Physics 2BL Homework Set 01
Taylor Problems: 3.10, 3.28, 3.36, 3.41
3.10: 1 Deck = 52 Cards
Thickness of one deck is measured to be: T = 0.590 ± 0.005 inches
(A) Thickness of one card = t = (1/52) T → t = 0.0113 ± 0.0001 inches
(B) N = Number of cards measured to obtain desired uncertainty
T=Nt
0.005inches
= N = 250 Cards
δT = N δt → N = δT/ δt =
0.00002inches
This is 4 decks and 42 cards or about 5 decks.
3.28: T = 2π L / g
L = 1.40 ± 0.01m ,
g = 9.81m·sec-2
(A) T = 2π (1.40m) /(9.81m ⋅ s −2 ) = 2.37sec
δT =
dT
1
δL = 2π ⋅ (L / g )−1 / 2 ⋅ (1 / g ) ⋅ δL
dL
2
= (π / g ) ⋅ g / L ⋅ δL
(
) (9.81m ⋅ s )/(1.40m) ⋅ (0.01m)
= π / 9.81m ⋅ s −2 ⋅
δT = 0.01sec
−2
T = 2.37 ± 0.01 sec
(B) TMeasured = 2.39 ± 0.01 sec
so, 2.36 sec ≤ TPredicted ≤ 2.38 sec
and 2.38 sec ≤ TMeasured ≤ 2.40 sec
These ranges just barely overlap since both share the endpoint at 2.38 seconds.
Thus, the predicted and measured values of the pendulum are consistent with
one another.
3.36: (A)
(12 ± 1) × [(25 ± 3) − (10 ± 1)]
= (12 ± 1) × (15 ± 3 2 + 12 ) = (12 ± 1) × (15 ± 3)
2
2
= 180 ± 180 (3 / 15) + (1 / 12 )
= 180 ± 40
(B)
16 ± 4 + (3.0 ± 0.1) (2.0 ± 0.1)
δq
1 4
1⎛1⎞ 1
1
q1 = 16 ± 4 , 1 =
= ⎜ ⎟ = ⇒ δq1 =
4
2 16 2 ⎝ 4 ⎠ 8
2
0.1
3 δq
q 2 = (3.0 ± 0.1) , 2 = 3
⇒ δq2 = 2.7 , round to 3
3.0
27
δ q3
2
0.1 2
q3 = (27 ± 3)(2.0 ± 0.1) ,
= ( 273 ) + ( 2.0
) ⇒ δ q3 = 6.6 , round to 7
54
3
q 4 = (4.0 ± 0.5) + (54 ± 7 ) = 58 ±
(0.5)2 + (7 )2
→ = 58 ± 7
(C)
(20 ± 2) exp[− (1.0 ± 0.1)]
q1 = exp[− (1.0 ± 0.1)] , δq1 = − exp[− (1.0 )] ⋅ (0.1) ⇒ δq1 = 0.37 ± 0.04
q 2 = (20 ± 2 )(0.37 ± 0.04 ) = 7.4
( 71.4 )δq2 = ( 202 )2 + ( 00..3704 )2
= 0.15 ⇒ δq 2 = 1.1
→ = 7.4 ± 1.1
3.41:
i (deg)
All ± 1
r (deg)
All ± 1
sin(i)
10
20
30
50
70
7
13
20
29
38
0.174
0.342
0.5
0.766
0.940
sin(r)
0.122
0.225
0.342
0.485
0.616
n=
sin i
sin r
1.42
1.52
1.46
1.58
1.53
δ sin i
δ sin r
δn
sin i
sin r
n
10%
5%
3%
1%
1%
14%
8%
5%
3%
2%
17%
9%
6%
3%
2%
δn
0.2
0.14
0.09
0.05
0.03
If we are told by the manufacturer that n = 1.50, we may check this against our
measured data. We check that n-δn ≤ 1.50 ≤ n+δn. This is true for each case
except i = 50°. Still, the lower bound is close to 1.50 (differing by 0.03). The
results are fairly consistent with the manufacturer's reported value. As the
angles increase, the uncertainties in the angles are constant so their fractional
uncertainties decrease. Thus, the fractional uncertainties in n also decrease as i
increases.