CHAPTER 8:VIBRATION AND TIME RESPONSE

SAMPLE PROBLEM 8/1,PAGE NUMBER:592

In [18]:
import math
# Given data 
W=25;# The weight of the body in lb
k=160;# lb/ft
v=2;# The downward velocity in  ft/sec
g=32.2;# The acceleration due to gravity in ft/sec^2

# Calculation
# (a)
delta_st=W*k**-1;# The static spring deflection in ft
delta_st=delta_st*12;# in
# (b)
omega_n=math.sqrt(k/(W/g));# The natural frequency of the system in rad/sec
f_n=omega_n*(1/(2*math.pi));# The natural frequency of the system in cycles/sec
# (c)
tau=1/f_n;# The system period in sec
print"\n(a)The static spring deflection,delta_st=%0.4f ft (or)%1.3f in \n(b)The natural frequency of the system,omega_n=%2.2f rad/sec \n   The natural frequency of the system,f_n=%0.3f sec \n(c)The system period,tau=%0.3f sec"%(delta_st/12,delta_st,omega_n,f_n,tau);
(a)The static spring deflection,delta_st=0.1562 ft (or)1.875 in 
(b)The natural frequency of the system,omega_n=14.36 rad/sec 
   The natural frequency of the system,f_n=2.285 sec 
(c)The system period,tau=0.438 sec

SAMPLE PROBLEM 8/2,PAGE NUMBER:592

In [19]:
import math

#Variable Declaration
m=8;# kg
s=0.2;# m
t_1=0;# s
t_2=2;# s
c=20;# N.s/m
k=32;# N/m

#Calculation
omega_n=math.sqrt(k/m);# rad/s
eta=c/(2*m*omega_n);# The damping ratio
omega_d=omega_n*math.sqrt(1-(eta**2));#The damped natural frequency in rad/s
x_2=0.256*math.exp(-1.25*t_2)*math.sin(((1.561*t_2)+0.896)*180/math.pi);# m
print"\nThe displacement in meters, x_2=%0.5f m"%x_2;
The displacement in meters, x_2=-0.01617 m

SAMPLE PROBLEM 8/4,PAGE NUMBER:608

In [20]:
import math

#Variable Declaration
m=50;#  kg
k=7500;# Number of springs
# x_B=0.002cos50t
omega=50;# rad/s
b=0.002;# m
k=7500;# The stiffness of the spring in N/m

#Calculation
omega_n=math.sqrt((4*k/m));# The resonant frequency in rad/s
X=b/(1-(omega/omega_n)**2);# m
print"\nThe amplitude of the steady-state motion of the instrument,X=%1.2e m (or) %0.3f mm"%(X,X*10**3);
The amplitude of the steady-state motion of the instrument,X=-6.32e-04 m (or) -0.632 mm

SAMPLE PROBLEM 8/6,PAGE NUMBER:608

In [21]:
import math

#Variable Declaration
m=100;# The mass of piston in lb
k=200;# The spring stiffness in lb/in
c=85;# The damping coefficient in lb-sec/ft
# A fluctuating pressure p=0.625sin30t in lb/in**2;
omega=30;# rad/s
a=80;# The top surface area of the piston in in**2
g=32.2;# The acceleration due to gravity in ft/sec**2

#Calculation
omega_n=math.sqrt((k*12)/(m/g));# The natural frequency of the system in rad/s
Zeta=c/(2*(m/g)*omega_n);# Damping ratio
#At steady state condition,
p=0.625;# lb/in**2
F_0=(p*a);# lb
X=(F_0/(k*12))/math.sqrt(((1-(omega/omega_n)**2)**2)+(((2*Zeta*omega)/omega_n)**2));# The steady-state amplitude in ft
phi=math.atan(((2*Zeta*(omega/omega_n))/(1-((omega/omega_n)**2))));#The phase angle in rad
F_trmax=X*math.sqrt(((k*12)**2)+((c**2)*(omega**2)));#The maximum force transmitted to the base in lb
print"\nThe steady-state displacement as a function of time,x_p=%0.5fsin(%2.0ft-(%1.3f))ft \nThe maximum force transmitted to the base,(F_tr)_max=%2.1f lb"%(X,omega,phi,F_trmax);
The steady-state displacement as a function of time,x_p=0.01938sin(30t-(-1.417))ft 
The maximum force transmitted to the base,(F_tr)_max=67.9 lb

SAMPLE PROBLEM 8/7,PAGE NUMBER:616

In [22]:
import math

#Variable Declaration
rbar=0.9;# Distance in m
k_o=0.95;# The radius of gyration in m
g=9.81; #The acceleration due to gravity in m/s**2

#Calculation
tau=2*math.pi*math.sqrt((k_o)**2/(g*rbar));# The period for small oscillations about the pivot in s
print"The period for small oscillations about the pivot is",round(tau,2),"s"
The period for small oscillations about the pivot is 2.01 s

SAMPLE PROBLEM 8/9,PAGE NUMBER:617

In [23]:
import math

#Variable Declaration
m=50;#The mass of cylinder in kg
r=0.5;#The radius of the cylinder in m
k=75;#The spring constant in N/m
c=10;#The damping coefficient in N.s/m
x=-0.2;# m
t=0;# s

#Calculation
#(a)
omega_n=math.sqrt((2*k)/(3*m));# The undamped natural frequency in rad/s
#(b)
Zeta=((c/(m*omega_n))/3);# The damping ratio
#(c)
omega_d=omega_n*math.sqrt(1-(Zeta)**2);# The damped natural frequency in rad/s
#(d)
tau_d=(2*math.pi)/omega_d;# The period of the damped system in s

# Result
print"The undamped natural frequency is",round(omega_n,0),"rad/s"
print"The damping ratio is",round(Zeta,4)
print"The damped natural frequency is",round(omega_d,3),"rad/s"
print"The period of the damped system is",round(tau_d,2),"s"
The undamped natural frequency is 1.0 rad/s
The damping ratio is 0.0667
The damped natural frequency is 0.998 rad/s
The period of the damped system is 6.3 s

SAMPLE PROBLEM 8/11,PAGE NUMBER:627

In [25]:
import math

#Variable Declaration
m_c=3;# Mass of the collar in kg
m_l=1.2;# Mass of the links in kg
k=1.5;# The stiffness of the spring in kN/m
g=9.81;# The acceleration due to gravity in m/s^2

#Calculation
P=(m_c*g)+(2*(1/2)*m_l*g);#The compression P in the spring in N
delta_st=P/(k*10**3);# The corresponding static deflection of the spring in m
# V_g=-5.89y J.....(1)
# V=750y**2  J.....(2)
# T=1.9y**2   .....(3)
# T_max=y_max with ydot=ydot_max.....(5)
omega_n=math.sqrt(750/1.9);
print"The natural frequency of vertical vibration,omega_n=",round(omega_n,2),"Hz"
The natural frequency of vertical vibration,omega_n= 19.87 Hz