INDIANA UNIVERSITY DEPARTMENT OF PHYSICS
P309: INTERMEDIATE PHYSICS LABORATORY
Introductory Laboratory 0: Buffon's Needle
Introduction: In geometry, the circumference of a circle divided by its diameter is a fundamental
constant, denoted by the symbol 𝜋𝜋. Curiously, this number has no simple form. Expressed as
a decimal, the digits of 𝜋𝜋,
𝜋𝜋 = 3.141592653589793238462643383279502884197169399375105820974944. ..
continue forever without settling into a regular pattern. This mysterious contrast between the
simple geometric explanation of the value of 𝜋𝜋, and the complexity of the numerical value has
inspired books, movies, and even cults. Because no simple mathematical form exists for π, we
must rely on methods which are able to to calculate 𝜋𝜋 iteratively where the approximation to its
value becomes better and better with each iteration. The first such estimate was made by
Archimedes roughly 200 years BC by calculating the circumference of polygons which fit just
inside and just outside of a circle. Currently, the world's best estimate of 𝜋𝜋 contains 1.24 trillion
decimal places. In this introductory lab we will explore an experimental method to calculate
𝜋𝜋 introduced by the French naturalist Georges Buffon in 1733.
Method: Buffon's method for estimating the value of 𝜋𝜋
relies on the following observation. Suppose you mark a
surface with parallel lines separated by a distance 𝑑𝑑.
Now imagine throwing a straight object (a needle, or a
toothpick) of length 𝑙𝑙 randomly onto that surface. It can
be shown that in the case where 𝑙𝑙 < 𝑑𝑑, the probability
2𝑙𝑙
that the needle lands such that it crosses a line is 𝑝𝑝 = 𝜋𝜋𝑑𝑑.
We can compute the probability 𝑝𝑝 experimentally by
repeatedly throwing a needle on to the grid. If we throw
the needle 𝑁𝑁 times and record 𝑛𝑛 hits and 𝑚𝑚 misses (𝑁𝑁 =
𝑛𝑛 + 𝑚𝑚), we can approximate 𝑝𝑝 as 𝑝𝑝 = 𝑛𝑛/𝑁𝑁. We can
2𝑙𝑙
then approximate 𝜋𝜋 as 𝜋𝜋 = 𝑑𝑑𝑑𝑑 or 𝜋𝜋 =
2𝑙𝑙𝑙𝑙
𝑑𝑑𝑑𝑑
. The more
throws of the needle we make the better our
approximation to 𝜋𝜋 becomes.
Last Updated: Prof. James A. Glazier, 8/31/15—Send corrections to jaglazier@gmail.com
INDIANA UNIVERSITY DEPARTMENT OF PHYSICS
P309: INTERMEDIATE PHYSICS LABORATORY
Questions:
[1] Can you derive the expression for the probability 𝑝𝑝? [Hint: draw
a graph with the distance the needle lands from a line on one axis
and the angle the needle lands with on the other axis. Which points
on this graph produce a hit?]
Let’s do this calculation explicitly (please look at the answer only
after you have tried doing the derivation yourself):
Consider a toothpick which lands between two lines. The center of the
toothpick must be a distance 𝑦𝑦 ≤
𝑑𝑑
2
𝑁𝑁𝑁𝑁𝑁𝑁𝑁𝑁𝑁𝑁𝑁𝑁𝑁𝑁 𝐿𝐿𝐿𝐿𝐿𝐿𝐿𝐿
𝑦𝑦
𝑦𝑦
𝜃𝜃
from the nearer of the two lines.
𝑑𝑑
𝑁𝑁𝑁𝑁𝑁𝑁𝑁𝑁 𝑁𝑁𝑁𝑁𝑁𝑁𝑁𝑁𝑁𝑁𝑁𝑁𝑁𝑁
𝐿𝐿𝐿𝐿𝐿𝐿𝐿𝐿
𝑙𝑙
2
𝑥𝑥
Call the angle of the toothpick with respect to the horizontal axis 𝜃𝜃.
