6.3 Solutions
281
(b) ω
=
2π
T =
2π
10
= 0.628 rad/s
b =
ω 2
0 − ω
2 =
2.236 2 − 0.628 2 = 2.146
r
2m
= b or r = 2mb = 2 × 4 × 2.146 = 17.17 Ns/m
(c) = bT = 2.146 × 10 = 21.46
6.54
d 2 x
dt 2 +
2dx
dt
+ 5x = 0
Let x = e λt . The characteristic equation then becomes λ 2 + 2λ + 5 = 0 with
the roots λ = −1 ± 2i
x = Ae
−(1−2i)t
+ Be
−(1+2i)t
or x = e
−t
[C cos 2t + D sin 2t]
where A, B, C and D are constants.
C and D can be determined from initial conditions. At t = 0, x = 5. Therefore
C = 5.
Also
dx
dt
= −e
−t
(C cos 2t + D sin 2t) + e
−t
(−2C sin 2t + 2D cos 2t)
At t = 0,
dx
dt
= −3
∴ −3 = −C + 2D = −5 + 2D
∴ D = 1
The complete solution is
x = e
−t
(5 cos 2t + sin 2t)
6.55 F = mg = kx
k =
mg
x
=
(1.0)(9.8)
0.2
= 49 N/m
Equation of motion is
m
d 2 x
dt 2 + r
dx
dt
+ kx = 0
( 1 )