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). Round in the tenths place.

Answer 1

"After finding the equation of motion"

what are you currently learning?!

you can jump in with \(x(t) = A \cos (\omega t + \phi)\) as the general solution to shm and fit it to these facts

or you can build a differential equation from forces or energy conservation...

Answer 2

We are learning how to solve in-homogeneous and homogeneous 2nd order linear differential equations. However the class I'm taking an engineering course but it is essentially half differential equations and half multi-variable calculus.

Answer 3

soz, that doesn't really tell me what you want to do. do you have the DE that you want to solve?

if not, do you need to find it?

to illustrate:

as it starts at equilibrium position (ie x=0 at t = 0), you could just say that \(x(t) = A \sin \omega t\) as that ticks all the shm boxes. so \(\dot x(t) = \omega A \cos \omega t\) ; and as it is "released from the equilibrium position with an upward velocity of 10 m/s", we know \(\dot x (0) \)

also we know/ should know that \(\omega = \sqrt{\dfrac{k}{m}}\); and we get k from the info that "400 Newtons stretches a spring 2 meters"

or we can start with \(m \ddot x + kx = 0\) and solve from there

i hope that helps....:p

Answer 4

the DE that needs to be solved is
\[x(t)=C _{1}\cos(\omega \times t)+ C _{2}\sin (\omega \times t)\]

Answer 5

ok

\(x(t)=C _{1}\cos(\omega t)+ C _{2}\sin (\omega t)\)
\(\dot x(t)=-\omega C _{1}\sin(\omega t)+ \omega C _{2}\cos (\omega t)\)

and you know that: \(x(0) = 0, \qquad \dot x(0) = -10\)

that still leaves us with the task of finding \(\omega\), though that has been covered above.

Answer 6

Where did I go wrong? If you can't read it let me know.

Answer 7

@IrishBoy123

Answer 8

how did you figure out k, the spring constant?

Answer 9

It gives you that m= 50 kg and the stretch is s=2m so
(mg)/s= k
(50)(9.8)/2=245

Answer 10

A force of 400 Newtons stretches a spring 2 meters.

Answer 11

take k from that :p

Answer 12

Do I need to convert the newtons to kg?

Answer 13

\(F = kx\)
\(k = \dfrac{F}{x}\)

and plug in the numbers

Answer 14

F in newtons, x in metres

Answer 15

so k=200 how do I get omega without a mass? Sorry to pester you so much.

Answer 16

we already saw that \(\omega = \sqrt{\dfrac{k}{m}}\)

they gave you the mass in the question

you're not pestering btw. this is interesting. i just hope that you are getting something from it :p

Answer 17

That makes sense because k is a constant you can m=50 instead of having to know how to convert from Newtons to kg.

Answer 18

I still got it wrong.

Answer 19

what do you make \(\omega\) out to be?

Answer 20

\[\omega=\sqrt{\frac{ 200 }{ 50 }}=\sqrt{4}=2\]

Answer 21

that looks good

Answer 22

my homework says I'm wrong

Answer 23

so... we have \(x(t)= A \sin 2t\)

so \(\dot x(t)= 2A \cos 2t\)

and from \(\dot x(0) = 10\), we can say \(2A \cos 0= 2A\)so \(A = 5\)

so we have \(x(t)= 5 \sin 2t\)

\(x(\frac{\pi}{12}) = 5 \sin \dfrac{\pi}{6}\)

Answer 24

which gives you x=2.5 that's what I put and it says it is incorrect.

Answer 25

you can also try x = -2.5

but this is the correct approach.

i can think of several different approaches to the problem that will give the same result.

Answer 26

Yeah it was x=-2.5 I'm assuming it is because the velocity is negative and A is technically equal to \[\pm5\]

Answer 27

it's convention. i stated it as -10 here and then thought i was over complicating it

Answer 28

Thank you for your help sorry it took so long but at least now I know how to do the problem.

Answer 29

it was good. thanks.