6.3 Solutions
283
Comparing this with the standard equation
d 2 x
dt 2 + 2b
dx
dt
+ ω
2
0 x = p cos ωt
b = 0.375; ω 0 =
√
20, p = 6, ω = 4
Z M =
ω 2
0 − ω 2
2 + 4b 2 ω 2 =
(20 − 16)
2
+ 4 × 0.375 2 × 4 2 = 5
(a) A =
p
Z m
=
6
5
= 1.2
(b) tan ε =
2bω
ω 2
0 − ω 2
=
2 × 0.375 × 4
(20 − 16)
= 0.75 → ε = 37 ◦
(c) Q =
ω 0 m
r
=
ω 0
2b
=
√
20
2 × 0.375
= 5.96
(d) F = pm = 6 × 2 = 12
W =
F 2
2Z m
sin ε =
12 2
2 × 5
sin 37 ◦ = 8.64 W
6.58 Q =
2π t c
T
= 2π t c f = 2π × 2 × 100 = 1256
6.59 (a) Energy is proportional to the square of amplitude
E = const.A
2
dE
E
=
2dA
A
=
2 × 5
100
= 10%
(b) E = E 0 e
−t/t c
∴
E
E 0
=
A 2
A 2
0
= e
−t/t c
∴
A
A 0
=
95
100
= e
−t/2t c
t
2t c
= ln
100
95
= 0.05126
t c =
3
2 × 0.05126
= 29.26 s
(c) Q =
2π t c
T
=
(2π)(29.26)
3.0
= 61.25
283
Comparing this with the standard equation
d 2 x
dt 2 + 2b
dx
dt
+ ω
2
0 x = p cos ωt
b = 0.375; ω 0 =
√
20, p = 6, ω = 4
Z M =
ω 2
0 − ω 2
2 + 4b 2 ω 2 =
(20 − 16)
2
+ 4 × 0.375 2 × 4 2 = 5
(a) A =
p
Z m
=
6
5
= 1.2
(b) tan ε =
2bω
ω 2
0 − ω 2
=
2 × 0.375 × 4
(20 − 16)
= 0.75 → ε = 37 ◦
(c) Q =
ω 0 m
r
=
ω 0
2b
=
√
20
2 × 0.375
= 5.96
(d) F = pm = 6 × 2 = 12
W =
F 2
2Z m
sin ε =
12 2
2 × 5
sin 37 ◦ = 8.64 W
6.58 Q =
2π t c
T
= 2π t c f = 2π × 2 × 100 = 1256
6.59 (a) Energy is proportional to the square of amplitude
E = const.A
2
dE
E
=
2dA
A
=
2 × 5
100
= 10%
(b) E = E 0 e
−t/t c
∴
E
E 0
=
A 2
A 2
0
= e
−t/t c
∴
A
A 0
=
95
100
= e
−t/2t c
t
2t c
= ln
100
95
= 0.05126
t c =
3
2 × 0.05126
= 29.26 s
(c) Q =
2π t c
T
=
(2π)(29.26)
3.0
= 61.25