# 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