Download Solution to Problem 11, p. 202, Boyce

Solution to Problem 11, p. 202, Boyce-DpPrima, 8th Ed.
11. A spring is stretched 10 cm by a force of 3 newtons. A mass of 2 kg is hung from the spring
and is also attached to a viscous damper that exerts a froce of 3 newtons when the velocity of mass
is 5 m/sec. If the mass is pulled down 4 cm below its equilibrium position and given an initial
downward velocity of 10 cm/sec, determine its position u at any time t. Find the quasi-frequency
µ and the ratio of µ to the natural frequency of the corresponding undamped motion.
Solution:
The equation of motion is
mü + γ u̇ + ku = 0.
(1)
The characteristic equation is mr2 + γr + k = 0 with roots
√
−γ ± γ 2 − 4km
r=
.
2m
It will turn out that the quantity γ 2 − 4km is negative, so the roots are complex of the form
√
γ
4km − γ 2
r=−
±i
.
2m
2m
The quasi-frequency is
√
µ=
4km − γ 2
.
2m
The natural frequency of the undamped motion is
s
ω0 =
k
.
m
The general solution is then,
γ
u(t) = e− 2m t (c1 cos(µt) + c2 sin(µt)) .
(2)
We only need to find m, γ, k, and determine c1 , c2 from the initial conditions. Also, we express
all terms in units of meters and kilograms.
Determinations of k, m, γ:
The spring is stretched 10 cm = 0.1 m by a force of 3 newtons.
Hooke’s law gives k × (distance stretched) = f orce. So, we get k ×0.1 = 3. This gives k = 30.
The mass is 2 kg, so m = 2.
The magnitude of the force due to damping is γ| u̇ | = 3, so γ × 5 = 3 or γ = 53 .
This gives
√
√
4km − γ 2
5991
3
µ=
=
= 3.87008,
γ=
= 0.15
2m
20
20
The initial conditions give:
u(0) = 5cm =
1
m,
20
u̇(0) = 10cm/sec = 0.1m/sec.
Using (2) we get
1
0.05 = u(0) = c1 , 0.1 = u̇(0) = −0.15c1 + c2 µ.
So,
c2 =
0.1 + (0.15)(0.05)
= 0.0277772.
3.87008
We can also write this as
u(t) = e−0.15t Rcos(µt − δ)
q
where R = c21 + c22 ∼ 0.0571977, and γ = arctan(c2 /c1 ) ∼ 0.50701.
Quasi-frequency: µ ∼ 3.87008,
q
√
k
Undamped frequency: ω 0 = m
= 15.
√
∼ 0.99925.
Ratio: √µ15 ∼ 3.87008
15
2