284
6 Oscillations
6.60 (a) E = E 0 e −t/t c
∴
t
t c
= ln
E 0
E
= ln 2 = 0.693
Put t = nT
∴ n = 0.693
t c
T
But −
E
E
=
3
100
=
T
t c
∴ n = 0.693 ×
100
3
= 23.1
(b) Q =
2π t c
T
= 2π ×
100
3
= 209.3
6.61 ω
= ω 0
1 −
1
4Q 2 =
9ω 0
10
∴ Q = 1.147
Q =
2π t c
T
or
T
2t c
=
π
Q
=
3.14
1.147
= 2.737
A
A 0
= e
−T /2t c = e
−2.737
= 0.065
6.62 ω
2
= ω
2
0 − b
2
(1)
where b =
r
2m
(2)
ω
= ω 0
1 −
b 2
ω 2
0
1/2
≈ ω 0
1 −
b 2
2ω 2
0
(3)
where we have expanded the radical binomially, assuming that b/ω 0 << 1.
Now ω
2
0 =
k
m
(4)
∴
b 2
2ω 2
0
=
r 2
8mk
(5)
Substituting (5) in (3)
ω
= ω 0
1 −
r 2
8mk
(for small damping)
6 Oscillations
6.60 (a) E = E 0 e −t/t c
∴
t
t c
= ln
E 0
E
= ln 2 = 0.693
Put t = nT
∴ n = 0.693
t c
T
But −
E
E
=
3
100
=
T
t c
∴ n = 0.693 ×
100
3
= 23.1
(b) Q =
2π t c
T
= 2π ×
100
3
= 209.3
6.61 ω
= ω 0
1 −
1
4Q 2 =
9ω 0
10
∴ Q = 1.147
Q =
2π t c
T
or
T
2t c
=
π
Q
=
3.14
1.147
= 2.737
A
A 0
= e
−T /2t c = e
−2.737
= 0.065
6.62 ω
2
= ω
2
0 − b
2
(1)
where b =
r
2m
(2)
ω
= ω 0
1 −
b 2
ω 2
0
1/2
≈ ω 0
1 −
b 2
2ω 2
0
(3)
where we have expanded the radical binomially, assuming that b/ω 0 << 1.
Now ω
2
0 =
k
m
(4)
∴
b 2
2ω 2
0
=
r 2
8mk
(5)
Substituting (5) in (3)
ω
= ω 0
1 −
r 2
8mk
(for small damping)