translation - 2.41

14. Write the differential equations for the system below.

Ks1	x1
M1
	s, k

M2				x2

Ks2

2.7 PRACTICE PROBLEM SOLUTIONS

1. Ks = 33.3 N/cm = 3333 N/m

2.

a) Keq = Kd1 + Kd2	b)	Kd1Kd2
	Keq	= -----------------------	+ Kd2
		Kd1

3.

·· ·	B		F
x + x	----		= –----
	M		M

4.

··		+ x	Ks	+ x	–Ks			=	F
x			------		--------				------
	1		1 M		2 M	1			M	1
			1
··		+ x	Ks	+ x	–Ks			=	0
x			------		--------
	2		2 M		1 M	2
			2

translation - 2.42

5.

··

· Kd1

Ks1 + Ks2

–Ks2

x1

+ x1 --------M

1

+ x1 ----------------------

M

1

+ x2

-----------M

1

= 0

··

· B

Ks2

–Ks2

F

x2

+ x2 M------

+ x2 -------M

+ x1

-----------M

2

= M------

2

2

2

6.

··

· Kd + B1

Ks

–Ks

x1

+ x1 ------------------M

1

+ x1 M------

+ x2 --------M

1

= 0

1

··

· B2

Ks

–Ks

–F

x2

+ x2 M------

+ x2

M------

+ x1

--------M

2

=

M------

2

2

2

7.

··

· Kd1

Ks1

–Ks2

x1

+ x1 --------M

1

+ x1 -------M

+ x2 -----------M

1

= 0

1

··

· B

··

· B

Ks

F

x1

+ x1 M------

+ x2

+ x2 M------

+ x2 M------

=

M------

2

2

2

2

translation - 2.43

8.

FBDs:

··

–K

d1

x·

M1x1

K

d2

( x·

– x·

)

K

d3

x·

1

M1

1

2

M2

2

–Ks1x1

Ks2( x1 – x2)

Ks3x2

··

F

M2x2

For M1:

+

·

·

·

··

∑F = – Kd1x1 – Ks1x1

– Kd2( x1

– x2)

– Ks2( x1 – x2)

= M1x1

··

·

( Kd1 + Kd2) + x1( Ks1 + Ks2)

·

x1

( M1) + x1

+ x2( –Kd2) + x2( –Ks2) = 0

··

·

Kd1

+ Kd2

+ x

Ks1

+ Ks2

·

–Kd2

+ x

–Ks2

= 0

x

+ x

1

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

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

+ x

-----------

-----------

1

M

1

1

M

1

2

M

1

2

M

1

For M2:

∑F = Kd2( x·1 – x·2) + Ks2( x1 – x2) + F – Kd3x·2 – Ks3x2 = M2x·2

+

··

( M2)

·

( Kd2

+ Kd3) + x2( Ks2

+ Ks3)

·

x2

+ x2

+ x1( –Kd2) + x1( –Ks2) = F

··

+

·

Kd2

+ Kd3

+ x

Ks2

+ Ks3

·

–Kd2

+

x

–Ks2

= F

x

x

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

2

---------------------- + x

-----------

-----------

2

2

M

2

M

2

1

M

2

1

M

2

9.

··

( M1)

·

( Kd1

+ B) + x1(

Ks1)

·

x1

+ x1

+ x2( –B) = 0

··

·

Kd1

+ B

+ x

Ks1

·

–B

= 0

x

+ x

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

-------

+ x

------

1

1

M

1

1

M

1

2

M

··

·

·

1

( M2)

x2

+ x2

( B) + x2( Ks2) + x1( –B) = F

··

·

B

+ x

Ks2

·

–B

=

F

x

+ x

------

-------

+ x

------

------

2

2

M

2

M

2

1

M

M

2

2

2

translation - 2.44

10.

··

·

( Kd1) + x1( Ks1) = FF

x1( M1) + x1

··

·

Kd1

+ x

Ks1

FF

x

+ x

--------

-------

= ------

1

1

M

1

1

M

1

M

1

··

( M2) + x2( Ks2) = F – FF

x2

··

+ x

Ks2

F – FF

x

-------

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

2

2

M

2

M

2

where,

FF

≤

sM2g

x·1 – x·2

F

=

M g

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

F

k 2

x·1 – x·2

if x·1 = x·2 if x·1 ≠ x·2

11.

(assuming no gravity

··

( M1)

·

·

( –Kd1) + x2( –Ks1) = F

x1

+ x1( Kd1) + x1( Ks1)

+ x2

··

· Kd1

Ks1

· –Kd1

–Ks1

F

x1

+ x1 --------M

1

+ x1 -------M

+ x2 -----------M

1

+ x2 -----------M

1

= M------

1

1

··

( M2)

·

+ Kd2)

·

x2

+ x2( Kd1

+ x2( Ks1 + Ks2) + x1( –Kd1) + x1( –K

··

· Kd1 + Kd2

Ks1 + Ks2

· –Kd1

–Ks1

x2

+ x2

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

M

2

+ x2

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

M

2

+ x1 -----------M

2

+ x1 -----------M

2

12.

if( x ≥ 0)

T = 0

if( x < 0)

T = –Ks2x

s1) = 0

= 0

·· ·	B		+ x	Ks1			=	T
x + x	------			-------				------
	M			M	1			M	1
	1