8.3 Solutions
381
8.60 v =
γ P
ρ
(Laplace formula)
∴
v H
v O
=
ρ O
ρ H
=
√
16 = 4
∴ v H = 4v O = 4 × 317 = 1268 m/s
8.61 IL = 10 log
I 2
I 1
= 10 log
10
0.4
= 14 dB
8.62 IL = 10 log
I 2
I 1
= 6
log
I 2
I 1
= 0.6
∴
I 2
I 1
= 3.98 or 4
8.63 The threshold of hearing intensity is taken as 10 −12 W/m 2 . Let r be the distance from the source at which the sound can be audible
I =
power
4πr 2 = 10
−12
∴ r =
power
4π × 10 −12 =
0.009
4π × 10 −12 = 2.677 × 10
4 m = 26.8 km
8.64 P max = kρ 0 v
2 A =
2π
λ
ρ 0 v( f λ)A = 2πρ 0 v f A
∴ A =
P max
2πρ 0 v f
=
2 × 10 −5
2π × 1.22 × 331 × 1000
= 7.9 × 10
−12 m
8.65 (a) A =
P max
2πρ 0 f v
A 1
A 2
=
ρ 2
ρ 1
v 2
v 1
=
4
3
×
2
3
=
8
9
(b) I =
1
2
p 2
max
ρ 0 v
I 1
I 2
=
ρ 2 v 2
ρ 1 v 1
=
4
3
×
2
3
=
8
9
381
8.60 v =
γ P
ρ
(Laplace formula)
∴
v H
v O
=
ρ O
ρ H
=
√
16 = 4
∴ v H = 4v O = 4 × 317 = 1268 m/s
8.61 IL = 10 log
I 2
I 1
= 10 log
10
0.4
= 14 dB
8.62 IL = 10 log
I 2
I 1
= 6
log
I 2
I 1
= 0.6
∴
I 2
I 1
= 3.98 or 4
8.63 The threshold of hearing intensity is taken as 10 −12 W/m 2 . Let r be the distance from the source at which the sound can be audible
I =
power
4πr 2 = 10
−12
∴ r =
power
4π × 10 −12 =
0.009
4π × 10 −12 = 2.677 × 10
4 m = 26.8 km
8.64 P max = kρ 0 v
2 A =
2π
λ
ρ 0 v( f λ)A = 2πρ 0 v f A
∴ A =
P max
2πρ 0 v f
=
2 × 10 −5
2π × 1.22 × 331 × 1000
= 7.9 × 10
−12 m
8.65 (a) A =
P max
2πρ 0 f v
A 1
A 2
=
ρ 2
ρ 1
v 2
v 1
=
4
3
×
2
3
=
8
9
(b) I =
1
2
p 2
max
ρ 0 v
I 1
I 2
=
ρ 2 v 2
ρ 1 v 1
=
4
3
×
2
3
=
8
9