𝜋𝜋
− 2 ≤ 𝜃𝜃 < 𝜋𝜋. Projecting in the horizontal direction, the length of the
toothpick from its center towards the nearer vertical line is 𝑥𝑥 =
𝑙𝑙
2
𝑙𝑙
cos(𝜃𝜃). If 2 cos(𝜃𝜃) = 𝑥𝑥 ≥ 𝑦𝑦, the toothpick crosses the line, otherwise it doesn’t.
Consider the simple extreme cases:
𝑙𝑙
If 𝜃𝜃 = 0, 𝑥𝑥 = 2 so the toothpick crosses the nearest line
𝑙𝑙
if 𝑦𝑦 ≤ 2.
𝑁𝑁𝑁𝑁𝑁𝑁𝑁𝑁𝑁𝑁𝑠𝑠𝑡𝑡 𝐿𝐿𝐿𝐿𝐿𝐿𝐿𝐿
𝑦𝑦
𝜃𝜃 =
𝑑𝑑
𝜋𝜋
2
𝑁𝑁𝑁𝑁𝑁𝑁𝑁𝑁𝑁𝑁𝑁𝑁𝑁𝑁 𝐿𝐿𝐿𝐿𝐿𝐿𝐿𝐿
𝜃𝜃 = 0
𝑁𝑁𝑁𝑁𝑁𝑁𝑁𝑁 𝑁𝑁𝑁𝑁𝑁𝑁𝑁𝑁𝑁𝑁𝑁𝑁𝑁𝑁
𝐿𝐿𝐿𝐿𝐿𝐿𝐿𝐿
𝑙𝑙
2
𝑥𝑥 = 0
𝑦𝑦
𝑥𝑥 =
𝑙𝑙
2
𝑑𝑑
𝑁𝑁𝑁𝑁𝑁𝑁𝑁𝑁 𝑁𝑁𝑁𝑁𝑁𝑁𝑁𝑁𝑁𝑁𝑁𝑁𝑁𝑁
𝐿𝐿𝐿𝐿𝐿𝐿𝐿𝐿
𝑙𝑙
2
𝜋𝜋
If 𝜃𝜃 = 2 , 𝑥𝑥 = 0 so the toothpick only crosses the nearest
line if 𝑦𝑦 = 0
In between, the toothpick crosses the line if
𝑙𝑙
2
cos(𝜃𝜃) ≥ 𝑦𝑦.
Last Updated: Prof. James A. Glazier, 8/31/15—Send corrections to jaglazier@gmail.com
INDIANA UNIVERSITY DEPARTMENT OF PHYSICS
P309: INTERMEDIATE PHYSICS LABORATORY
Let’s plot the phase space for the problem (whether the toothpick crosses the line as a function
of 𝑦𝑦 and 𝜃𝜃.
𝑑𝑑
𝑑𝑑
All four quadrants are the same (0 ≤ 𝜃𝜃 ≤ 𝜋𝜋 vs −𝜋𝜋 ≤ 𝜃𝜃 ≤ 0 and 0 ≤ 𝑦𝑦 ≤ 2 and − 2 ≤ 𝑦𝑦 ≤ 0) so
let’s just consider the quadrant where both 𝜃𝜃 and
𝑦𝑦 are positive.
In this quadrant, if
𝑙𝑙
2
cos(𝜃𝜃) ≥ 𝑦𝑦 then the
toothpick crosses the line.
We can draw this result as shown. To the left of
the curved blue line, the toothpick crosses the
nearest line. To the right it does not cross. Then
the probability that the toothpick crosses the
nearest line is just the area of the region to the
left of the blue line divided by the total area of
the rectangle bounded by the dashed and dotted
red lines:
The area of the rectangle is:
𝜋𝜋 𝑑𝑑 𝜋𝜋𝜋𝜋
𝐴𝐴𝑅𝑅𝑅𝑅𝑅𝑅𝑅𝑅𝑅𝑅𝑅𝑅𝑅𝑅𝑅𝑅𝑅𝑅 = × =
.
