Chapter 6 Oscillatory motion

Example 6.1 Page no 218

In [6]:
#Given
A =0.05                         #Amplitude of SHM in m
T =6.0                           # period in sec
Xo=A
X =0.025

#Calculation
import math
w =2*3.14/T
ph= math.asin (Xo/A)
t=(math.asin (X/Xo)-ph)/w 
t1= abs (t)

#Result
print"Timeto move from equilibrium position is",round(t1,1),"sec"

Timeto move from equilibrium position is 1.0 sec

Example 6.2 Page no 219

In [8]:
#Given
A =0.06                     #Amplitude of SHM in m
T =31.4

#Calculation
w =2*3.14/T
V=A*w

#Result
print"Maximum velocity is",V,"m/s"

Maximum velocity is 0.012 m/s

Example 6.3 Page no 219

In [20]:
#Given
A=0.05                   #m
T=2                      #S

#Calculation
import math
x=A*math.sin(math.pi/3.0)
v=A*math.pi*math.sqrt(1-(x**2/A**2))

#Result
print"Displacement is",round(x*10**2,1),"cm"
print"Velocity is", round(v*10**2,2),"cm/s"

Displacement is 4.3 cm
Velocity is 7.85 cm/s

Example 6.4 Page no 219

In [11]:
#Given
l =1                         # length of pendulum in m
m =1                       #mass of pendulum in kg
g =9.8

#Calculation
T =2*3.14* math.sqrt (l/g)

#Result
print"Period of oscillation is",round(T,1),"sec"

Period of oscillation is 2.0 sec

Example 6.5 Page no 220

In [26]:
#Given
m1 =8                           #mass suspended in kg
l =0.32
m =0.50
g =9.8

#Calculation
import math
k=m1*g/l
T=2*math.pi* math.sqrt (m/k)

#Result
print"K=",k,"N m"
print"Time period of the oscillation is",round(T,2),"sec"

K= 245.0 N m
Time period of the oscillation is 0.28 sec

Example 6.6 Page no 220

In [31]:
#Given
K=98.0                        #N/m
x=20                         #cm
g=9.8                       #m/sec**2

#Calculation
F=K*x/100.0
m=K*x/(100*g)

#Result
print"Restoring force is",F,"N"
print"Mass is",m,"Kg"

Restoring force is 19.6 N
Mass is 2.0 Kg

Example 6.7 Page no 220

In [35]:
#Given
T=57
l=79

#Calculation
import math
g=4*math.pi**2*l*T**2/(100*100**2)

#Result
print"Value of g is", round(g,2),"m/s**2"

Value of g is 10.13 m/s**2

Example 6.8 Page no 221

In [39]:
#Given
m=4
T=55

#Calculation
import math
K=(2*math.pi*math.sqrt(m)*55/100.0)**2

#Result
print"Stiffness factor is", round(K,2),"N/m"

Stiffness factor is 47.77 N/m

Example 6.9 Page no 221

In [45]:
#Given
K=89.2               # N/m
T=1                     #S

#Calculation
import math
M=(T*math.sqrt(K)/(2*math.pi))**2

#Result
print"Mass is", round(M,2),"Kg"

Mass is 2.26 Kg

Example 6.10 Page no 221

In [49]:
#Given
m=4                         #Kg
g=9.8                       #m/sec**2
x=16.0
m1=0.5

#Calculation
import math
K=m*g*100/x
T=2*math.pi*math.sqrt(m1/K)

#Result
print"Time period is", round(T,2),"Sec"

Time period is 0.28 Sec

Example 6.11 Page no 222

In [55]:
#Given
T=50.0                               #S

#Calculation
t=1/(2*T)
t1=1/t
t2=t1/2.0

#Result
print"Time is", t2,"Second"

Time is 50.0 Second

Example 6.12 Page no 222

In [61]:
#Given
a=200.0
b=2.3

#Calculation
y=b/a
D=1/y

#Result
print"(i) Damping constant is", y
print"(ii) Decay modulus is",round(D,1),"Second"

(i) Damping constant is 0.0115
(ii) Decay modulus is 87.0 Second

Example 6.13 Page no 223

In [64]:
#Given
a=20                     #cm
b=2.0
x=100.0
y=2.3

#Calculation
import math
w=a/b
w1=y/x

#Result
print"Logarithmic decrement of the system is", w1

Logarithmic decrement of the system is 0.023

Example 6.14 Page no 223

In [70]:
#Given
n=2
m=0.3
Q=60.0

#Calculation
import math
K=16*math.pi**2*m
p=2*math.pi*n*m/Q

#Result
print"Force constant is",round(K,2),"N/m"
print"Mechanical resistance is",round(p,5),"Kg/m"

Force constant is 47.37 N/m
Mechanical resistance is 0.06283 Kg/m

Example 6.15 Page no 223

In [77]:
#Given
Q=10**4
b=500

#Calculation
import math
a=math.log(10)*Q/(b*math.pi)

#Result
print"Time interval is", round(a,2),"Seconds"

Time interval is 14.66 Seconds

Example 6.16 Page no 224

In [6]:
#Given
Q=2*10**3
v =240

#Calculation
import math
w =2*3.14*v
r= math.exp (2)
T=Q/w
t =2* T*log(r)

#Result
print"Time to become for new amplitude is",round(t,1),"sec"

Time to become for new amplitude is 5.3 sec

Example 6.17 Page no 224

In [87]:
#Given
A0=10.0
T=200

#Calculation
import math
y=math.log(10)/A0
t=1/(2*y)
Q=2*math.pi*t*T
t1=t*math.log(A0)

#Result
print"(i) Relaxation time is", round(t,3),"Second"
print"(ii) Quality factor is", round(Q,0)
print"(iii) Time is",t1,"Second"

(i) Relaxation time is 2.171 Second
(ii) Quality factor is 2729.0
(iii) Time is 5.0 Second

Example 6.18 Page no 225

In [93]:
#Given
E=10**3
w=256

#Calculation
import math
t=E/(2*math.pi*w)
n=E/(2*math.pi)

#Result
print"Time is", round(t,2),"Sec"
print"Number of oscillations is",round(n,0)

Time is 0.62 Sec
Number of oscillations is 159.0

Example 6.20 Page no 226

In [20]:
#Given
Ao =0.1                                 #amplitude at minimum frequency in mm
A =100                             #maximum amplitude
w =100

#Calculation
Q=A/Ao
T=Q/w
hw =1/(2* T)

#Result
print"(a) Quality factor is",Q
print"(b) Energy decay time is",T,"sec"
print"(c) Half width of power resonance curve is",hw,"rad/sec"

(a) Quality factor is 1000.0
(b) Energy decay time is 10.0 sec
(c) Half width of power resonance curve is 0.05 rad/sec

Example 6.21 Page no 226

In [110]:
#Given
m =0.1                          #suspended mass in kg
k =100                        #force constant in N/m
Fo =2
p =1
W =50
a=1000
a1=2500

#Calculation
import math
Wo= math.sqrt (k/m)
f=Fo/m
d=p /(2* m)
B=f/ (math.sqrt((a-a1)**2+(k*W**2)))
delta = math.atan(2*d*W/( Wo**2-W**2) )*180/3.14+180

#Result
print"Amplitude of oscillation is",round(B,4),"m"
print"Relative phase is",round(delta,1),"degree"

Amplitude of oscillation is 0.0126 m
Relative phase is 161.6 degree

Example 6.22 Page no 227

In [113]:
#Given
Q=50
a=1.4

#Calculation
B=1/a

#Result
print"Value of B/Bmax is",round(B,2)

Value of B/Bmax is 0.71