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).

Question

anonymous · Answer

no one else has taken a crack the question, so i'll give it a try, i can get the first bit tho.

F=k*deltaX
400=k*2
k = 200n/m

s = DeltaV/time
v = DeltaA/time; A being acceleration

A = g + spring deceleration

spring decleration;
ma=F=k*deltaX
50(a) = 200x
a=(200x)/50
a=40x

with;
A = g + spring deceleration
A = -9.8 + -40x

with;
v = DeltaA/time;
v = (-9.8 + - 40x)/t

with;
s = DeltaV/time
s = x
x = (+10m/s  +  ((-9.8 + - 40x)/t)) / t
$$x =\frac{ 10  +  (\frac{-9.8 + - 40x}{t}) } {t}$$
becomes
$$x=\frac{-9.8+10t}{t^2 + 40}$$
(if I did my algebra correctly)

*i'm not 100% this is correct, its been a while since i've done physics stuff like this. check it.

anonymous · Answer

i messed up somewhere... 

...the spring makes it a cycle.... and I was using linear variables when I should be using rotational motion. so, w, alpha and theta...

anonymous · Answer

haha, then that makes sense why I have no idea what else to do, i don't know differential equations ^_^

anonymous · Answer

Thank you for trying. This helps me at least to head in the right direction.