What is the Distance Between Two Masses When a Spring is Fully Stretched?

[cissablecat23]

A mass, m1 = 9.04 kg, is in equilibrium while connected to a light spring of constant k = 122 N/m that is fastened to a wall
A second mass, m2 = 7.31 kg, is slowly pushed up against mass m1, compressing the spring by the amount A = 0.220 m The system is then released, and both masses start moving to the right on the frictionless surface. When m1 reaches the equilibrium point, m2 loses contact with m1 and moves to the right with speed v. Determine the value of v.

potential energy= kinetic energy
0.5kA^2=0.5mv^2
v= sqrt (kA^2/(m1+m2))
v=0.601 m/s

How far apart are the masses when the spring is fully stretched for the first time?

This would involve finding the time that this occurs in order to get the distance. With the variables I have.. i don't know what to do next.. i have speed.. and masses..and an amplitude..Help?

 

[Leong]

Simple Harmonic Motion, Newton's 2nd Law, Hooke's Law, Law of Conservation of Energy

New amplitude of the block- m1-spring system :
\frac{1}{2}m_{1}v^2=\frac{1}{2}kA^2
A=0.164 m with v is the speed of m1 and m2 at the equlibrium point.
Simple harmonic motion of the system :
x=Asinwt with A=0.164 m
-kx=m_{1}a
a=-\frac{k}{m_{1}}x
a=-w^2x
w=\sqrt{\frac{k}{m_{1}}}
x=Asinwt
Left direction : Positive
When m1 is fully stretched for the first time, x=0.164 m, then
sin\sqrt{\frac{k}{m_{1}}}t=1
t=0.428 s
m1 takes 0.428 s to be fully strecthed relative from the equilibrium point.
At this time interval, m2 has gone s=ut=0.257 m from the equilibrium point. so the distance between the two masses is 0.257m -0.164 m =0.093 m

 

[wais]

To find the distance between the masses when the spring is fully stretched, we can use the equation for the displacement of a mass on a spring: x = A*cos(ωt), where A is the amplitude, ω is the angular frequency (ω = √(k/m)), and t is the time.

Since we know the amplitude (A = 0.220 m) and the angular frequency (ω = √(k/m) = √(122/9.04+7.31) = 3.641 rad/s), we can plug these values into the equation to find the time when the spring is fully stretched.

0.220 = 0.220*cos(3.641t)
cos(3.641t) = 1
3.641t = 0
t = 0 seconds

Therefore, the spring is fully stretched at t = 0 seconds. To find the distance between the masses at this time, we can plug in t = 0 into the equation for displacement:

x = A*cos(ω*0)
x = 0.220*cos(0)
x = 0.220*1
x = 0.220 m

Therefore, the masses are 0.220 m apart when the spring is fully stretched for the first time.