PHY-200
Quiz 1(1)
Quiz 1(1): Experimental Physics Lab-I
Solution
1. Suppose you measure three independent variables as,
x = 20.2 ± 1.5,
y = 8.5 ± 0.5,
θ = (30 ± 2)◦ .
Compute the following quantity,
q = x(y − sin θ).
Find the uncertainty in q and quote your results?
Solution:
Given values are:
x = 20.2 ± 1.5,
y = 8.5 ± 0.5,
θ = (30 ± 2)◦ .
Using the values given above, q becomes,
q = x(y − sin θ) = 20.2(8.5 − sin 30),
= 20.2(8.5 − 0.5) = 161.6
The uncertainty to the inferred quantity q can be find out using the following relationship,
√(
∆q =
∂q
∆x
∂x
)2
(
+
∂q
∆y
∂y
)2
(
+
∂q
∆θ
∂θ
)2
.
(1)
The first expression on the R.H.S of Equation (1) can be obtained by differentiating q
w.r.t x and multiplying with ∆x,
( )
∂q
∆x = (y − sin θ)∆x = (8.5 − sin 30)(1.5),
∂x
= (8.5 − 0.5)1.5 = 12
Likewise differentiating w.r.t y yields,
( )
∂q
∆y = x∆y = (20.2)(0.5) = 10.1
∂y
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PHY-200
Quiz 1(1)
and w.r.t θ gives,
(
∂q
∂θ
)
(
)
2π
∆θ = (−x cos θ)∆θ = (−20.2 × 0.866)
,
180
= 0.6103
Substituting the above expressions in Equation (1) results in,
q =
√
(12)2 + (10.1)2 + (0.6103)2 ,
= 15.69
Finally, the value of q alongwith its uncertainty can be quoted as,
q = (162 ± 16).
2. A group of students measures g, the acceleration due to gravity, with a compound
pendulum and obtain the following values in units of ms−2 .
9.81, 9.79, 9.84, 9.81, 9.75, 9.79, 9.82
(a) Calculate the mean and the standard deviation. Find the standard uncertainty
and quote your results.
(b) How many measurements would you expect to lie outside the 95% confidence
interval? How many actually do?
Solution:
(a) The best approximated value of acceleration due to gravity g is,
∑7
68.61
i=1 gi
=
= 9.80 m/s2 .
<g> =
n
7
(2)
The deviations di can be calculated as,
di = gi − < g > .
The calculated deviations for the measured values of g are tabulated in Table (I).
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PHY-200
Quiz 1(1)
g (m/s2 ) deviations di (m/s2 ) d2i (m2 /s4 )
9.81
0.01
1 × 10−4
9.79
-0.01
1 × 10−4
9.84
0.04
1.6 × 10−3
9.81
0.01
1 × 10−4
9.75
-0.05
2.5 × 10−3
9.79
-0.01
1 × 10−4
9.82
0.02
4 × 10−4
TABLE I: Table for calculated deviations.
The standard deviation is,
√
√
∑7
2
4.9 × 10−3
i=1 di
s =
=
= 0.026 m/s2 .
n
7
The standard uncertainty can be calculated as,
√
√
n
7
σ =
(s) =
(0.0264) = 0.0285 m/s2 ,
n−1
6
and standard uncertainty in the mean value is,
σ
0.0285
σm = √ = √
= 0.01 m/s2 .
n
7
To quote the final value of the acceleration due to gravity g alongwith its uncertainty, we will write the expression as,
g = (9.80 ± 0.01) m/s2 .
(b) For type A evaluations, the probability distribution function associated with the
measurement is a Gaussian probability distribution function as shown in Figure
(1). The coverage probability or the confidence of interval for any arbitrary
measurand µ corresponding to different standard uncertainties is given as,
(< µ > ±1σ) → (68%)
(< µ > ±2σ) → (95%)
(< µ > ±3σ) → (99%)
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PHY-200
Quiz 1(1)
p(x)
The shaded area between
µ−σ and µ+σ is 0.68
of the total area
µ−3σ µ−2σ µ−σ
µ
µ+σ µ+2σ µ+3σ
x
2σ
FIG. 1: A Gaussian probability distribution function.
Now as the above expressions predict, the 95% confidence of interval corresponds
to (< g > ±2σ). Considering this, the range of interval becomes,
g = (9.80 ± 2(0.01)) m/s2 = (9.80 ± 0.02) m/s2 ,
g = [9.78, 9.82] m/s2 .
The values (in m/s2 ) that lie within 95% confidence of interval are,
9.79, 9.81, 9.81, 9.79, 9.82,
and values (in m/s2 ) which lie outside this interval or lying within 5% confidence
of interval are,
9.84, 9.75.
3. Suppose you measure the mass m of a steel ball (unit in g) using an analog weighing
balance. The reading shown on the balance is depicted in Figure (2).
(a) What is the best approximation of m?
(b) Find the standard uncertainty associated with the reading?
(c) What probability distribution function you will use?
Solution:
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PHY-200
Quiz 1(1)
0
20
10
30
40
grams (g)
FIG. 2: An analog reading displayed on an analog balance.
(a) The uncertainty is so inherent in nature that even if you take a single reading
and don’t repeat the experiment, you get an uncertainty. In Figure (2), the scale
is an analog scale and the probability distribution function associated with it is
a triangular pdf. This is type B uncertainty which comes in due to the limited
resolution of the measuring instrument. For evaluating type B uncertainties, one
has to use his/her judgment.
Using this idea of judgement, the best approximated value of m being displayed
on an analog weighing balance is 25.5 g.
(b) The standard uncertainty associated with the reading can be find out by calculating the second moment (variance) of the probability distribution function. The
mathematical relationship for which is given as,
∫ ∞
2
σ =
(x − µ)2 f (x)dx,
−∞
where µ is the mean value where the function is maximum. The standard uncer√
tainty is just the square root of the variance i.e., (u = σ 2 ).
The value of (∆ = 0.2 g) is chosen such that the value couldn’t be greater than
25.70 g and less than 25.30 g. The scale uncertainty becomes,
∆
0.2
uscale = √ = √ = 0.08 g.
6
6
The final value of m alongwith its uncertainty can be quoted as,
g = (25.5 ± 0.1) g.
(c) The probability distribution function (pdf) will be a triangular pdf with (∆ =
0.2). This shows that probability is maximum at 25.50 g, gradually slopes down,
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PHY-200
Quiz 1(1)
eventually goes to zero at 25.30 g and 25.70 g. The (pdf) associated with reading
of an analog weighing balance is sketched in Figure (3).
1/∆ Probability density / (1/g)
25.30
25.50
25.70
Mass / g
∆=0.2
FIG. 3: The probability distribution function associated with an analog reading.
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