Question

Just need a quick check. ODE

\[y"+16y'+320=0\] I'm getting \[r=-8+4i~,~-8-4i\] Is that right?

Answer 1

@hartnn do you mind?

Answer 2

@dan815 could you?

Answer 3

y=e^rx
y'=re^rx
y''=r^2*e^rx
y''+16y'+320=0
e^rx(r^2+16r+320)=0
(r+8)^2-64+320=0
r=+/-sqrt-256 -8

Answer 4

r=-2^(9/2) i -8
r=2^(9/2)-8

Answer 5

I see my mistake... thank you

Answer 6

ok np :)

Answer 7

I more question in a second if you don't mind

Answer 8

okk

Answer 9

so r=-8+-16i

Answer 10

lol why?

Answer 11

right! it was 8/2

Answer 12

my bbaadd

Answer 13

oh, I saw that I thought 4 times 320 was 320 that was my error lol

Answer 14

i forgot my middle finger is 8 not 9, my finger counting skills have begun to fail me

Answer 15

lol as have all of ours

Answer 16

ok so if I can just check my polar forms

Answer 17

u wanna write in re^ix form?

Answer 18

yep for the general sln I have to

Answer 19

okk

Answer 20

r=mag of the vector <8,4>

Answer 21

and thet = tan^-1(4/-8)

Answer 22

radius is 5 sqrt10

Answer 23

that looks big

Answer 24

how'd you get 4?

Answer 25

I like that better

Answer 26

oh my bad i was lookign at ur top solution on post

Answer 27

in that case 5root10 looks about right!

Answer 28

still I'm getting arctan(-2)...???

Answer 29

yeah thats right

Answer 30

whats wrong with that

Answer 31

but I don't know the answer to that

Answer 32

oh lol

Answer 33

u cant use calculator??

Answer 34

no no calculators allowed

Answer 35

well thats not nice umm

Answer 36

ughhh did I do something wrong setting it up then? hmm

Answer 37

when is cos theta half of sin tehta then?

Answer 38

I guess can I give you the whole problem to look at?

Answer 39

ok

Answer 40

A mass weighing 2lb stretches a spring 1.2in. The mass is pushed 6in. downward and released. The motion takes place in a medium with damping constant 1lb s/ft. Find the position of the mass at any time t\(\ge\)0

Answer 41

so
w=2lb
m=2/32=1/16
k=w/x=2/.1=20
x=1.2in=.1ft
\(\gamma\)=1 lb s/ft

Answer 42

my eq should be of the form: \[my"+\gamma y'+ky=g(t)\]

Answer 43

so i set it up: \[\frac{1}{16}y"+y'+20y=0\]
which becomes \[y''+16y'+320y=0\]

Answer 44

re u sure k=w/x

Answer 45

is that some thing arbritray they gave u

Answer 46

no hookes law mg=kL

Answer 47

oh really

Answer 48

wait what

Answer 49

that's what the book said...

Answer 50

wait do u mean F=-kx

Answer 51

mg − kL = 0.

Answer 52

according to the book

Answer 53

by hookes law u mean like the acceleration relationship wrt to postion right

Answer 54

yea

Answer 55

is it supposed to be negative?

Answer 56

well its usually written in negative unless they refine L

Answer 57

I mean is k supposed to be negative in the end?

Answer 58

but u cannot get K from that eqution though

Answer 59

k has to be suppplied

Answer 60

we have to find k

Answer 61

oh hoho

Answer 62

well then!

Answer 63

m dy^2/dt= -k*(Y_o-y) + D*(dy/dt)

Answer 64

Y_o is equlibrium pos

Answer 65

wha???

Answer 66

that is your normal equation with D for damping

Answer 67

I've never seen that

Answer 68

see how the damping is taking away from -kdeltaposition that is gonna slow down our accerlation

Answer 69

m dy^2/dt= -k*(Y_o-y)
this is
F=-kx

Answer 70

i just added damping term to it

Answer 71

but we aren't supposed to?

Answer 72

fibonaccci!

Answer 73

ill hit u -.-

Answer 74

I''m sorry, we never covered that in class!!! I'm a wee bit confused

Answer 75

we only covered mg=kx

Answer 76

okay fine lets get something straight here hookes law

Answer 77

is ma=kdeltax isnt it?

Answer 78

are talking about the samething

Answer 79

http://people.math.umass.edu/~tanguay/331/331spr10hw17sol.pdf
They say it the same way here

Answer 80

and yes, the elongation of the spring

Answer 81

how can u say
mg=kx
firstly its
k*deltaX
and the whole point is delta x is changing wrt to the acceration is experiences unless the k constant is the particular enlogation L when it experiences a acceraltion = gravity

Answer 82

but that's what it tells me to do for spring constant at equilibrium

Answer 83

ok ok i think u are right !! baha

Answer 84

it's how they do it in the book

Answer 85

lol

Answer 86

god... I'm correcting my test so I can go beg for points at like 9a

Answer 87

but I keep screwing something up

Answer 88

grrrrrr

Answer 89

i see ur mistake i think

Answer 90

what? please tell me

Answer 91

mg=kL
2*9.81/1.2=k

Answer 92

wait thats in lb

Answer 93

whats the gravity thing in lb?

Answer 94

32 not 9.81

Answer 95

okay

Answer 96

and change that 1.2in to feet

Answer 97

where'd the 2 come from?

Answer 98

u said 1.2in.