Calculus IV - HW 4

2. (Problems 7 from B&D) A mass weighing 3 lb stretches a spring 3 in. Suppose the mass is pushed upward, contracting the spring a distance of 1 in., and is then set in motion with a downward velocity of 2 ft/s. Assuming there is no damping, find the position u of the mass at any time t. Determine the frequency, period, amplitude, and phase of the motion.

1

Solution:

To

find

the

mass,

we

note

that

w

=

mg

so

3

lb

=

m

?

32

ft s2

implying

m=

3 32

lb?s2 ft

.

Since

the

mass

weighing

3

lb

stretches

the

spring

3

in.,

we

have

that

FS

=

3

lb

=

k?

1 4

f

t

implying

k

=

12

lb ft

.

Since

there

is

no

damping

or

external

force,

we have = 0 and F (t) = 0. Thus the equation we get is

3

1

u + 12u = 0; y(0) = - , y (0) = 2

32

12

If

we

multiply

through

by

32 3

,

we

get

u + 128u = 0

whichhas characteristic equation r2 + 128 = 0 which has roots r= ? -128 = ?i ? 8 2. Thus the general solution is u(t) = A sin(8 2t) + B cos(8 2t). Plugging

in

our

initial conditions

we

seey(0)

=B

=

-

1 12

.

Taking the derivative, we get

y (t) = A8 2 cos(8 2t) - B8 2 sin(8 2t). In particular y (0) = A8 2 = 2.

Therefore A = 1 . Thus the solution is 42

u(t) =

1

1

sin(8 2t) - cos(8 2t).

42

12

Here 0 = 8

2 rad/s and

T

=

2 0

=

s.

42

As usual,

R=

1 2 +

42

-

1 12

2=

1 32

+

1 144

=

11 288

.

Finally

=

tan-1

-1/12 1/4 2

= - arctan( 3 ) 2.0113 since we 2

want sin() > 0 and cos() < 0.

3. (Problems 11 from B&D) A string is stretched 10 cm by a force of 3N. A mass of 2 kg is hung from the spring and is also attached to a viscous damper that exerts a force of 3 N when the velocity of the mass is 5 m/s. If the mass is pulled down 5 cm below its equilibrium position and given an initial downward velocity of 10 cm/s, determine its position u at any time t. Find the quasi frequency ? and the ratio of ? to the natural frequency of the corresponding undamped motion.

Solution: We are given m = 2 kg. Furthermore, we know a force of 3N stretches

the spring 10 cm implying FS = 3N = 0.1k cm.

Thus

k

=

30

N m

.

Furthermore,

Fd

=

3N

=

?

5

m s

.

Thus

=

3 N ?s 5m

and

our

equation

is

3 2u + u + 30u = 0; u(0) = .05, u (0) = 0.1

5

Page 2

Dividing by two we get,

3 u + u + 15u = 0

10

which

has

characteristic

equation

r2

+

3 10

r

+

15

=

0

which

has

roots

3 r=- ?

20 Thus the general solution is

9 100

-

4(1)(15)

= -0.15 ? i3.87008

2

u = Ae-0.15t sin(3.87008 t) + Be-0.15t cos(3.87008 t).

We can plug in our initial conditions and see that u(0) = B = 0.05. Taking the derivative, we get

u = -0.15Ae-0.15t sin(3.87008 t) + 3.87008Ae-0.15t cos(3.87008 t)

-0.15Be-0.15t cos(3.87008 t) - 3.87008Be-0.15t cos(3.87008 t) so

u (0) = 3.87008A - 0.15B = 3.87008A - 0.15(0.05) = 0.1 implying A 0.02778. Thus the solution is

u(t) = 0.02778e-0.15t sin(3.87008 t) + 0.05e-0.15t cos(3.87008 t).

Then we see that the quasi frequency is ? = 3.87008 rad/s. Recall that the natural

frequency is 0 =

k m

=

15

rad/s.

Thus

the

ratio

is

? 0

=

3.87008 15

0.99925.

