PHYS 131 Final: Phys 131 Final Study Guide 3-4
SHM
-
displacement
✗
,
as
a
function
of
time
-
mass
on
a
spring
1
↳
described
by
sinusoidal
function
(
cosine
)
}
-
uniformly
rotating
fmÉÉ☒
/
Wheel
-
A
6
✗
A
vertical
mass
spring
system
Energy
in
sum
:
total
energy
is
conserved
11111
1/1/11
YB
:-O
F-
=
W
(
+
QI
{
It
{
"
t
"
=¥%H§
"
"
"
"
"
"
"
±
"
"
"
"
"
"
"
"
"
"
"
"
°
"
KE
:
K=tzMV2
from
released
state
-
-
I
-
_n
-
-
Kmax
When
Vmax
111=01
AY
/
•
FSP
eqim
Ya
:O
↳
izmvmaxz
=
2-
MCAW
)2
tEtot=K+#
=
-
-
-
-
-
-
Ent
-
-
-
-
Fnety
-
-0
↳
constant
when
calculate
A
:
tw=mg
RE
Wex
-1=0
A-
-
Imax
'
PEIU
)
w
wifmorfylradl
energy
is
always
F-
×
.
First
time
mass
goes
through
ea
'm
positive
-
draw
graph
-
p
g
A
kilt
Vmax
1-
=
-114
,
-112
.
.
.
solve
for
-1=2*1
w
-
Vmax
__
IRA
'
KE
Max
when
a-
-
negative
✗
=
eq
.
position
f
:O
.
-5-5=2
.
5M
0
i
:
0.5-0=0
m
what
fraction
of
Max
displacement
,
A.
displacement
are
the
kinetic
&
elastic
PE
equal
K=V
?
=
positive
Utv
=
20=21
'zkX4
2(#kX4=¥✗A2
✗
=
A-
V2
-
-
-1=2*5-1
°;¥_=JE
my
=/
9¥42
I
*
solve
for
angular
frequency
cut
/
ex
.
I
=
0.454M
,
Ñ=
0.166m
/
S
,
Ñ=
-0.34m
/
S2
F=k×→mg=k×
✗
=
my
-
g
att
-
Aw
>
coslw
-1+01
w=_q¥[
I
✗
1-11
__
Acoslwt
-101
✗
=
.
/
9,3¥
)
?
9.8
=/
0.03217nF
1
phase
shift
10
)
FI-whatisthe-pedofthe-NK.TT
'
'
It
=
Acosta
A=Éo
-
spring
length
-_
1.2mm
it
=
-
Awsinlo
)
A=-Ñ#
A-
1.9-1.0=0
-4m
-1=3-5
•
•
w
sin
10
)
1<=240
NIM
Y
-
-
-
-
-
-
-
-
-
-
-
-
-
-
-
0.454
=
-0.161
W=fEn
m=Iwz
I
I
0.8655in
101
put
-0.1b£
calculator
in
W=
2¥
=
2¥51
,
/
{¥{¥=(
0.8651104541
rad
mode
m=
241
=
54.7kg
spee
use
consv
.
of
1
D=
tan
-
if
-0¥
(0-865110.4541)=-0.3995
(
2117312
energy
E+o+=k+U=Iz
KAZ
rod
Find
amplitude
y
-1.2-1.0=0
-2m
*
MVZ
+
¥KyZ=
#
KAZ
{
It
)
=
A
COSIO
)
0.454
=
ACOSC
-0.39951
v=JÉ_y=JÉÉ
M
go.gg#547K9
A-
-0.45£
=
0.4928m
COSI
-
0.3995
)
-
Document Summary
Displacement , as a function of time (cid:8869) described by sinusoidal function ( cosine ) } - uniformly rotating fm / Energy in sum : total energy is conserved. {" t "= %h """" " " " "" " """" """"" " Kmax when vmax 111=01 (cid:8869) izmvmaxz = 2- mcaw)2 tetot=k+# (cid:8869) constant when eqim ya :o. - - fnety --0 from released state tw=mg. 1- = -114 , -112 . solve for -1=2*1 w. Vmax __ ira" energy is always positive kilt vmax. I = 0. 454m , att - aw> coslw-1+01 w=_q [ It = acosta a= o it = - awsinlo) a=- # w sin 10) 1 d= tan-if -0 (0-865110. 4541)=-0. 3995 rad mode m= 241. = 54. 7kg spee use consv . of (2117312 energy e+o+=k+u=iz kaz y-1. 2-1. 0=0 -2m *mvz + kyz= # kaz { it) = acosio) 0. 454 = acosc-0. 39951 v=j _y=j . Umping causes an oscillator to lose } ff_g 1 .
