Please help, finding the acceleration of g from an excel graph

[awilliams315]

Homework Statement

I have to calculate g or find the acceleration of g using the slope of the trend line from the y vs. t^2 graph.

I had to use excel and did this for lab with data and use the equation excel came up with after i finished the graph, and I have no idea how to find the acceleration.

ALSO, I had to use each set to find g for each one, so the data for the falling object is totally separate from the pendulum data, I just am required to find g for each one.

Homework Equations

the equations I have to use i guess from excel look like this for timing the fall of a dropped object to determine acceleration of g.

y=0.676x + 0.082
R^2 = 0.966

then the second set of equations for finding the acceleration of g using a simple pendulum were these

y= 0.976x + 0.044
R^2 = 0.992

The Attempt at a Solution

I have tried everything, and it is supposed to be within a 25% discrepency of the book value of 9.8 m/s^2 for the acceleration of gravity. Everything I have tried has come not even close to that value.

in the lab we were given these equations under the timing the fall of a dropped object.

s= 1/2at^2
y=mx
slope= 1/2a

for this one I tried (1/2)(0.676)= .338 not a bit close to the 9.8

and these equations for the simple pendulum.

T= 2(pie) square root (L/g)
T^2 = 4(pie^2) (L/g) rearranged to L= (g/ 4(pie^2)) * T^2
slope = g / (4 (pie^2))

for this I tried slope being 0.976 (4*3.14^2) = g
0.976(39.4384)=g
38.49=g - which is was off.

I just can't figure out what formula I am supposed to use and what parts of the y= equations i am supposed to use to find this

 

[gneill]

For the falling object part of the lab, what was the experimental setup? Was the object simply allowed to drop vertically through the air, and timings made for different amounts of vertical drop? In what units were things measured (time, distance). Were there any calibration or conversions to apply to the instrument readings?

 

[Delphi51]

Welcome to PF.
It looks like you did an experiment, dropping an object and measuring its vertical position y at various times t. And graphed y vs t.
The theory for this is y = ½gt², so you would expect the graph to be curved in the shape of a parabola. The usual procedure is to work with the theory and the data together, deducing that if you graphed y vs t² you should get a straight line graph. And that its slope would be ½g.
You can see that if you replace t² with x in y = ½gt² and compare it with the equation of a straight line, y = mx + b.
Do graph y vs t² and let us know what it looks like and (if it is close to a straight line) what you get for the slope.

 

[awilliams315]

For the falling part of the lab, we were to tape a meter stick on the wall and label the following heights, 0.6, 0.9, 1.2, 1.5, and 1.8 m. We were then to do five trials at each height by placing a nickel against the stick and timing the fall from each height onto the floor with a stopwatch. The following is the data table

5 timed trials in seconds
height (m)... 1... 2...3... 4... 5...ave...time^2...max time sq - ave time squared...ave time squared - min time squared


0... 0.00...0.00...0.00... 0.00... 0.00... 0.00...0.00.... 0.00......... 0.00
0.6... 0.75...0.72...0.75... 0.72... 0.78... 0.74...0.55.... 0.05......... 0.04
0.9... 0.78...0.97...0.87... 0.91... 0.91... 0.89...0.79.... 0.15......... 0.18
1.2... 0.90...0.99...1.09... 0.91... 0.91... 0.96...0.92.... 0.27......... 0.11
1.5... 1.00...1.06...0.97... 0.88... 1.06... 0.99...0.99.... 0.14......... 0.21
1.8... 1.18...1.19...1.09... 1.10... 1.15... 1.14...1.30.... 0.11......... 0.12

 

[awilliams315]

Delphi,

We were required to have a linear treadline which I have and also is supposed to be displacement vs. time squared, then adding custom error bars horizontally according to the max, min avg time squared on the data table i just provided.

I don't know if I am overlooking something very simple, but I have never in my life done physics, and am taking this class online, which seems to make it a lot harder. thanks

the slope for the falling object was

y=0.676x + 0.082
R^2 = 0.966

 

[Delphi51]

Did you graph displacement vs. time squared ?
On this graph got a slope of 0.676?
Unreasonable answer!

I can't follow the table you entered. Maybe just tell very clearly what you got for two heights and how you put it on the graph. Say you dropped from height 0.6 m and got an average time of 0.35 s. Then dropped from 0.9 m and got time of 0.43 s. Then you would get

The slope is 5.1 - pretty close to half of g.

 

[Delphi51]

Oops, bungled the graph. . . . wait a minute. Should be okay now.

 

[awilliams315]

the teacher gave us step by step instructions for excel, the 1-5 trials were to be added up and averaged as shown under ave. Then the time squared was the average squared. I used height (m) such as 0.6, 0.9, etc as y axis, and 0.55, 0.79, etc for the time squareed x axis.

 

[gneill]

Did the instructions say anything about correcting for human reaction time? Your reflexes in manning the stopwatch could add a consistent bias to the time readings. In fact, given your supplied data, if you subtract a bias of 0.45 seconds from the average times, your plotted data will shift nicely to overlap the theoretical plots for both the t and t² plots.

 

[Delphi51]

What do we do if we suspect a reaction time error?
Fudging results to match theory is . . . very crude and disappointing. Also a very bad habit. No special equipment needed; you could easily repeat the experiment at home.

Practising will improve measuring skill. Check by measuring a pendulum swing (measure 100 swings to get a very accurate time for one swing) and keep at it until you can measure within a tenth of a second.