4. (Problems 17 from B&D) A mass weighing 8 lb stretches a spring 1.5 in. The mass is also attached to a damper with coefficient . Determine the value of for which the system is critically damped; be sure to give the units for .

Solution: Recall that a system is critically damped when 2 - 4mk = 0. To

convert

from

weight

to

mass,

we

note

w

= mg

so

m=

8 32

lb?s2 ft

.

For

k,

we

see

that

the system reaches equilibrium after the spring stretches 1.5 in. Thus Fs +w = 0 so

Fs

=

-w

=

-8

lb.

However,

Fs

=

-kL

where

here

L

=

1.5 12

=

1 8

f t.

Thus

k

=

64

lb ft

.

Page 3

Thus we have

8 lb ? s2

= 4mk = 4

32 f t

lb

lb2 ? s2 lb ? s

64 = 64

=8 .

ft

f t2

ft

5. Optional (Problems 19 from B&D) Assume that the system described by the equation mu + u + ku = 0 is either critically damped or overdamped. Show that the mass can pass through the equilibrium position at most once, regardless of the initial conditions. Hint: Determine all possible values of t for which u = 0.

Solution: The characteristic equation of mu + u + ku = 0 is mr2 + r + k = 0

with roots

2 - 4mk

r=- ?

2m

2m

Since the system is critically damped or overdamped, by definition 2 - 4mk 0.

Let

=

-

2m

and

=

2-4mk 2m

.

Thus

the

equation

has

solutions

of

the

form

u(t) = Ae(+)t + Be(-)t

For the mass to pass through the equilibrium point at time t means that u(t) = 0.

Thus we set Ae(+)t + Be(-)t = 0

= Ae(+)t = -Be(-)t

=

e(+)te-(-)t

=

B -

A

=

e2t

=

B -

A

which

has

exactly

one

solution

if

-

B A

>

0

and

zero

solutions

if

-

B A

0.

Thus, the mass can pass through the equilibrium point at most once.

Section 3.8

6. (B&D # 5) A mass weighing 4 lb stretches a spring 1.5 in. The mass is displaced 2 in. in the positive direction from its equilibrium position and released with no initial velocity. Assuming that there is no damping and that the mass is acted on by an external force of 2 cos(3t) lb, formulate the initial value problem describing the motion of the mass.

Page 4

Solution: We can immediately assume the following.

w = 4 lb

L

=

1.5

in.

=

1 8

ft.

=

0

s?lb. ft.

F (t) = 2 cos(3t) lb

u(0)

=

2

in.

=

1 6

ft.

u (0)

=

0

ft. s

Some further computation gives

m k

= =

w g mg L

=

= 4 lb

32

m s2

1 lb?s2 8m

4

1

8

lb ft.

=

32

lb ft.

It follows that the initial value problem modeling the motion of the mass is

1

1

u + 32u = 2 cos(3t); u(0) = , u (0) = 0.

8

6

7. (B&D # 6) A mass of 5 kg stretches a spring 10 cm, The mass is acted on by an external force of 10 sin(t/2) N (newtons) and moves in a medium that imparts a viscous force of 2 N when the speed of the mass is 4 cm/s. If the mass is set in motion from its equilibrium position with an initial velocity of 3 cm/s, formulate the initial value problem describing the motion of the mass.

Solution: We can immediately assume the following.

m = 5 kg

L = 10 cm = .1 m

=

2N

4

cm s

=

2N

.04

m s

=

50

kg s

F (t) = 10 sin(t/2) N

u(0) = 0 m

u (0)

=

3

cm s

=

.03

m s

Some further computation gives

k

=

mg L

5

kg?9.8

m s2

.1 m

=

490

N m

It follows that the initial value problem modeling the motion of the mass is

5u + 50u + 490u = 10 sin(t/2); u(0) = 0, u (0) = .03.

Page 5

 ................
................

In order to avoid copyright disputes, this page is only a partial summary.

Google Online Preview   Download