2 2
4
.
The area to the left of the blue line is:
𝐴𝐴𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐
𝜋𝜋
𝜋𝜋
2
𝜋𝜋
𝜃𝜃 =
2
𝜃𝜃 = 0
𝜃𝜃
𝑇𝑇𝑇𝑇𝑇𝑇𝑇𝑇ℎ𝑝𝑝𝑝𝑝𝑝𝑝𝑝𝑝
𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐 𝑙𝑙𝑙𝑙𝑙𝑙𝑙𝑙
𝑦𝑦 = 0
2𝑙𝑙
2𝑙𝑙
⇒ 𝜋𝜋 =
,
𝜋𝜋𝜋𝜋
𝑝𝑝𝑝𝑝
𝑑𝑑
𝑦𝑦 =
2
𝜋𝜋
𝑙𝑙
𝑙𝑙 2
= � cos(𝜃𝜃) 𝑑𝑑𝑑𝑑 = � cos(𝜃𝜃) 𝑑𝑑𝑑𝑑
2 0
0 2
Which was what we wanted to show.
𝑙𝑙
𝑦𝑦 =
2
(eq. 1)
2
𝑙𝑙
𝑙𝑙
𝜋𝜋
𝑙𝑙
= (− sin(𝜃𝜃)) � = �sin � � − sin(0)� = .
2
2
2
2
0
So the probability of crossing is:
𝑙𝑙
𝐴𝐴𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐
4𝑙𝑙
2𝑙𝑙
2
𝑝𝑝 =
=
=
=
.
𝐴𝐴𝑅𝑅𝑅𝑅𝑅𝑅𝑅𝑅𝑅𝑅𝑅𝑅𝑅𝑅𝑅𝑅𝑅𝑅 𝜋𝜋𝜋𝜋 2𝜋𝜋𝜋𝜋 𝜋𝜋𝜋𝜋
4
.
If we rearrange we can calculate 𝜋𝜋 from 𝑝𝑝:
𝑝𝑝 =
𝑇𝑇𝑇𝑇𝑇𝑇𝑇𝑇ℎ𝑝𝑝𝑝𝑝𝑝𝑝𝑝𝑝 𝑑𝑑𝑑𝑑𝑑𝑑𝑑𝑑
𝑛𝑛𝑛𝑛𝑛𝑛 𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐 𝑙𝑙𝑙𝑙𝑙𝑙𝑙𝑙
(eq. 2)
(eq. 3)
(eq. 4)
Last Updated: Prof. James A. Glazier, 8/31/15—Send corrections to jaglazier@gmail.com
𝑦𝑦
INDIANA UNIVERSITY DEPARTMENT OF PHYSICS
P309: INTERMEDIATE PHYSICS LABORATORY
[2] If the distance 𝑙𝑙 and 𝑑𝑑 are measured with a precision 𝜎𝜎𝑙𝑙 and 𝜎𝜎𝑑𝑑 , and the probability 𝑃𝑃 is known
with uncertainty 𝜎𝜎𝑃𝑃 , what is the uncertainty 𝜎𝜎𝜋𝜋 in the result to 𝜋𝜋?
Remember our basic relationship for error propagation:
𝜎𝜎𝑓𝑓(𝑥𝑥1 ,𝑥𝑥2 ,𝑥𝑥3 ,… ) = ��
2
1
2
2
𝜕𝜕𝜕𝜕
𝜕𝜕𝜕𝜕
�
𝜎𝜎𝑥𝑥21 + �
�
𝜎𝜎 2 + ⋯ � .
