6.3 Solutions
259
θ
2
=
v 2
gL
= 0.01
∴ θ =
√
0.01 = 0.1 rad
6.13 a = A sin ωt 0
b = A sin 2ωt 0
c = A sin 3ωt 0
a + c = 2A sin 2ωt 0 cos ωt 0
a + c
2b
= cos ωt 0
ω =
1
t 0
cos
−1
a + c
2b
f =
1
2π t 0
cos
−1
a + c
2b
6.14 (a) ω =
k
m
k = mω
2
=
4π 2 m
T 2 =
4π 2 × 4
2 2
= 39.478 N/m
(b) F max = mω 2 A = k A = 39.478 × 2 = 78.96 N
6.15 x = a sin ωt
y = b cos ωt
∴
x 2
a 2 +
y 2
b 2 = sin
2
ωt + cos
2
ωt = 1
Thus the path of the particle is an ellipse.
6.16 (a) To show that ∇ × F = 0.
∇ × F =
i
j k
∂
∂ x
∂
∂ y
∂
∂z
−K x 0 0
= 0
(b) U = −
Fdx = −
(K ix) (− ˆ
i dx) =
1
2
K x 2