differential equations - 3.56

3.9 PRACTICE PROBLEM SOLUTIONS

1.

			B
x( t)	=	–-----------FMe–M----t		–	F--t + FM--------
		B	2		B	B	2

2.

homogeneous:

guess

yh = e

At

·

At

··

2

e

At

yh = Ae

yh = A

A2eAt + ( 100s–2) eAt

= 0

A2

= –100s–2

A

= ± 10js–1

yh

= C1 cos ( 10t + C2)

particular:

guess

yp

= A

·

= 0

··

= 0

yp

yp

( 0) + ( 100s–2) A = 9.81ms–2

( 100s–2) A = 9.81ms–2

9.81ms–2

= 0.0981m

A = ----------------------

100s–2

yp = 0.0981m

initial conditions:

y =

yh + yp = C1 cos ( 10t + C2) + 0.0981m

y' =

–10C1 sin ( 10t + C2)

for d/dt y0 = 0m:

0 = –10C1 sin ( 10( 0) + C2)

C2 = 0

for y0 = 0m:

0 = C1 cos ( 10( 0)

+ ( 0) ) + 0.0981m

–0.0981m = C1 cos ( 0)

C1

= –0.0981m

y( t) = ( –0.0981m) cos ( 10t) + 0.0981m

differential equations - 3.57

3.

case 1:	x( t)	=	– 0.0115e–5t cos ( 8.66t – 0.524) + 0.010
case 2:	x( t)	=	– 0.010e–10t – 0.10te–10t + 0.010
case 3:	x( t)	= 775 10–6e–37.32t – 0.0108e–2.679t + 0.010

4.

Vo( t)		=	–8.465e–0.6t sin ( 1.960t + 1.274) + 8.095
or	( t)	=	–8.465e–0.6t cos ( 1.960t – 0.2971) + 8.095
Vo

5.

V0( t) = –8.331e–0.6t cos ( 1.96t – 0.238) + 8.095

6.

a)	b)
x1( t)	= e–t – 1		–0.5t	cos ( 0.866t – 0.524) + 10.81
	x( t) = –12.485e
x2( t) = 2e–t – 2
τ = 1	ζ = 0.5	ω n	= 1

7.

case 1:	x( t)	=	–0.00117e–5t sin ( 8.66t – 1.061) + 0.0101 sin ( t – 0.101)
case 2:	x( t)	= ( 1.96		10–3) e–10t + ( 9.9 10–3) te–10t + ( 9.9 10–3) sin ( t + 0.20)
case 3:	x( t)	=	( 3.5	10–3) e–2.679t – ( 18 10–6) e–37.32t + ( 9.4 10–3) sin ( t + 0.382)

8.24.37 rad/sec

9.fn=3.77Hz, damp.=.575

differential equations - 3.58

10.

·· · ( ) x + 8.048x + 77.88x = F t

11.

Given

K

s

K

=

36

N

M =

28N

= 2.85kg

d

= 6N---

s

---

----------------

m

m

N

9.81-----

kg

The typical transfer function for a mass-spring-damper systems is,

·· ·

Kd

+ x

Ks

=

F

x + x

-----

-----

----

M

M

M

The second order parameters can be calculated from this.

··

·

( 2ζω

n)

+ x( ω

2

x

+ x

n) = y( t)

N

kgm

36----------

Ks

---

ms2

rad

ω

n

=

=

36m

12.63s

–2

= 3.55

= 0.6Hz

-----

---------------- =

---------------- =

--------

M

2.85kg

2.85kg

s

Kd

s

-----

6N---

ξ

M

m

=

-----------

=

--------------------------------------------

= 0.296

2ω

n

rad

--------

2.85kg

2( 3.55) s

ω

d

= ω n

1 – ξ

2

= 3.39

rad

--------

s

If pulled and released the system would have a decaying oscillation about 0.52Hz

A critically damped system would require a damping coefficient of....

Kd

-----

Kd

Ns

M

K

= 20.2

ξ

=

-----------

=

--------------------------------------------

= 1.00

d

-----

2ω

n

2( 3.55)

rad

m

--------

2.85kg

s

12.

θ ( t) = – 66 10–6e–39.50t – 3.216e0.1383t + 1.216e–0.3368t + 2.00

differential equations - 3.59

13.

y( t)	= 0.05e	–2.6t	cos ( 12.74t)	= 0.05e	–2.6t	sin		π
								12.74t + --
								2

14.

x
4
					t(s)
0	τ1	= 1	2	3	4

Given the equation form,

·	1
x + --τ		x = A

The values at steady state will be

·	= 0	x = 4
x

So the unknown ‘A’ can be calculated.

		1		A = 4
	0 +	1--	4 = A

·	1
x +	1--x = 4

·

x + x = 4

15.

Key points:

First-order:

find initial final values

find time constant with 63% or by slope use these in standard equation

Second-order:

find damped frequency from graph find time to first peak

use these in cosine equation

differential equations - 3.60

16.

For the first peak:

b

=

–σ

tp

------

e

∆ x

π

2

ω

d ≈

=

e

–σ

0.5

---

-----

tp

10

ln

2

= –σ

0.5

-----

10

2

σ

= –2 ln

-----

= 3.219

1

10

ξ

For the damped frequency:

=

-----------------------------

π

2

2π

-------

+ 1

ω

d

=

=

2π

tpσ

-----

1s

These values can be used to find the damping coefficient and natural frequency

σ	= ξω				ω			3.219
				n		n = ------------
									ξ
ω	d	= ω		n 1 – ξ		2
2π		=	3.219		1 – ξ			2
			------------
				ξ	1 – ξ 2
	2π			2
	3.219------------			=	-------------ξ	2
	2π			2	=	1			ξ = ------------------------------	1	= 0.4560
	3.219------------			+ 1		----				2π	2
						ξ	2				+ 1
									------------3.219
ω	n	=	3.219		=	3.219			= 7.059
			------------	ξ		---------------
						0.4560

This leads to the final equation using the steady state value of 10

··	·	2
x + 2	ξω nx + ω	nx = F
··		·	( 7.059)	2	x = F
x + 2	( 0.4560) ( 7.059) x +
··	·
x + 6.438x + 49.83x = F
( 0) + 6.438( 0)		+ 49.83( 10) = F			F = 498.3

···

x+ 6.438x + 49.83x = 498.3