𝜕𝜕𝑥𝑥1 𝑥𝑥̅ 1 ,𝑥𝑥̅ 𝑠𝑠 ,…
𝜕𝜕𝑥𝑥2 𝑥𝑥̅ 1 ,𝑥𝑥̅ 𝑠𝑠 ,… 𝑥𝑥2
(eq. 5)
In this case, the function
𝜋𝜋(𝑝𝑝. 𝑙𝑙, 𝑑𝑑) = 𝑓𝑓(𝑝𝑝, 𝑙𝑙, 𝑑𝑑) =
so
𝜕𝜕𝜕𝜕 2𝑙𝑙
1
𝜋𝜋
= �− 2 � = − ,
𝜕𝜕𝜕𝜕 𝑑𝑑
𝑝𝑝
𝑝𝑝
Plug in to eq. 5:
𝜎𝜎𝜋𝜋(𝑝𝑝.𝑙𝑙,𝑑𝑑)
2𝑙𝑙
,
𝑝𝑝𝑝𝑝
𝜕𝜕𝜕𝜕
2
𝜋𝜋
=
= ,
𝜕𝜕𝜕𝜕 𝑑𝑑𝑑𝑑 𝑙𝑙
𝜕𝜕𝜕𝜕 2𝑙𝑙
1
𝜋𝜋
= �− 2 � = −
𝜕𝜕𝜕𝜕 𝑝𝑝
𝑑𝑑
𝑑𝑑
1
2
𝜋𝜋 2
𝜋𝜋 2
𝜋𝜋 2
= ��− �
𝜎𝜎𝑝𝑝2 + � �
𝜎𝜎𝑙𝑙2 + �− �
𝜎𝜎𝑑𝑑2 � ,
𝑝𝑝 𝑝𝑝̅ ,𝑙𝑙,̅ 𝑑𝑑�
𝑙𝑙 𝑝𝑝̅ ,𝑙𝑙,̅ 𝑑𝑑�
𝑑𝑑 𝑝𝑝̅ ,𝑙𝑙,̅ 𝑑𝑑�
(eq. 6)
(eq. 7)
(eq. 8)
We can simplify this equation by looking at the relative error rather than the absolute error.
Divide through by 𝜋𝜋� and eliminate extra – signs:
1
2
𝜎𝜎𝜋𝜋(𝑝𝑝.𝑙𝑙,𝑑𝑑)
12
12
12
= �� � 𝜎𝜎𝑝𝑝2 + � � 𝜎𝜎𝑙𝑙2 + � � 𝜎𝜎𝑑𝑑2 � ,
𝜋𝜋�
𝑝𝑝̅
𝑙𝑙 ̅
𝑑𝑑̅
Put 𝜎𝜎…2 inside |… |2𝑝𝑝̅,𝑙𝑙,̅ 𝑑𝑑� :
(eq. 9)
1
𝜎𝜎𝑝𝑝 2
𝜎𝜎𝜋𝜋(𝑝𝑝.𝑙𝑙,𝑑𝑑)
𝜎𝜎𝑙𝑙 2
𝜎𝜎𝑑𝑑 2 2
= �� � + � � + � � � .
𝜋𝜋�
𝑝𝑝̅
𝑙𝑙 ̅
𝑑𝑑̅
(eq. 10)
This equation for propagation of relative error is a true as long as the individual errors are
uncorrelated
with
each
other
(for
the
more
complicated
cases,
see
https://en.wikipedia.org/wiki/Propagation_of_uncertainty). It is also the reason that working with
relative error is easier than working with absolute error.
