Question

A force of 400 Newtons stretches a spring 2 meters. A mass of 50 kilograms is attached to the end of the spring and is initially released from the equilibrium position with an upward velocity of 10 m/s. After finding the equation of motion, calculate x(t = pi/12).

I know this is a second order differential equation but I don't know where to start.
k=245.25
w(frequency)=2.2147 hz

The general formula is x(t)=c1cos(wt) +c2sin(wt)
and I think x(0)=10 but help would be really appreciated.

Answer 1

By definition, velocity is the derivative of displacement with respect to time, so in this case to find the eqn of motion you need to find dx/dt

If you remember, d/dx ( cos(a*x)) = - sin(a*x)*a while d/dx(sin(a*x)) = cos(a*x)*a

Answer 2

So, to find your solution complete the differentiation and then apply the initial value of t = pi/12 to find your answer.

Answer 3

The general formula is x''+(k/m)x=0
And I know it should come out to x(t)=c1cos(wt) +c2sin(wt) But then I have to plug in intitial conditions and I'm not sure what they are

Answer 4

oh i see what you're saying haha, you need to solve your second order differential.

Answer 5

Yep, and there's something about using A=sqrt((c1^2+c2^2))
where x(t)=Asin(wt+(phi)

Answer 6

If I knew the initial conditions, I can solve for c1 and c2 and then solve A and then hopeful plug in t=(pi/12)

Answer 7

hmmm unfortunately I do not remember the physics eqns as well it has been a few years. If the spring is going through the air, you know that it's velocity is equal to 0 at the point of maximum extention

Answer 8

is the spring on the ground sliding or oriented vertically?

Answer 9

upward velocity so vertically

Answer 10

ok that's what i thought, I just wanted to make sure. Yeah, so one condition would be that dx/dt = 0 at its maximum extention

Answer 11

I guess the other one is x(0)=10 judging by the problem?

Answer 12

initially released from the equilibrium position with an upward velocity of 10 m/s

Answer 13

yeah that is the initial condition

Answer 14

so you know that at the peak of motion, dx/dt is = 0 m/s and you know that initially dx/dt = 10 m/s

Answer 15

and the values for x depend on how you define your axis, since the spring is being stretched downward 2 meters first

Answer 16

excuse me, it's being compressed, so at x = 0 dx/dt = 10 m/s like you said, and then at x = maximum extention dx/dt = 0. I misread the statement.

Answer 17

So wait what is the maximum extention?

Answer 18

Normally there should be something like
x(0)=10
x(0)'= (some #) That way, the sin and cos will cancel out c1 or c2 with a zero.

Answer 19

well reasonably, if k = 245.55 N/m that means it goes 1 M with 245.55 N of force, except in this case we also have to consider gravity force acting against it

Answer 20

you're right about the initial conditions, I'm just thinking of a way to deal with the x'(0) condition because I haven't worked with springs in years haha

Answer 21

yeah the diff eq is very easy to solve I'm just trying to reason what the max height is. I'm guessing a force balance is going to be the best bet

Answer 22

No it's okay because you're doing a really good job! I really didn't think anyone could help me out with this because it's for one of my engineering courses with cal 3

Answer 23

haha, I'm actually a Chemical engineer lol, in grad school

Answer 24

unfortunately we don't ever deal with springs, mostly quantum mechanics of particles and stuff like that, so the last time i did a balance like this was prob 3 years ago

Answer 25

Wow, I'm civil engineering. I'm only a sophmore in undergrad though

Answer 26

springs are stupid, but anyway, I think if this is for a calc course there probably won't be a force balance involved and I'm probably overthinking this. I mentioned earlier the second derivative test, that finds the max easily but it won't work for your function.

Answer 27

I think this class has it's stupid moments. But I'm looking at an example from class and it mentioned that x(0) might also equal -frequency or (2.2147) in this case. Does that even make sense?

Answer 28

Wait nevermind.

Answer 29

x(0)= -10 and x(0)= 2 (the given maximum they give)

Answer 30

lol

Answer 31

wait i think i know what they want you to do, there's no way they want you to do the mass balance I'm thinking of

Answer 32

Or x(0)= -2
x(0)'= -10 since it's velocity

Answer 33

ok, so normally f = xk, however in this case that would yield a k value of 200 instead of 245.55 that the problem gives

Answer 34

Oh that changes my frequency (w) The example from class didn't have a given force. So it prolly does mean that k=200

Answer 35

wait, is the k = 245.55 given in the problem statement?

Answer 36

No

Answer 37

oh ok, yeah your k is equal to 200

Answer 38

I used w^2=(k/m)

Answer 39

But that problem from class didn't start with a force

Answer 40

So i prolly gotta use the 400 to find k so it's balenced

Answer 41

if you can find the initial force applied to the block, you can then find how high it went if you ignore gravity

Answer 42

if you assume that the data given was generated from the system in question then you can ignore gravity since it would be included for in the data

Answer 43

I got that c1= -2 and c2= -5

Answer 44

this problem in truth is very obtuse, and they left out a lot of assumptions

Answer 45

yeah i'd just go with that haha

Answer 46

I should be able to find A and solve the Asin(wt+(phi))

Answer 47

yeah, sorry for making this so complicated, i didn't know how many assumptions needed to be made about the systems and the data

Answer 48

No way! You totally helped me step this out

Answer 49

haha glad I could, yeah idk about calc 3 but engineering is all about conservation eqns and this would be solved a lot easier using energy

Answer 50

using potential and kinetic energy to solve the system would be easier, and in a practical setting where you can choose how to solve your problems it'd be what i chose. They do give the same answer though.

Answer 51

anyways best of luck

Answer 52

Thanks!