Upload
justme8
View
219
Download
0
Embed Size (px)
Citation preview
8/12/2019 Spring Mass
1/12
A mass is hooked on a flexible spring suspended vertically from a rigidsupport. Suppose the amount of elongation on the spring until it attain
equilibrium is s. Then at this point, the weight of the mass will have the same
magnitude as the restoring force. Once we pull the spring from the point of
equilibrium and release it, then a restoring force will be applied on it.rovided that there is no other external force acting on the system ! i.e.
retarding force", then we can equate the force acting on the spring and therestoring force. #esides, without damping ! retarding force" the spring$ mass
system will continue to oscillate continuously.
Since , so
So we will obtain the following expression for the force acting on the spring%
.
The usual case for the spring mass system is%
&" 'e pull the mass downwards ! the spring and release.
(" 'e push the mass upwards !the spring compresses"
mg ks= -mx ks= -&&
kx mg- +
mx kx= -&&
8/12/2019 Spring Mass
2/12
Solving the second order O)* !",
#y using
& & ( (x c y c y= +
mx kx= -&&
8/12/2019 Spring Mass
3/12
+irst try%
#y substituting the expression into the equation , we obtain%
Thus the two roots of a is%
(a m k= -
aty e=
8/12/2019 Spring Mass
4/12
8/12/2019 Spring Mass
5/12
, but let
So our solution is for the (nd order O)* is%
onsider case &"
- -
& (
i t i
x c e c e-
= +
- k
m=
ka im
= \
8/12/2019 Spring Mass
6/12
f the initial condition is such that , then , so
.
#esides, if we start the whole spring mass oscillation system such that .
!/" /x =( ) ( & &cos- sin - cos - sin - c t i t c t i = + - -
& (c c= -
!/" /x =max0!/"x v=
8/12/2019 Spring Mass
7/12
Since , .
Thus.
max&
(-
vc
i=
&0! " (-cos-x t c i =
&( sin-x c i t=
8/12/2019 Spring Mass
8/12
So the solution is% .
onsider case ("
f the initial condition is such that , then
max! " sin--
vx t =
& (c A c= -!/"x A=
8/12/2019 Spring Mass
9/12
t is clear that when its position is at its amplitude then the velocity of the
block is /.
Since , we get%
So
0!/" /x =
& (c c=
- -
& (0- i t i tx ic e ic e-= -
8/12/2019 Spring Mass
10/12
An alternative method can be used%
#ut in real life situations, we usually do not have free motion ! i.e. retarding force isalways present". 1et2s only consider the first order resistive force which is .
So the force
( ) ( cos- sin - cos- sin -( (cos-
A A t i t t i
A t
= + + -
=
retardingF Cx= -
8/12/2019 Spring Mass
11/12
( ) ( )
( ) ( )
( )
( ) ( )
- -
& (
& (
& ( & (
& (
( ( & & (
(
&
(
((( (
& ( & ( & ( & (
cos- sin - cos- sin -
cos- sin -
cos- sin -
sin- 3
,
)sin3
cos3
n3
4 (
i t i t
c e c ec t i t c t i t
c c t i t c c
A t A t
D tA
c A c ci
D
A
A
A A c c i c c c c i
-= +
+ - -
- + +
+
+
- = + =
=
=
=
= + = - + + = - =
8/12/2019 Spring Mass
12/12
& (
&
&
(
(
& (
( 5( (
& (
5( (
( 5( (
& (
/
5 - /
5 - /
5 5 4-
(! "
4- 5 4- /, , ! "
(
4- /, ! "
5 4- 4- /, , ! "
(
a t a t
a t
t
a t a
x Cx kx
x x
a
t c e c e
ia x t e c e c e
x t ce
a x t e c e c e
-
-
-
+ + =
+ + =
+ + =
- -=
= +
-- < = = +
- = =
- - > = = +
&& &
&