[3] Inferred vs. Relative Error:
From your measurements, you can now calculate both the measured relative error and the
theoretical or inferred relative error for 𝜋𝜋:
1
Given your measured values of {𝑝𝑝1 , 𝑝𝑝2 , … 𝑝𝑝𝑁𝑁 } calculate 𝑝𝑝̅ = 𝑁𝑁 ∑𝑁𝑁
𝑖𝑖=1 𝑝𝑝𝑖𝑖 , and 𝜎𝜎𝑝𝑝 =
�
1
𝑁𝑁−1
2
̅
̅
∑𝑁𝑁
𝑖𝑖=1(𝑝𝑝𝑖𝑖 − 𝑝𝑝̅ ) , similarly calculate 𝑙𝑙 , 𝜎𝜎𝑙𝑙 and 𝑑𝑑 , 𝜎𝜎𝑑𝑑 for your measured values of {𝑙𝑙1 , 𝑙𝑙2 , … 𝑙𝑙𝑀𝑀 }
Last Updated: Prof. James A. Glazier, 8/31/15—Send corrections to jaglazier@gmail.com
INDIANA UNIVERSITY DEPARTMENT OF PHYSICS
P309: INTERMEDIATE PHYSICS LABORATORY
and �𝑑𝑑1 , 𝑑𝑑2 , … 𝑑𝑑𝑄𝑄 �.
First look at
𝜎𝜎𝑝𝑝 𝜎𝜎𝑝𝑝
𝑝𝑝
,
𝑝𝑝
and
𝜎𝜎𝑑𝑑
𝑑𝑑
: Is one of these much larger than the others? If so it is the dominant
source of error (and you can essentially neglect the others). In your lab write of give these
individual values, discuss their relative significance and why they have the relationship they do.
You should make and present this analysis in every experiment you do in this course!
Now calculate the theoretical or inferred relative error using eq. 10:
1
𝜎𝜎𝑝𝑝 2
𝜎𝜎𝜋𝜋
𝜎𝜎𝑙𝑙 2
𝜎𝜎𝑑𝑑 2 2
= �� � + � � + � � � .
𝜋𝜋�
𝑝𝑝̅
𝑙𝑙 ̅
𝑑𝑑̅
What is this value?
(eq. 10)
Separately, you will calculate the measured error as follows:
2𝑙𝑙 ̅
From {𝑝𝑝1 , 𝑝𝑝2 , … 𝑝𝑝𝑁𝑁 }, 𝑙𝑙 ̅ and 𝑑𝑑̅ , calculate {𝜋𝜋1 , 𝜋𝜋2 , … 𝜋𝜋𝑁𝑁 }, using eq. 6 �𝜋𝜋𝑖𝑖 = 𝑝𝑝 𝑑𝑑��. Your estimated
value of
𝜋𝜋� =
𝑁𝑁
1
� 𝜋𝜋𝑖𝑖 .
𝑁𝑁
𝑖𝑖=1
.
The measured relative error is now:
𝑁𝑁
𝜎𝜎𝜋𝜋′ 1
1
= �
� (𝜋𝜋𝑖𝑖 − 𝜋𝜋�)2 .
𝜋𝜋�
𝜋𝜋� 𝜋𝜋�(𝑁𝑁 − 1)
𝑖𝑖=1
𝑖𝑖
(eq. 11)
(eq. 12)
You expect that the inferred and measured relative errors should be about the same (they need
not be exactly equal). Compare the two values and their difference. Is this difference significant?
Why or why not? If one relative error is much bigger than the other, you should think carefully
about why and explain this discrepancy in your report.
[4] Which uncertainties in your experiment would you label as systematic? Which as statistical?
[Hint—look up the concept of counting error in Wikipedia or your statistics notes and textbook.
Also consider measurement errors like the accuracy and precision of your ruler, or the possibility
of interaction between your toothpicks (what would the effect of clumping toothpicks be on your
result]?
[5] If you make 𝑁𝑁 throws of a single toothpick {𝑝𝑝1 , 𝑝𝑝2 , … , 𝑝𝑝𝑁𝑁 } resulting in 𝑛𝑛 hits and 𝑚𝑚 misses,
what is the uncertainty in 𝑝𝑝? Consider each throw to be a measurement of the value of 𝑝𝑝 which
results in either a hit (𝑝𝑝𝑖𝑖 = 1) or a miss (𝑝𝑝𝑖𝑖 = 0) and apply the formulas for estimating the error
on the mean of a distribution from repeated measurements (see statistics handouts from first class)
to show that:
Last Updated: Prof. James A. Glazier, 8/31/15—Send corrections to jaglazier@gmail.com
INDIANA UNIVERSITY DEPARTMENT OF PHYSICS
P309: INTERMEDIATE PHYSICS LABORATORY
𝜎𝜎𝑝𝑝2 =
𝑝𝑝(1 − 𝑝𝑝)
.
