Question

A mass weighing 2 lbs stretches a spring 6 inches. At t=0 the mass is released from a point 8 inches below the equilibrium position with an upward velocity of 4/5ft/s. Determine the equation of free motion.

Answer 1

Stuck. I don't get how to write the equation of free motion

Answer 2

not sure if it means: Acos(wt-d) or m*d^2/dt^2 +k/m*y=0

Answer 3

the latter, the differential equation, is the equation of free motion. the former is a standard solution

Answer 4

ah so I just need to plug in the values? Not sure where to plug in though ... :/

Answer 5

is x'(0) = 4/5ft/s?

Answer 6

well, if you need to "determine the equation of free motion", i would have though that you need to derive the differential equation that you mention. as this is in the vertical plane, you might wish to establish why g does not feature, ie why the DE is the same as for spring-mass system in horizontal plane.

then, actually solving the DE, using a standard solution, will involve plugging in initial values for displacement and veclocity and so on.

Answer 7

Ok I know how to solve the DE and get general solution but I don't know how to write the DE given this information

Answer 8

this will guide the derivation is that is what you think you need to do.

then indeed you take the boilerplate solution x(t) = A cos (wt + phase) and differentiate to get v(t) etc and plug in values

you will get spring constant k from the equilibrium position.

Answer 9

sorry, the link i meant to provide
http://www.ux1.eiu.edu/~cfadd/1150/15Period/Vert.html
with helpful drawings

Answer 10

displacement is 2 inches (8-6?). mass is 2 lbs.

Answer 11

if x(t) = Acos(wt-d)
v(t) = ??
a(t) = ??
w = root (k/m)

Answer 12

The link you sent makes sense but don't know about the rest...

Answer 13

Acos(wt-d) = height?

Answer 14

or length?

Answer 15

Do I take Acos(wt-d) and div once for v(t) and twice for a(t)?

Answer 16

Do i need to plug in values for A,w,t,d and x(t) before i take those derivatives?

Answer 17

x (t) = Acos(wt-d) is displacement from mean position as function of time

yes, differentiate for v(t) and again for a(t)

but you will need omega which is root (k/m)

Answer 18

w=root (k/m)

Answer 19

so I need to set up F=-km to solve for k? But what is F?

Answer 20

hint
"A mass weighing 2 lbs stretches a spring 6 inches. "

get k from that??

Answer 21

F=Ma or F=-km

Answer 22

so i set up ma=-km

Answer 23

m=2 a=? k=? 2a=-2k

Answer 24

at equilibrium position
mg = kx
you're doing this in imperial which is not my game, but **if** g in imperial is 32 then
2 * 32 = k * 6

Answer 25

yes, from wiki:
g = acceleration of gravity (9.81 m/s2, 32.17405 ft/s2)
so that works.

Answer 26

f=-kx? I thought it was -km lol

Answer 27

I thought i assume g is turned of or something

Answer 28

is x the initial stretch i.e. 6 in? because that is when it finally stops and is at rest?

Answer 29

in the equilibrium position, gravity and spring forces are equal and opposite. so at that position you can say mg = kx and calculate k for the spring

"A mass weighing 2 lbs stretches a spring 6 inches. " => 2 * g = k * 6

Answer 30

k=32/3 w= 4/sqrt3

Answer 31

ok ty for that i feel i am almost there

Answer 32

what is d in the Acos(wt-d)? and is t initial time =0?

Answer 33

so diff Acos(4/sqrt(3)*t(0)-d)

Answer 34

or do i leave t alone and that is just the variable

Answer 35

it is the phase shift. if everything started at t = 0, x = A, then it would be a pure cosine curve. however, that is not the case.
you will need
\[x = A \cos (\omega t - \delta); \ \dot x = - \omega A \sin (\omega t - \delta)\]
at t = 0, which is where the initial values are, you have
\[x(0) = A \cos ( - \delta); \ \dot x(0) = - \omega A \sin ( - \delta); \ - \omega \tan ( - \delta) = \frac{\dot x(0)}{x(0)}\]

Answer 36

\[\tan (- \delta) = - \frac{1}{\omega} \times \frac{\dot x(0)}{x(0)} = - \sqrt{\frac{3}{16}} \times \frac{\frac{-4}{5}}{8} = \frac{\sqrt{3}}{40}\]

not paying great attention to arithmetic so please check the numbers...we should try and agree

Answer 37

Thank you so much I get it now!