Frequency Response

Frequency response
Question 2
a. ,
m ̈x + kx = 10 cos 10 t .................. ..................................Equation 1
1000 ̈x + 2000x = 10 cos 10 t
The complimentary solution is shown below for displacement
u ( t )=c1 sos ( ω0 t ) + c2 sin ( ω0 t ) ........................ ............................Equation 2
ω0 is the natural frequency. it shall be assumed that ω ≠ ω0
The particular solution will be given as shown below
up ( t )= A cos ( ωt ) + B sin ( ωt ) .................. ..............................Equation 3
Differentiating equation 3 we have
f 0 cos ( ωt ) =¿ ( cos ( ωt ) )∗( −mω2 A+ Ak ) + ( sin ( ωt ) ) ∗( −m ω2 B+ Bk ) ... . ¿..Equation 4
Grouping the like terms and equating the coefficients
( cos ( ωt ) )∗(−mω2 A+ Ak ) ¿ f 0 cos ( ωt )
−m ω2 A + Ak= f 0
A= f 0
−mω2 +k
(−mω2 B+Bk ) + ( sin ( ωt ) )= 0
B=0
Particular equation now becomes
up ( t )= f 0
k−mω2 cos ( wt )
Now the whole solution can be written as shown below

u(t )=Rcos(ω0 t−δ )+ F 0
m ( ω0
2 −ω2 ) cos(ωt)
ω0= √ k
m
= √ ( 2000
100 )
= 4.4721
u(t )=c 1cos (10 t)+c 2sin (5 t)+ 10
2 ( 10 ) ( 10 ) tsin( 10t )
After feeding the initial conditions
u(t )=0.1 cos( 10t )+ 0.01sin (5 t)+ 10
2 ( 10 ) ( 10 ) tsin(10 t )
Now evaluating the value of R
R= √ ( ( 0.1 ) 2 + ( 0.01 ) 2 )
= 0.0101
δ=tan−1
( 0.01
0.1 )
= 5.71
The displacement now becomes
u(t )=0.0101 cos(10 t +5.71)+ ( 1
20 ) tsin(10 t)
b. ,
c. The displacement is
d. u(t )=0.0101 cos(10 t +5.71)+ ( 1
20 ) tsin(10 t) + 0.05