𝑁𝑁 − 1
(eq. 13)
As above, I’ll present this derivation (please look at the answer only after you have tried it
yourself):
𝑁𝑁
1
(eq. 14)
𝑝𝑝̅ = � 𝑝𝑝𝑖𝑖 ,
𝑁𝑁
𝑖𝑖=1
and
𝑁𝑁
1
𝜎𝜎𝑝𝑝2𝑖𝑖 =
� (𝑝𝑝𝑖𝑖 − 𝑝𝑝̅ )2 .
(eq. 15)
𝑁𝑁 − 1 𝑖𝑖=1
Now, expand the square on the right-hand side of equation 15:
𝜎𝜎𝑝𝑝2𝑖𝑖 =
𝑁𝑁
1
� 𝑝𝑝𝑖𝑖2 − 2𝑝𝑝𝑖𝑖 𝑝𝑝̅ + 𝑝𝑝̅ 2 .
𝑁𝑁 − 1 𝑖𝑖=1
Since 𝑝𝑝𝑖𝑖 𝜖𝜖{0,1}, 𝑝𝑝𝑖𝑖2 = 𝑝𝑝𝑖𝑖 , so:
𝑁𝑁
1
� 𝑝𝑝𝑖𝑖 − 2𝑝𝑝𝑖𝑖 𝑝𝑝̅ + 𝑝𝑝̅ 2 .
𝜎𝜎𝑝𝑝2𝑖𝑖 =
𝑁𝑁 − 1 𝑖𝑖=1
(eq. 16)
(eq. 17)
Expand −2𝑝𝑝𝑖𝑖 𝑝𝑝̅ = −𝑝𝑝𝑖𝑖 𝑝𝑝̅ − 𝑝𝑝𝑖𝑖 𝑝𝑝̅:
𝑁𝑁
1
𝜎𝜎𝑝𝑝2𝑖𝑖 =
� 𝑝𝑝𝑖𝑖 − 𝑝𝑝𝑖𝑖 𝑝𝑝̅ − 𝑝𝑝𝑖𝑖 𝑝𝑝̅ + 𝑝𝑝̅ 2 .
(eq. 18)
𝑁𝑁 − 1 𝑖𝑖=1
Factor out −𝑝𝑝𝑖𝑖 in the first two terms:
𝑁𝑁
1
𝜎𝜎𝑝𝑝2𝑖𝑖 =
� 𝑝𝑝𝑖𝑖 (1 − 𝑝𝑝̅) − 𝑝𝑝𝑖𝑖 𝑝𝑝̅ + 𝑝𝑝̅ 2 .
(eq. 19)
𝑁𝑁 − 1 𝑖𝑖=1
𝑝𝑝̅ is a constant, so pull 1 − 𝑝𝑝̅, 𝑝𝑝̅ and 𝑝𝑝̅ 2 out of the summation:
𝑁𝑁
𝑁𝑁
𝑁𝑁
1
𝜎𝜎𝑝𝑝2𝑖𝑖 =
�(1 − 𝑝𝑝̅ ) � 𝑝𝑝𝑖𝑖 − 𝑝𝑝̅ � 𝑝𝑝𝑖𝑖 + 𝑝𝑝̅ 2 � 1� . (eq. 20)
𝑁𝑁 − 1
𝑖𝑖=1
𝑖𝑖=1
𝑖𝑖=1
𝑁𝑁
∑𝑁𝑁
𝑖𝑖=1 𝑝𝑝𝑖𝑖 = 𝑁𝑁𝑝𝑝̅ and ∑𝑖𝑖=1 1 = 𝑁𝑁, so:
1
{(1 − 𝑝𝑝̅)𝑁𝑁𝑝𝑝̅ − 𝑝𝑝̅𝑁𝑁𝑝𝑝̅ + 𝑝𝑝̅ 2 𝑁𝑁}.
