1261
*22–72.
SOLUTION
Since
the system is underdamped.
F
rom Eq. 22–32
Appling the initial condition at
and .
(1)
(2)
Solving Eqs
. (1) and (2) yields:
Ans.y
=
0.622[e–0.939tsin (11.9t
+
1.49)]
f=85.5° =1.49 rad
D=0.622 ft
11.888 cos f–0.9392 sin u=0
0=De–0
C
11.888 cos (0 +f)–0.9392 sin (0 +f)
D
since DZ0
Dsin f=0.62
0.62 =D
C
e–0sin (0 +f)
D
y=0t=0, y=0.62 ft
y=De–
A
c
2m
B
tcvdcos (vdt+f)–c
2msin (vdt+f)d
y=y
#=Dce–
A
c
2m
B
tvdcos (vdt+f)+a–c
2mbe–
A
c
2m
B
tsin (vdt+f)d
y=Dce–
A
c
2m
B
tsin (vdt+f)d
c
2m=0.7
2(0.3727) =0.9392
vd=vnB1–ac
ccb2
=11.925B1–a0.7
8.889 b2
=11.888 rad>s
c6cc
cc=2mvn=2(0.3727)(11.925) =8.889 lb #s>ft
vn=Ak
m=A53
0.3727 =11.925 rad>s
c=0.7 lb #s>ft
k=53 lb>ft
m=12
32.2 =0.3727 slug
The block, having a weight of 12 lb, is immersed in a liquid
such that the damping force acting on the block has a
magnitude of where is in . If the block
is pulled down 0.62 ft and released from rest, determine the
position of the block as a function of time.The spring has a
stiffness of Assume that positive displacement
is downward.
k=53 lb>ft.
ft>svF =10.7 ƒvƒ2lb,
k
22–73.
The bar has a weight of 6 lb.If the stiffness of the spring is
and the dashpot has a damping coefficient
determine the differential equation which
describes the motion in terms of the angle of the bar’s
rotation. Also,what should be the damping coefficient of the
dashpot if the bar is to be critically damped?
u
c=60 lb #s>ft,
k=8lb>ft
SOLUTION
a
[1]
From equilibrium . Also, for small , and hence
.
From Eq. [1]
Ans.
By comparing the above differential equation to Eq. 22-27
Ans.
A
cd#p
B
c=2
9
2km =2
9
2200(1.55) =3.92 lb #s>ft
£
9
A
cd#p
B
c
2m≥2
–k
m=0
m=1.55 k=200 vn=A200
1.55 =11.35 rad>sc=9cd#p
1.55u
#+540u
#+200u=0
1.5528u
$+180(3u
#)+40(5u)=0
y
#
2=3u
y2=3uy1=5uu40yst –15 =0
1.5528u
$+180y
#
2+40y1+40yst –15 =0
+©MA=IAa; 6(2.5) –(60y
#
2)(3) –8(y1+yst)(5) =c1
3a6
32.2 b(5)2du
$
CA
k
B
c
2ft3ft
1263
22–74.
k
c
v0
k
A bullet of mass mhas a velocity of just before it strikes
the target of mass M. If the bullet embeds in the target, and
the vibration is to be critically damped, determine the
dashpot’s critical damping coefficient, and the springs’
maximum compression.The target is free to move along the
two horizontal guides that are “nested” in the springs.
v0
SOLUTION
.Thus, the natural frequency of the system is
When the system is critically damped
Ans.
