8.3 Solutions
361
8.9 f 1 =
1
2L 1
F
μ 1
f 2 =
1
2L 2
F
μ 2
∴
μ 2
μ 1
=
(L 1 f 1 ) 2
(L 2 f 2 ) 2 =
(0.05 × 4800) 2
(2.0 × 32) 2 14
8.10 y = 5 sin π(0.02x − 4.00t) = 5 sin 2π(0.01x − 2.00t) (given equation) (1)
y = A sin 2π
x
λ
− ft
(standard equation)
(2)
Comparing (1) and (2)
A = 5 cm, f = 2 Hz
1
λ
= 0.01 or λ = 100 cm
v = f λ = 2 × 100 = 200 cm/s
8.11 y = 4 sin
1
2
π x cos 20π t (standing wave)
(1)
y = 2A sin kx cos ωt (standard equation)
(2)
Comparing (1) and (2)
(a) 2A = 4 or A = 2 cm, k =
π
2
, ω = 20π
v =
ω
k
=
20π
π/2
= 40 cm/s
(b) λ =
2π
k
=
2π
π/2
= 4 cm
Distance between nodes =
λ
2
=
4
2
= 2 cm
(c)
∂ y
∂t
= −(4)(20π) sin
1
2
π x sin 20π t
∂ y
∂t
x=1.0, t=9/4 = −80π sin
π
2
sin 45π = 0
8.12 The wave is of the form
y = A sin(kx − ωt + φ)
(a) ω = 2π f = (2π)(250) = 500π rad/s
k =
ω
v
=
500π
375
=
4π
3
m
−1