𝜎𝜎𝑝𝑝2𝑖𝑖 =
(eq. 21)
𝑁𝑁 − 1
But 𝑝𝑝̅𝑁𝑁𝑝𝑝̅ = 𝑝𝑝̅ 2 𝑁𝑁, so the second and third terms cancel, and:
1
{(1 − 𝑝𝑝̅ )𝑁𝑁𝑝𝑝̅ } .
𝜎𝜎𝑝𝑝2𝑖𝑖 =
(eq. 22)
𝑁𝑁 − 1
1
Remember that the uncertainty of the mean squared 𝜎𝜎𝑝𝑝̅ is 𝑁𝑁 times the uncertainty of the individual
measurements squared 𝜎𝜎𝑝𝑝2𝑖𝑖 , so:
𝜎𝜎𝑝𝑝̅2 =
𝜎𝜎𝑝𝑝2𝑖𝑖 1 𝑁𝑁𝑝𝑝̅ (1 − 𝑝𝑝̅) 𝑝𝑝̅(1 − 𝑝𝑝̅ )
=
=
.
𝑁𝑁 𝑁𝑁 − 1
𝑁𝑁 − 1
𝑁𝑁
(eq. 23)
Last Updated: Prof. James A. Glazier, 8/31/15—Send corrections to jaglazier@gmail.com
INDIANA UNIVERSITY DEPARTMENT OF PHYSICS
P309: INTERMEDIATE PHYSICS LABORATORY
[6] If we assume that 𝑙𝑙 and 𝑑𝑑 are perfectly well known (i.e., 𝜎𝜎𝑙𝑙 = 0, 𝜎𝜎𝑑𝑑 = 0), how many throws
𝜎𝜎
(N) do you need to measure 𝜋𝜋 with 1% uncertainty (i.e. for 𝜋𝜋�𝜋𝜋 = 0.01)? Call the uncertainty 𝛿𝛿 ≡
𝜎𝜎𝜋𝜋
�
𝜋𝜋
As above, I’ll present this derivation (please look at the answer only after you have tried it
yourself):
From equation 10,
𝜎𝜎𝑝𝑝 2
𝜎𝜎𝜋𝜋2
𝜎𝜎𝑙𝑙 2
𝜎𝜎𝑑𝑑 2
𝛿𝛿 2 = 2 = � � + � � + � � .
𝜋𝜋�
𝑝𝑝̅
𝑙𝑙 ̅
𝑑𝑑̅
(eq. 24)
Since we assume 𝜎𝜎𝑙𝑙 = 0, 𝜎𝜎𝑑𝑑 = 0:
𝜎𝜎𝑝𝑝 2
𝛿𝛿 2 = � � .
𝑝𝑝̅
From equation 23, 𝜎𝜎𝑝𝑝̅2 =
𝛿𝛿 2 =
Solve for 𝛿𝛿:
𝑁𝑁 =
𝑝𝑝̅ (1−𝑝𝑝̅ )
𝑁𝑁−1
, so:
𝑝𝑝̅(1 − 𝑝𝑝̅ )
1 − 𝑝𝑝̅
=
.
2
𝑝𝑝̅ (𝑁𝑁 − 1) 𝑝𝑝̅(𝑁𝑁 − 1)
𝑁𝑁 − 1 =
So:
(eq. 25)
(eq. 26)
1 − 𝑝𝑝̅
,
𝛿𝛿 2 𝑝𝑝̅
(eq. 27)
1 1 − 𝑝𝑝̅
+ 1.