The equation that describes the critically dampened system is
When ,.Thus,
Then,
(1)
Taking the time derivative,
(2)
Since linear momentum is conserved along the horizontal during the impact, then
Here, when ,.Thus, Eq. (2) gives
And Eqs. (1) and (2) become
(3)
(4) v=ca m
m+Mbv0de–vn t(1 – vn t)
x=ca m
m+Mbv0dte–vn t
B=am
m+Mbv0
v=am
m+Mbv0
t=0
v=am
m+Mbv0
mv0=(m+M)v
A
;
+
B
v=Be–vn t(1 – vn t)
v=x
#=Be–vnt – Bvn te–vn t
x=Bte–vn t
A=0
x=0t=0
x=(A+Bt)e–vn t
c=cc=2mTvn=2(m+M)B2k
m+M=28 (m+M)k
vn=Bkeq
mT
=B2k
m+M
mT=m+M
1264
Since , then
Substituting this result into Eq. (3)
Ans. =cm
eA1
2k(m
+
M)dv0
xmax=ca m
m+Mbv0daAm+M
2kbe–1
t=1
vn
=Bm+M
2k
1 – vn t=0
am
m+Mbv0Z0
e
A
2k(m
+
M)
1265
22–75.
get of mass M.If the bullet embeds in the target, and the
shpot’s damping coefficient is , determine
springs’ maximum compression. The target is free to
ve along the two horizontal guides that are“nested” in
springs.
06cVcc
0
ince the springs are arranged in parallel, the equivalent stiffness of the single spring
ystem is .Also, when the bullet becomes embedded in the target,
.Thus, the natural circular frequency of the system
he equation that describes the underdamped system is
(1)
hen .Thus, Eq. (1) gives
ince .Then .Thus, Eq. (1) becomes
(2)
aking the time derivative of Eq. (2),
(3)
ince linear momentum is conserved along the horizontal during the impact, then
hen ,.Thus, Eq. (3) gives
s. (2) becomes
(4)x=ca m
m
Mbv0
v
de–(c>2mT) t sin vdt
C=am
m+Mbv0
vd
am
m+Mbv0=Cvd
v=am
m+Mbv0
t=0
v=am
m+Mbv0
mv0=(m+M)v
A
;
+
B
v=Ce–(c>2mT)t
B
vd cos vdt–c
2mT
sin vdt
R
v=x
#=C
B
vde–(c>2mT)t cos vdt–c
2mT
e–(c>2mT)t sin vdt
R
x=Ce
–(c>2mT)t sin vd t
f=0C Z 0, sin f=0
0=C sin f
t=0, x=0
x=Ce
–(c>2mT)t sin (vdt+f)
vn=
C
keq
mT
=B2k
m+M
T=m+M
keq =2k
k
c
v0
k
28k(m+M)–c2 e
1267
*22–76.
Determine the differential equation of motion for the
damped vibratory system shown. What type of motion
occurs? Take ,,.m=25 kgc=200 N #s>mk=100 N>m
SOLUTION
(1)
Here,, and . Substituting these values into
Eq. (1) yields
Ans.
Comparing the above differential equation with Eq. 22–27, we have ,
and .Thus,.
=Ak
m=A12
1=3.464 rad>sk=12 N>mc=16 N #s>m
m=1kg
y
$+16y
#+12y=0
25y
$+400y
#+300y=0
k=100 N>mc=200 N #s>mm=25 kg
my
$+2cy
#+3ky =0
+c©Fx=0; 3ky +mg +2cy
kk k
m
n
v
1268
22–77.
SOLUTION
Using Table 22–1,
Ans.Lq +Rq +(1
C)q=E0cos vt
Draw the electrical circuit that is equivalent to the
mechanical system shown. Determine the differential
equation which describes the charge qin the circuit.
k
m
FF0cos vt
c
Cb
1269
22–78.
SOLUTION
F
or the block,
Using
Table 22–1,
Ans.Lq
$+Rq
#+(2
C)q=0
mx
$+cx#+2k=0
Draw the electrical circuit that is equivalent to the
mechanical system shown. What is the differential equation
which describes the charge qin the circuit?
kk
c
m
Cb
22–79.
Draw the electrical circuit that is equivalent to the
mechanical system shown. Determine the differential
equation which describes the charge qin the circuit.
SOLUTION
k
m