Period of oscillation for a pendulum

[Johnahh]

Homework Statement

The aim of the experiment is to find out if the period of oscillation is proportional to the square root of the length of the pendulum.
We used 5 different lengths and got the average time for 1 oscillation for each length
These are the results:
Length=48cm, Time=1.33s
Length=43.5cm, Time=1.26s
Length=35cm, Time=1.15s
Length=25cm, Time=0.95
Length=15cm, Time=0.73
With this data I have drawn a graph with a straight line as we can tell period is proportional to length
Question:
Calculate the percentage of error and identify main sources of error.

Homework Equations

T= 2∏√L/g
T= period
L= length
g= acceleration due to gravity

I am told 2∏/gradient should give me a result near to 9.81ms but my results are coming out with a massive error percentage.
my graph is drawn with x=√L and y=T

The Attempt at a Solution

the gradient for my graph is 1.25s-0.25s / 6.5-1.6. which is 1 / 4.9 = 0.20
2∏ / 0.20 = 31.41
9.81-31.41 / 9.81 x 100 = 220% error?
surely this cannot be correct as the error percentage should be at maximum around 3%.

I must be doing a calculation wrong but i cannot figure out what, any help would be appreciated.

Thanks

 

[BruceW]

I am told 2∏/gradient should give me a result near to 9.81ms

This isn't true. The units are not right, and the value is not right either. You know the equation T= 2∏√L/g (remember that g is inside the square root). And you plotted T and √L, so you can work out what the gradient should be.

 

[Johnahh]

Thanks for the reply Bruce, I hope i have figured it out:
g=4∏^2L / T^2

so g = 4 x 9.86 x 0.48m / 1.76 = 10.75 m/s/s

9.81 - 10.75 / 9.81 = 9.5% error.

Is this correct Bruce?

 

[Delphi51]

Using g=4∏^2L / T^2 you are calculating g from just the first point. The idea is to use some measure of all the points to get an accurate value of g out of the experiment.
You should be calculating g from the slope on your graph.
Comparing your T= 2∏√L/g to good old y = mx + b, you see that y corresponds to T, x corresponds to √L and the slope m corresponds to 2∏/√g
slope = 2∏/√g converts to g = (2∏/slope)².
If you use your slope of 0.2, it works out badly.
I don't agree with a slope of 0.2 for that data. As a rough check, do slope = ΔT/Δ√L for the Δ's of the first and last data points: (√L,T) = (.69, 1.33) and (.39,.73).
I get slope = (1.33-.73)/(.69-.39) = .6/.3 = 2.
But don't use that value; go back to your graph and get the slope of the line of best fit through all the points on the graph.
My compliments on your data; I see it makes a very nearly perfectly straight line! You must have measured very carefully.

 

[Johnahh]

Hi Delphi thanks for your reply, they are very close as I used the same length 3 times and got the average, but thank you :)
ok I see that a value of 2 works as it would be ∏^2 which is 8.96
that makes a error of 0.5% which is very nice.
I can see my error before not making my lengths into meters, but there is still a problem I do not know how to measure the slope for each point?

Also i have never seen y = mx + b?

Thanks

 

[Delphi51]

You don't measure the slope for each point.
You measure the rise and run for the straight line of the graph. Slope is rise/run.

y = mx + b is a grade 9 math concept. It is the standard equation for a straight line on a y vs x graph. b is the y-intercept, m is the slope. Once you know b and m, then given any value of x you can calculate the y value of the (x,y) point on the graph. So y = mx+b holds the same information as the straight line does. In physics we use formulas a lot and each one represents a straight line on some graph. Most of them were originally found by graphing. In your experiment, you could have deduced that formula for the period of a pendulum from your experimental data through knowing that y = mx + b is the formula for the straight line. The physics formula is nothing more or less than the equation for the line that best fits the experimental data.

 

[Johnahh]

Thanks a lot for clearing that up.
So as my line of best fit runs through the first and last point i can indeed use the gradient of 2.
meaning that my experiment had a overall error of 0.5%.

 

[Delphi51]

Yes, you are quite correct.
But I fear your misunderstanding of how to use the slope arose from someone showing you that shortcut! Better to do it with the rise and run. After all, the line of best fit never EXACTLY hits a data point.

BTW, there is something in the question that disturbs me a little.
The purpose of the experiment was to measure g at your location (it does vary; the local college here requires students to use the value 9.83 which is what you measure there - so you probably don't know what exactly to compare your value with). Yet you are asked to find the % difference between your measurement and the accepted value. To me, the "% error" is an estimate of how accurately your experiment was done. I would ask my students to estimate how accurately they measured the times - they would repeat one a few times and see how much the times vary. And the length measurement is probably accurate to within half of one of the divisions on your ruler. These estimates would give you error bars both horizontally and vertically on the graph. A very important question at this point is whether or not the line of best fit actually fits to within the error bars. Then, how much could the slope of the line vary and still fit? That variation in the range of possible slopes would determine how accurately the final calculation of the experimental value of g is. After doing all that work, you can say whether or not the data fits a formula and if so something like "local g = 9.82±.03". Sorry if I'm distracting you from your assignment; just trying to contribute to your education.