𝛿𝛿 2 𝑝𝑝̅
(eq. 28)
[7] If we assume that l and d are both measured to be 4.0 ± 0.1cm, how many throws (what value
of 𝑁𝑁) does it take to make the uncertainty resulting from your measurement of 𝑃𝑃 equal to the
uncertainty from the measurements of 𝑙𝑙 and 𝑑𝑑? [Hint, see point 2 above and eq. 28]. At this point
we refer to the measurement as systematics-limited since the statistical error is no longer the largest
contribution to the relative error. Repeating the measurement further makes little sense until the
systematic errors can be reduced via improved methods and instruments.
[8] In the experiment, you can choose to make the distance 𝑑𝑑 which separates the lines either
nearly equal to 𝑙𝑙 or much larger than 𝑙𝑙. What are the advantages and disadvantages of each? Which
choice of
𝑙𝑙
𝑑𝑑
gives you the smallest relative error for a given number of repetitions (𝑁𝑁)? [Hint,
what value of 𝑝𝑝 gives the smallest possible value for
𝑙𝑙
1−𝑝𝑝̅
𝑝𝑝̅
in equation 28? Why is this the
optimum?What ratio of 𝑑𝑑 gives this value of 𝑝𝑝 (use equation 3)?]
Last Updated: Prof. James A. Glazier, 8/31/15—Send corrections to jaglazier@gmail.com
INDIANA UNIVERSITY DEPARTMENT OF PHYSICS
P309: INTERMEDIATE PHYSICS LABORATORY
Procedure:
Count out 20 toothpicks and measure their length. Compute the average length 𝑙𝑙 of the
toothpicks in your sample and estimate the uncertainty in this length. Next draw a set of
parallel lines on a piece of paper at a distance d apart. Be sure to choose a distance 𝑑𝑑 such that
𝑑𝑑 > 𝑙𝑙. A good choice is to pick a value for d that will give you a large hit probability, do pick d
as close to l as you can while ensuring that 𝑙𝑙 < 𝑑𝑑. Estimate your uncertainty in d. Devise a
way to throw your toothpicks such that they land with random positions and orientations on
the paper. Toss any toothpicks that do not land on the paper again. After all 20 sticks have
been thrown, record the number of hits. Let's call this set of 20 toothpicks a trial. Repeat
this process as many times as you can. Finish at least 𝑁𝑁 = 50 trials.
Analysis:
𝑛𝑛
[1] Using the estimate 𝑝𝑝 = 𝑁𝑁, compute your estimate of 𝜋𝜋 after each trial (include all preceding
trials in the calculation). Plot the value of 𝜋𝜋 as a function of the trial number. Do you see the
trend you expect?
[2] Make a histogram which shows the number of times a trial yielded, 1, 2, 3, . . . , 20 hits on
a line. Calculate the mean and standard deviation of this distribution. For a “counting experiment”
like this one, the standard deviation of the distribution is roughly equal to the square­root of the
mean. How does this expectation compare with your measurements?
Calculate the uncertainty in the mean of this distribution. Based on this calculation what do you
𝜎𝜎𝑝𝑝
estimate for 𝑝𝑝̅ and for the error in 𝑝𝑝̅ (remember that 𝜎𝜎𝑝𝑝̅ = )? What value of 𝜋𝜋 does this value
√𝑁𝑁
of 𝑝𝑝̅ yield? What is the uncertainty in 𝜋𝜋, 𝜎𝜎𝜋𝜋 from this measurement?
[3] Use the results derived in Question 4 to estimate 𝑝𝑝 and its uncertainty. How does this
compare to the result in part [2] above? What result do you get for 𝜋𝜋 using this method?
[4] Exchange your data with as many lab groups as you can. Using all of the data available to
you, what is your estimate for 𝜋𝜋? What is the uncertainty in your experimental calculation? Which
contributions to the uncertainties matter most? The limited statistics? The measurement of 𝑙𝑙? 𝑑𝑑?
Last Updated: Prof. James A. Glazier, 8/31/15—Send corrections to jaglazier@gmail.com