Chapter4-Simple Harmonic motion

Ex1-pg40

##Solutions to Problems In applied mechanics
##A N Gobby
import math
##initialisation of variables
t=3.##sec
m=20.##per mint
a=4.##ft
##CALCULATIONS
T=2.*math.pi/t##ft/s
V=T*a##ft/s
F=(T)**2*a##ft/s^2
##RESULTS
print'%s %.2f %s'%('th acceleration x must be a maximum= ',F,' ft/s^2')

th acceleration x must be a maximum=  17.55  ft/s^2

Ex2-pg42

##Solutions to Problems In applied mechanics
##A N Gobby
import math
##initialisation of variables
a=10.##ft/s
x=1/12.##ft/s
g=32.2##ft
##CALCULATIONS
P=2.*math.pi*math.sqrt(x/a)##sec
L=(P)/(2.*math.pi/math.sqrt(g))/2.##ft
##RESULTS
print'%s %.2f %s'%('the simple pendulum = ',L,' ft')

the simple pendulum =  0.26  ft

Ex3-pg43

##Solutions to Problems In applied mechanics
##A N Gobby
import math
##initialisation of variables
w=20.##lbf
p=12.##ft/s
v1=15.##ft/s
g=32.2##ft
v2=10.##ft/s
d1=6.##in
d2=9.##in
a=10.82##in
##CALCULATIONS
Um=(v2*p)/math.sqrt(a**2-d2**2)##sec^-1
P=2.*math.pi/Um##sec
V=w*a##in/s
M=w**2*a/p##ft/s
F=(w/g)*M##lbf
##RESULTS
print'%s %.2f %s'%('the velocity= ',a,' in')
print'%s %.2f %s'%('periodic time= ',p,' sec')
print'%s %.2f %s'%('the maximum velocity= ',V,' in/s')
print'%s %.2f %s'%('maximum acceleration= ',F,' lbf')

the velocity=  10.82  in
periodic time=  12.00  sec
the maximum velocity=  216.40  in/s
maximum acceleration=  224.02  lbf

Ex4-pg44

##Solutions to Problems In applied mechanics
##A N Gobby
import math
##initialisation of variables
w=4.##lbf
h=40.##lbf/ft
d=2.##in
g=32.2##ft/s
##CALCULATIONS
P=(d*math.pi)*math.sqrt(w/(h*g))##sec
V=(d*math.pi*d)/(P*12.)##ft/s
M=(d*math.pi/P)**2*(d/12.)##ft/s
##RESULTS
print'%s %.2f %s'%('the period of vibration= ',P,' sec')
print'%s %.2f %s'%('Maximum veloity= ',V,' ft/s')
print'%s %.2f %s'%('Maximum acceleration= ',M,' ft/s')

the period of vibration=  0.35  sec
Maximum veloity=  2.99  ft/s
Maximum acceleration=  53.67  ft/s

Ex5-pg45

##Solutions to Problems In applied mechanics
##A N Gobby
import math
##initialisation of variables
w=80.##lbf
p=4.##ft
d=20.##stroke/min
d1=3.##in
p1=0.6##sec
h=2.##ft/s
g=32.2##ft/s
t=60.##sec
##CALCULATIONS
P=t/d##sec
U=2.*math.pi/d1##sec^-1
V=U*math.sqrt(h**2-(3./4.)**2)##ft/s
K=(w*V**2/(h*g))##lbf
M=U**2*h##ft/s^2
M1=(w/g)*M##lbf
D=h*math.cos(U*0.6*180./math.pi)##ft
D1=h-D-2##ft
##RESULTS
print'%s %.2f %s'%('the kinetic energy of the crosshead= ',K,' lbf')
print'%s %.2f %s'%('the maximum acceleration of force on crosshead= ',M1,' lbf')
print'%s %.2f %s'%('the distance from end of the path= ',D1,' ft')

the kinetic energy of the crosshead=  18.73  lbf
the maximum acceleration of force on crosshead=  21.80  lbf
the distance from end of the path=  1.93  ft

Ex6-pg46

##Solutions to Problems In applied mechanics
##A N Gobby
import math
##initialisation of variables
d=3.##in
v=40.##ft/s
a=3000.##ft/s^2
p=5.31##in
##CALCULATIONS
U=math.sqrt(a/(d/12.))##sec^-1
E=(U*60./(2.*math.pi))##rev/min
P=2./U##sec
W=U*(p/12.)##ft/s
M=U**2*(p/12.)##ft/s^2
##RESULTS
print'%s %.2f %s'%('the velocity of acceleration against time during one complete= ',M,' ft/s^2')

the velocity of acceleration against time during one complete=  5310.00  ft/s^2