error. If you rerun the experiment a number of times, or a run it more carefully, or with better equipment, you should get closer to the expected result.
This experiment was carried out through a book I was reading on, if you didn't read my background information. So i carefully followed it and wanted to study this aspect of newton's law. 1m/s^2 more than the original is a huge difference. Thanks everyone for helping me.  25. 1m/s^2 more than the original is a huge difference.
This is true, but these types of experiments require a precision that is difficult to achieve sometimes. That is why it (along with others) has to be repeated many times to keep narrowing down on the true value.  26. Originally Posted by KALSTER
1m/s^2 more than the original is a huge difference.
This is true, but these types of experiments require a precision that is difficult to achieve sometimes. That is why it (along with others) has to be repeated many times to keep narrowing down on the true value.
Thanks for explaining. You guys helped me dig even deeper.  27. Originally Posted by Lightz Originally Posted by MagiMaster
I'm guessing these results were gathered from a high-school (or similar) class experiment? In that case, 10.8 and 9.8 are close enough to be within the experimental error. If you rerun the experiment a number of times, or a run it more carefully, or with better equipment, you should get closer to the expected result.
This experiment was carried out through a book I was reading on, if you didn't read my background information. So i carefully followed it and wanted to study this aspect of newton's law. 1m/s^2 more than the original is a huge difference. Thanks everyone for helping me.
Sorry, I missed that. Anyway, an experiment from a textbook would probably be in the same category. Try it a few more times and see what the average looks like. 10.8 is about 110% of 9.8. 10% isn't that huge of an error in this type of experiment, especially with just one run.  28. The mass of the cord will intoduce an error.  