The Slope in Pendulum Equation

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nasi112

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Hi all. I hope that someone can help me.

It is asked to find the slope of the graph [MATH]T^2 \ vs \ L[/MATH]
First, in the table it is written [MATH]T_{10}(s)[/MATH]. I don't understand what it means by the subscript [MATH]10[/MATH]
If I let [MATH]y = T^2[/MATH] and [MATH]x = L[/MATH], I can say the slope is [MATH]\frac{4{\pi}^2}{g_{exp}}[/MATH], where [MATH]g_{exp}[/MATH] is the experimental gravitational acceleration

If I follow the same logic of linear equation, the slope will be

[MATH]m = \frac{{T_2}^2 - {T_1}^2}{L_2 - L_1} = \frac{{13.66}^2 - {11.72}^2}{0.4 - 0.3} = 492.372[/MATH]
but when I calculate another point, I get different slope

[MATH]m = \frac{{14.98}^2 - {13.66}^2}{0.5 - 0.4} = 378.048 [/MATH]
So, what am I doing wrong? And how to find the slope from the values in the table?
 
I presume some context you omitted might explain the subscript. Is 10 the length of the pendulum, perhaps?

If you plot T^2 against L, you should see that it is not exactly linear, but close. A good idea would be to use the first and last points to get a better sense of the overall slope; any answer is really only an approximation anyway. (They really should tell you that; but it goes without saying if you are learning about real data as opposed to fake math data.) The proper way is to use linear regression, which perhaps you haven't learned and therefore are not expected to do.
 
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[MATH]T_{10}[/MATH] is notation for 10 oscillations (periods) of the pendulum.

Dividing the [MATH]T_{10}[/MATH] values by 10 gives a good average measurement value for a aingle period, [MATH]T[/MATH].

Construct a scatterplot of [MATH]T^2[/MATH] as a function of [MATH]L[/MATH] and perform a linear regression. Most graphing calculators will do this, as will spreadsheets.
 
Dr.Peterson said:
I presume some context you omitted might explain the subscript. Is 10 the length of the pendulum, perhaps?

If you plot T^2 against L, you should see that it is not exactly linear, but close. A good idea would be to use the first and last points to get a better sense of the overall slope; any answer is really only an approximation anyway. (They really should tell you that; but it goes without saying if you are learning about real data as opposed to fake math data.) The proper way is to use linear regression, which perhaps you haven't learned and therefore are not expected to do.
Click to expand...
Thanks a lot Dr.Peterson for the reply.

If I follow your way to find the slope by taking the first and last points, I get this

[MATH]m = \frac{18.25 - 11.72}{0.7 - 0.3} = 16.325[/MATH]
or I should square the [MATH]T[/MATH] like this?

[MATH]m = \frac{{18.25}^2 - {11.72}^2}{0.7 - 0.3} = 489.26025[/MATH]
I have a linear regression in my calculator

[MATH]y = A + Bx[/MATH]
if i don't use square for the [MATH]T[/MATH], I get
[MATH]A = 7.198[/MATH][MATH]B = 15.48[/MATH] (slope)

So, [MATH]15.48 = \frac{4{\pi}^2}{g_{exp}}[/MATH]
[MATH]g_{exp} = \frac{4{\pi}^2}{15.48} = 2.55[/MATH]
if i use square for the [MATH]T[/MATH], I get
[MATH]A = -3.69284[/MATH][MATH]B = 463.379[/MATH] (slope)

So, [MATH]463.379 = \frac{4{\pi}^2}{g_{exp}}[/MATH]
[MATH]g_{exp} = \frac{4{\pi}^2}{463.379} = 0.085[/MATH] (I feel that this gravitational accelearation does not make sense)

Which method above was correct to approximate the slope?

Is the unit of the slope [MATH]s^2/m[/MATH]?

Is the percentage error, [MATH](1 - \frac{2.55}{9.81}) \cdot \ 100[/math]% [math] = 74[/math]% (too high) why?




skeeter said:
[MATH]T_{10}[/MATH] is notation for 10 oscillations (periods) of the pendulum.

Dividing the [MATH]T_{10}[/MATH] values by 10 gives a good average measurement value for a aingle period, [MATH]T[/MATH].

Construct a scatterplot of [MATH]T^2[/MATH] as a function of [MATH]L[/MATH] and perform a linear regression. Most graphing calculators will do this, as will spreadsheets.
Click to expand...
Thanks a lot skeeter for the reply.

Do you mean I have to do this to find a good average value for a single period?

[MATH]T_{av} = \frac{11.72 + 13.66 + 14.98 + 16.08 + 18.25}{10} = 7.469 \ s[/MATH]
[MATH]y = A + Bx[/MATH]
Is [MATH]T_{av} = A[/MATH] in this linear equation?

What should I do with this period to proceed next?

I have performed a linear regression by using my calculator. You can see the result above when I was showing Dr.Peterson my results. Can you see them and tell me if there is something wrong?
 
no, divide each individual value of [MATH]T_{10}[/MATH] in the table to make a new table of [MATH]L[/MATH] vs [MATH]T^2[/MATH] ...

EA33A52F-3FDE-4D90-9101-79D29DEB129C.png
 
Thanks a lot skeeter. Now, it is getting closer

[MATH]g_{exp} = \frac{4{\pi}^2}{4.63379} = 8.52[/MATH]
percentage error [MATH]= (1 - \frac{8.52}{9.81}) \cdot 100[/MATH]% [MATH]= 13[/MATH]% (this makes all sense)

It is obvious that the unit of [MATH]g_{exp}[/MATH] is [MATH]m/s^2[/MATH], but what about the unit of the slope?
 
nasi112 said:
Thanks a lot skeeter. Now, it is getting closer

[MATH]g_{exp} = \frac{4{\pi}^2}{4.63379} = 8.52[/MATH]
percentage error [MATH]= (1 - \frac{8.52}{9.81}) \cdot 100[/MATH]% [MATH]= 13[/MATH]% (this makes all sense)

It is obvious that the unit of [MATH]g_{exp}[/MATH] is [MATH]m/s^2[/MATH], but what about the unit of the slope?
Click to expand...
Unit of slope → T^2/L → s^2/m
 
Thanks a lot guys. Your help was very important because I was thinking in this question for 3 days.

Thanks Dr.Peterson, skeeter, and Khan.
 
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