input output equations - 6.20

6.5SUMMARY

•The differential operator can be manipulated algebraically

•Equations can be manipulated into input-output forms and solved as normal differential equations

6.6PRACTICE PROBLEMS

1.Develop the input-output equation for the mechanical system below. There is viscous damping between the block and the ground. A force is applied to cause the mass the accelerate.

x

F

M

B

2. Find the input-output form for the following equations.

·	=	3
y + y + x
·	=	0
x + x + y

3. Find the input-output form for the following equations.

··	·	+ 2x	·	– x2	= 0
x1	+ x1		1 – x2

··· ·

–x1 – x1 + x2 + x2 + x2 = F

4.The following differential equations were converted to the matrix form shown. Use Cramer’s rule to find an input-output equation for ‘y’.

··

F

y + 2x =

-----

2

( 2)

y

F

10

( D )

=

-----

·

··

( 7D + 4) ( 9D

2

+ 3)

x

10

7y + 4y + 9x + 3x = 0

0

input output equations - 6.21

5. Find the input output equation for y2. Ignore the effects of gravity.

Ks1

M1 y1

Ks2

M2 y2

F

6.Find the input-output equations for the systems below. Here the input is the torque on the left hand side.

θ 1

θ 2

τ	Ks1	J1	Ks2
		B1

J2

B2

7.Write the input-output equations for the mechanical system below. The input is force ‘F’, and the output is ‘y’ or the angle theta (give both equations). Include the inertia of both masses, and

input output equations - 6.22

gravity for mass ‘M’.

K	θ

R JM

K						y
M
F

8. The applied force ‘F’ is the input to the system, and the output is the displacement ‘x’.

K1 = 500 N/m

								K2 = 1000 N/m
		x
					M = 10kg
w
					F

a)Find x(t), given F(t) = 10N for t >= 0 seconds.

b)Using numerical methods, find the steady-state response for an applied force of F(t) = 10cos(t + 1) N ?

a)Solve the differential equation to find the explicit response for an applied force of F(t) = 10cos(t + 1) N ?

d)Set the acceleration to zero and find an approximate solution for an applied force of F(t) = 10cos(t + 1) N. Compare the solution to the previous solutions.

6.7PRACTICE PROBLEM SOLUTIONS

1.

input output equations - 6.23

·· ·	B		=	F
x + x	----			----
	M			M

2.

···

x+ 2x = –3

···

y+ 2y = 3

3.

d

4

+ 2

d

3

+ 3

d 2

x

+

d

x

+ x1

=

d

F + F

----

x

----

x

----

----

----

dt

1

dt

1

dt

1

dt

1

dt

d

4

+ 2

d

3

+ 3

d 2

x

+

d

x

+ x2

=

d

2

d

F + 2F

----

x

----

x

----

----

----

F +

----

dt

2

dt

2

dt

2

dt

2

dt

dt

4.

					F				( 2)
				-----
				10																			F		2
										2
y =				0			( 9D				+ 3)										=		10( 9D + 3)								=			F( 0.9D2 + 0.3)
																																	---------------------------------------------------
				2							( 2)										D2	( 9D2 + 3) – 2( 7D + 4)											9D4 + 3D2 – 14D – 8
			( D )
		( 7D + 4) ( 9D2 + 3)
y( 9D4 + 3D2 – 14D – 8)																= F( 0.9D2 + 0.3)
d 4	y( 9)			+		d			2		( 3)	+			d						1					=	d			2			+ F( 0.3)
----							----			y							----				y( –14) + y( –8)							----			F( 0.9)
dt						dt									dt												dt
d 4	y +		d				2	1			+	d					1			–14		+ y	–8	=		d		2	F		1	+ F		1
----				----			y		--			----						y		--------9			-----				----				-----			--
dt			dt					3				dt											9			dt					10			3

5.

d

4

+

d

2

y

Ks1M2 + Ks2M2 + Ks2M1

+ y

Ks1Ks2

=

----

y

2

----

2

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

2

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

dt

dt

M1M2

M1M2

d 2

F

M 1

+ F

Ks1 + Ks2

----

M--------------1M2

----------------------M 1 M 2

dt

input output equations - 6.24

6.

d		4	θ 2
----
dt

d		4	θ 1
----
dt

d 3

J1B2

+ J2B1

d

2

J1Ks2

+ J2Ks2

+ B1B2

+

----

θ

2

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

+

----

θ

2

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

+

J1J2

dt

J1J2

dt

d

B1Ks2

+ B2Ks2

= τ

Ks2

----

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

---------

dt θ 2

J

1

J

2

J

1

J

2

d 3

J1B2

+ J2B1

d

2

J1Ks2

+ J2Ks2

+ B1B2

+

----

θ

1

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

+

----

θ

1

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

+

J1J2

dt

J1J2

dt

d

B1Ks2 + B2Ks2

d 2

1

d

B2

Ks2

----

θ 1

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

=

----

τ

----

----

τ

---------

+ τ

---------

dt

J1J2

+

dt

J1J2

J1J2

dt

J1

input output equations - 6.25

7.

θ

∑M = – R( K( θ R) )

··

+ R( K( – y – θ R) ) = JMθ

JM

R2Kθ

+ RKy + R2Kθ = –JMθ··

R( K( θ R) )

R( K( – y –

θ R) )

θ

2R2K + JMD2

= y

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

–RK

K( – y – θ

R)

y

∑F = K( – y – θ R)

··

– F – Mg = My

M

K( y + θ R) + F + Mg = –MyD2

θ ( KR) + F + Mg = y( – MD2 – K)

F+Mg

θ ( KR) + y( MD2 + K)

= – F – Mg

for the theta output equation;

θ ( KR) +θ

2R2K + JMD2

( MD

2

+ K) = – F – Mg

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

–RK

θ ( –K2R2) +θ

( 2R2K + JMD2) ( MD2 + K)

= FKR + MgKR

θ ( –K2R2 + 2R2MKD2 + 2R2K2 + JMMD4 + JMKD2)

= FKR + MgKR

d

4

θ

JMM

+

d

4

θ

(

2

M + JM)

+ θ

(

K

2

2

=

FKR + MgKR

----

-----------

----

2R

R )

dt

KR

dt

3

d

4

d

4

2R

3

K

K

3

R

2

R

2

2

R

2

θ

θ

+

KR

= F

K

+

MgK

+

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

+ θ

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

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

----

----

-------

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

dt

dt

JM

M

JMM

JMM

JMM

for the y output equation;

–RK

2

y

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

( KR)

+ y( MD + K) = – F – Mg

2R2K + J

M

D2

JMD4M

JMD2

2

MD

2

+

+ K2R

2

+

2

=

y 2R

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

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

– R

K

K

K

JMD2

JMD2

2

2

– F

2R +

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

– Mg

2R +

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

d

4 JMM

d

2

JM

K

K

2

2

----

y

-----------

+

----

y 2R M +

-----

dt

K

K

+ y( R K) =

dt

d

2

–JM

2

2

----

F

---------

+ F( –2R )

+ ( –2MgR )

2

dt

K

d

4

d

2KR2

+

1

R2K2

----

y

+

y

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

+ y

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

=

----

----

dt

dt

JM

M

JMM

–2KR2

–2gKR2

d 2

–1

F

-----

+ F

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

+

----

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

dt

M

JMM

JM