1.41
Consider a spring mass damper system, like the one in Figure 1.9, with the following
values: m =10 kg, c = 3 N/s and k = 1000 N/m. a) Is the system overdamped,
underdamped or critically damped? b) Compute the solution if the system is given initial
conditions x0 = 0.01 m and v0 = 0.
Solution: a) Using equation 1.30 the damping ratio is
z=
c
2 km
=
3
2 10 ×1000
= 0.015 < 1
Thus the system is underdamped.
b) Using equations (1.38) the amplitude and phase can be calculated from the initial
conditions:
A=
( v0 + zw n x0 )2 + ( x0w d )2
w d2
=
1
(0.015 ×10 × 0.01)2 + ( 0.01×9.999 )2 = 0.01 m
9.999
2
x0w d
-1 1- z
f = tan
= tan
= 1.556 rad
v0 + zw n x0
z
-1
So the solution is Ae-zw nt sin (w d t + f ) = 0.01e-0.15t sin ( 9.999t +1.556 ) m.
Note that for any system with v0 = 0 the phase is strictly a function of the damping ratio.
Also note that the code given below can be used to generate many problems for
homework or quizzes by just rearranging the numbers, always making sure to keep the
damping low enough to be underdamped. Mathematica and Matlab can also be used to
write very simple codes to generate a lot of problems.
1.43
Consider the system x + 4 x + x = 0 for x0 = 1 mm, v0 = 0 mm/s. Is this system
overdamped, underdamped or critically damped? Compute the solution and determine
which root dominates as time goes on (that is, one root will die out quickly and the other
will persist.
Solution: From equation (1.30) the damping ratio is
z=
c
4
=
= 2 >1
2 km 2 1×1
Hence the system is overdamped.
Given x + 4 x + x = 0 where x0 = 1 mm, v0 = 0
x = aert Þ x = arert Þ x = ar 2 ert
Let
Substitute these into the equation of motion to get:
ar 2 ert + 4arert + aert = 0
Þ r 2 + 4r + 1 = 0 Þ r1,2 = -2 ± 3
So
-2 + 3 ) t
-2 - 3 ) t
x = a1e (
+ a2 e (
( -2+ 3 ) t
( -2- 3 ) t
x˙ = ( - 2 + 3 ) a1 e
+ ( - 2 - 3 ) a2 e
Applying initial conditions yields,
x0 = a1 + a2
Þ x0 - a2 = a1
(1)
v0 = ( - 2 + 3 ) a1 + ( - 2 - 3 ) a2
(2)
Substitute equation (1) into (2)
v0 = - 2 + 3 (x 0 - a2 ) + - 2 - 3 a2
(
)
= ( - 2 + 3 )x
(
)
- 2 3 a2
-v0 + - 2 + 3 x0
a2 =
2 3
Substituting the value of a2 into equation (1), and solving for a1 yields,
v0
Solve for a2
a1 =
0
(
)
v 0 + ( 2 + 3 ) x0
2 3
v0 + 2 + 3 x0 ( -2 + 3 ) t -v0 + - 2 + 3 x0 ( -2 - 3 ) t
\ x(t) =
e
+
e
2 3
2 3
The response is dominated by the root: -2 + 3 as the other root dies off very fast.
(
)
(
)
1.49
Calculate the constants a1 and a2 in terms of the initial conditions and thus verify
equations (1.42) and (1.43) for the overdamped case.
Solution: From Equation (1.41)
x (t ) = e
-zw n t
(a e
2
wn z -1 t
1
+ a2 e
2
-w n z -1 t
)
taking the time derivative yields:
2
x˙ (t ) = (-zw n + w n z -1)a1e
(-zw
n
)
+ w n z 2 -1 t
(-zw
+ (-zw n - w n z -1)a2 e
2
n
)
-w n z 2 -1 t
Applying initial conditions yields,
x0 = x ( 0 ) = a1 + a2
Þ x0 - a2 = a1
(
) (
(1)
)
v0 = x ( 0 ) = -zw n + w n z 2 - 1 a1 + -zw n - w n z 2 - 1 a2
Substitute equation (1) into equation (2) and solve for a2
(
v = ( - zw
)
-1 ) x
(
(2)
)
v0 = - zw n + w n z 2 - 1 (x 0 - a2 ) + - zw n - w n z 2 - 1 a2
0
n
+ wn z
2
- 2w n z - 1 a2
2
0
Solve for a2
a2 =
-v0 -zw n x 0 + w n z 2 - 1 x0
2w n z 2 - 1
Substitute the value for a2 into equation (1), and solve for a1
a1 =
1.83
v0 + zw n x0 + w n z 2 - 1 x0
2w n z 2 - 1
Consider the punch fixture of Figure P1.82. If the system is giving an initial velocity of
10 m/s, what is the maximum displacement of the mass at the tip if the mass is 1000 kg
and the bar is made of steel of length 0.25 m with a cross sectional area of 0.01 m2?
Solution: First compute the frequency:
EA
wn =
=
lm
( 2.0 ´10 )( 0.01) (N/m )m
11
1000 ( 0.25)
2
kg × m
From equation (1.9) the maximum amplitude is
2
= 2.828 ´103 rad/s
Amax =
w 2n x02 + v02
wn
=
v0
10 m/s
=
= 0.0035 m ,
w n 2828 1/s
or about 0.35 mm, not much.
1.86
The frequency of oscillation of a person on a diving board can be modeled as the
transverse vibration of a beam as indicated in Figure 1.26. Let m be the mass of the diver
(m = 100 kg) and l = 1.5 m. If the diver wishes to oscillate at 3 Hz, what value of EI
should the diving board material have?
Solution: From equation (1.67),
w 2n =
3EI
ml 3
and
æ 2p rad ö
= 6p rad/s
è 1 cycle ø
w n = 3Hz ç
Solving for EI
w n2 ml 3 ( 6p ) (100 ) (1.5)
EI =
=
= 3.997 ´104 Nm 2
3
3
2
1.89
3
Calculate the natural frequency of the system in Figure 1.32(a) if k1 = k2 = 0. Choose m
and nonzero values of k3, k4, and k5 so that the natural frequency is 100 Hz.
Solution: Given: k1 = k2 = 0 and w n = 2p (100) = 628.3 rad/s
From Figure 1.29, the natural frequency is
wn =
k5 k3 + k5 k4 + k3 k4
m(k3 + k4 )
and
æ
kk ö
keq = ç k5 + 3 4 ÷
è
k3 + k4 ø
Equating the given value of frequency to the analytical value yields:
w 2n = (628.3) =
2
k5 k3 + k5 k 4 + k3 k4
m( k3 + k 4 )
Any values of k3, k4, k5, and m that satisfy the above equation will do. Again, the answer
is not unique. One solution is
k3 = 1 N/m, k 4 = 1 N/m, k5 = 50,000 N/m, and m = 0.127 kg