Question

. b

Answer 1

can you rewrite this Acos((root(k/m))t) using the equation tool?

Answer 2

yes

Answer 3

yes\[x(t)=A \cos ((\sqrt{k/m}) t)\]

Answer 4

did you guys get what I got

Answer 5

is that t supposed to be an exponent?

Answer 6

oh no

Answer 7

Can you tell us more about the question? It seems like you are using outside info that is not included in the problem.

I understood what you typed, but where are you getting that from?
You don't have that in the given prompt Is a formula in your text?
It seems like you need to do more work because you didn't sue the F = -kx or F = ma info at all.

Answer 8

let me see your derivative work

Answer 9

Yea so the problem is typed and it is asking to guess and equation

Answer 10

You also didn't set up a differential equation
The goal of this problem is to write a differential equation

Answer 11

so I guessed x(t)=Acos(bt)

Answer 12

she is using the simple harmonic motion

Answer 13

that's what I needed to know @nincompoop

Answer 14

then got the second deriv d^2x/dt^2= -Ab^2 cos(bt)

Answer 15

Right but that simple harmonic motion stuff has other assumptions in it

Answer 16

correct

Answer 17

what do you mean

Answer 18

So I concluded : x(t)=Acos((√(k/m))t)

Answer 19

I wouldn't plug anything in yet - get the diff eq first
Using algebra, how do we get "a" alone in -kx = ma?

Answer 20

ma = -k x

Answer 21

right

Answer 22

where a = (d^2 x) or second derivative of x

Answer 23

taking the second deriv makes sense right ?

Answer 24

yes because acceleration is 2nd derivative of displacement

Answer 25

the acceleration is the second derivative of position function
x'(t) = v(t)
v'(t) = a(t)

Answer 26

right but does it make sense to find the acceleration even thought the problem asks for displacement ?

Answer 27

Here position and displacement mean the same thing

Answer 28

right but I am still a but confused about why acceleration is being used..?

Answer 29

(we solved this as a group and I was not sure about that part)

Answer 30

because it is still in terms of acceleration
m"a" = -kx

Answer 31

That's given in the problem
F = ma has "a" as acceleration already
like nicompoop said

Answer 32

They want you to use F = -kx and F = ma to get the differential equation

Answer 33

I just found you ways to save on car insurance and math problem
http://farside.ph.utexas.edu/teaching/301/lectures/node138.html#eshm

Answer 34

right but the question asks for displacement but we solves for acceleration.. ?

Answer 35

The answer is for displacement... But the diff eq is for the 2nd derivative of displacement
and that's acceleration.
You should be getting answers for 2 parts. The first part is for the diff eq itself
The 2nd part is for the equation (x = ) that solves that diff eq

Answer 36

oh so the equation that I got was incorrect?

Answer 37

Let me just write out all your steps so I can see the logic you used

Answer 38

ok

Answer 39

What are you solving for when you plug in x(0) = A?

Answer 40

I guess it's supposed to be k?
You used x = Ab^2 *cos(t) to start so I assume you want k?

Answer 41

You solved for b, but you already had b as a constant in x = Ab^2*cos(t)?

Answer 42

wait sorry what is your question

Answer 43

nvm; I understand what they want

Answer 44

ok

Answer 45

They want you to derive that simple harmonic motion equation
That's why you solved for b

Answer 46

so the answer I wrote initially is the displacement equation then

Answer 47

right

Answer 48

I assume a = -Ab^2*cos(bt) is given elsewhere in your course

Answer 49

no we assumed that

Answer 50

since the graph is like a cos graph

Answer 51

You do need to make an assumption on "a" there or you can't get to x
You probably need another term

Answer 52

oh okay

Answer 53

I'm thinking you didn't actually need to do all that

Answer 54

but does it make sense

Answer 55

I would assume you need to start with x = Acos(sqrt(k/m)t + C) + D
Kind of, but I don't know enough physics to say whether a = -Ab^2*cos(bt) is a standard assumption. I assume it is because it does get you pretty close to simple harmonic motion

Answer 56

oh ok

Answer 57

Let's talk about the last sentence of the directions though
They do say you have to differentiate ..... And you never did that in your work

Answer 58

You should actually be starting with x = Acos(b*t) or similar
And then plugging in to show the diff eq works

Answer 59

oh and how would I of done that

Answer 60

Any ideas on how to start taking the derivative of A*cos(sqrt(k/m)*t)?

Answer 61

yeah
I call it the chain-rule

Answer 62

wait, that's not exponent.. laughing out loud I keep forgetting

Answer 63

I agree :)
How would we apply that here?

Answer 64

O.O

Answer 65

It's still chain rule (function inside of a function)

Answer 66

I shall find my calculator lol its 3 am here ahaha function

Answer 67

The exponent rule is power rule for x^n terms
*or t^n terms here

Answer 68

just product rule

Answer 69

SQRT can broken down into two products
but you can do it however you're comfortable with

Answer 70

You could use product rule too but A, sqrt(k/m) are all numbers

Answer 71

I think Ill just leave the answer to what I originally got because I need to sleep x.x

Answer 72

Sure but it's still chain rule overall
It's function inside of a function

Answer 73

HAHAH let us assume you got the right answer
read the link I gave you though. you're going to need it

Answer 74

ok thanks!

Answer 75

yo let us!

Answer 76

Good luck! :-]