Forced Oscillations and Resonance

## **Problem**

   A 1 kg object is attached to spring that would stretch the spring 1 m by a force of 1 N. There is no damping in the system and a forcing function of the form

F
(
t
# )
2
cos
⁡
(
t
)

{\displaystyle F(t)=2\cos(t)}

is attached to the object and the system will experience resonance. If the object is initially at the equilibrium position, and at rest, determine the displacement at any time

t
≥
0

{\displaystyle t \geq 0}

.

## **Solution**

The first thing we need to do is find the spring constant

k

{\displaystyle k}

.

F
s

=
k
x

{\displaystyle F\_s = kx}

# k

1

N

1

m

=
1

N
m

{\displaystyle k={\frac {1\,N}{1\,m}}=1\,{\frac {N}{m}}}

With no damping, the differential equation to solve for this problem follows the profile

y
″

k
m

# y
F
(
t
)

,

y
(
0
# )

y

0

,

y
′

(
0
# )

v

0

.

{\displaystyle y''+{\frac {k}{m}}y=F(t)\,,\,y(0)=y\_{0}\,,\,y'(0)=v\_{0}.}

Therefore,

y
″

# y
2
cos
⁡
(
t
)

,

y
(
0
# )
0

,

y
′

(
0
# )

{\displaystyle y''+y=2\cos(t)\,,\,y(0)=0\,,\,y'(0)=0.}

**Solving the complementary solution**

Solve the second-order differential equation

y
″

# y
0

{\displaystyle y''+y=0}

. The characteristic equation is

r

2

# 1
0

{\displaystyle r^{2}+1=0}

; thus, the roots are pure-imaginary:

# r
±
i

{\displaystyle r=\pm i}

.

Therefore, the complementary solution is

y

c

(
t
# )

C

1

cos
⁡
(
t
)

C

2

sin
⁡
(
t
)
.

{\displaystyle y\_{c}(t)=C\_{1}\cos(t)+C\_{2}\sin(t).}

**Solving the particular solution**

Let's solve the particular solution using undetermined coefficients. For this type of differential equation, the ansatz (educated guess) is

y

p

(
t
# )
t
[
A
sin
⁡
(
t
)
B
cos
⁡
(
t
)
]
.

{\displaystyle y\_{p}(t)=t[A\sin(t)+B\cos(t)].}

Taking the two derivatives yield

y

p

′

(
t
# )
A
sin
⁡
(
t
)
B
cos
⁡
(
t
)
t
[
A
cos
⁡
(
t
)
−
B
sin
⁡
(
t
)
]

{\displaystyle y\_{p}'(t)=A\sin(t)+B\cos(t)+t[A\cos(t)-B\sin(t)]}

y

p

″

(
t
# )
(
2
A
−
B
t
)
cos
⁡
(
t
)
−
(
2
B
A
t
)
sin
⁡
(
t
)
.

{\displaystyle y\_{p}''(t)=(2A-Bt)\cos(t)-(2B+At)\sin(t).}

Now substituting the derivatives into the original differential equation yields

2
A
cos
⁡
(
t
)
−
2
B
sin
⁡
(
t
# )
2
cos
⁡
(
t
)
.

{\displaystyle 2A\cos(t)-2B\sin(t)=2\cos(t).}

Therefore,

# A
1

{\displaystyle A=1}

 and

# B
0

{\displaystyle B = 0}

, which means

y

p

(
t
# )
t
sin
⁡
(
t
)
.

{\displaystyle y\_{p}(t)=t\sin(t).}

**Final solution**

The complete solution is

y
(
t
# )

y

c

(
t
)

y

p

(
t
)

{\displaystyle y(t)=y\_{c}(t)+y\_{p}(t)}

:

y
(
t
# )

C

1

cos
⁡
(
t
)

C

2

sin
⁡
(
t
)
t
sin
⁡
(
t
)
.

{\displaystyle y(t)=C\_{1}\cos(t)+C\_{2}\sin(t)+t\sin(t).}

Since the initial position is at equilibrium

y
(
0
# )

C

1

cos
⁡
(
0
)

C

2

sin
⁡
(
0
)
(
0
)
sin
⁡
(
0
# )

{\displaystyle y(0)=C\_{1}\cos(0)+C\_{2}\sin(0)+(0)\sin(0)=0.}

This means

C
1

=
0

{\displaystyle C\_1=0}

.

The derivative of the complete solution is

y
′

(
t
# )
−

C

1

sin
⁡
(
t
)

C

2

cos
⁡
(
t
)
sin
⁡
(
t
)
t
cos
⁡
(
t
)
.

{\displaystyle y'(t)=-C\_{1}\sin(t)+C\_{2}\cos(t)+\sin(t)+t\cos(t).}

Since the initial velocity is at rest

y
′

(
t
# )
−

C

1

sin
⁡
(
0
)

C

2

cos
⁡
(
0
)
sin
⁡
(
0
)
(
0
)
cos
⁡
(
0
# )

{\displaystyle y'(t)=-C\_{1}\sin(0)+C\_{2}\cos(0)+\sin(0)+(0)\cos(0)=0.}

This means

C

2

=
0

{\displaystyle C\_{2}=0}

.

Consequently the solution of this scenario is just

y
(
t
# )
t
sin
⁡
(
t
)
.

{\displaystyle y(t)=t\sin(t).}