# Oscillating spring derivation

Tags:
1. Oct 11, 2011

### Villyer

1. The problem statement, all variables and given/known data

This was a highly simplistic lab we did, and that I have been taking further for my personal enjoyment (I always enjoy deriving things).

The lab was if you have a spring hanging from a hook with a spring constant k, place an object of mass m on the spring so it just barely touches it and exerts no force on it, then drop the spring, what is the maximum stretch of the spring? (We had to place an egg there and see if the mass will just kiss the egg and go back up, but that's irreverent)

So known variables are k, m, and g.

After doing this lab I was thinking of a way to represent the movement with respect to time, and so I got to my derivations.

2. Relevant equations

Eg = mgh
Es = kx2/2
Ek = mv2/2
Fnet = ma
Fg = mg
Fs = kx

3. The attempt at a solution

Throughout this I will be using x to represent the stretch of the spring, which is also the positive displacement from the starting point.

The resting position of the spring is found when the net force on the mass is 0, or when Fs = FG
kx = mg
x = mg/k
(This distance x is given in displacement from the highest point, the point of release)

The lowest distance is when all of the energy from gravity has been transferred into spring energy.
Eg = Es
mgh = kx2/2
mgx = kx2/2 (the height fallen is the same as the stretch of the spring, so h = x)
x = 2mg/k
(This distance is noticed to be twice the resting distance)

The maximum velocity is going to be achieved when the object is at its equilibrium point, since that is when the acceleration switched from positive to negative.
Et = Eg + Ek + Es
mghmax = mghcurrent + mv2/2 + kx2/2
mghmax - mg(hmax - x) - kx2/2 = mv2/2
2mgx - kx2 = mv2
2mg(mg/k) - k(mg/k)2 = mv2
(mg)2/k = mv2
v = g√(m/k)

From here, the two pieces of valuable information I learned were that the maximum velocity was v = g√(m/k), and so the velocity will oscillate between this positive and negative value, and that the displacement will oscillate between 0 and 2mg/k, or between mg/k and -mg/k, depending on your reference point (I choose the latter).

The motion of the spring is sinusoidal (or so I assume, I honestly don't know. This may be the flaw in my work, as I have no backing to this claim) and since the starting position of the mass is at mg/k, I choose a cosine graph to represent it.
Thus x(t) = (mg/k)cos(u), where u is my function of time. The velocity graph will have a maximum value of g√(m/k), and it is the derivative of the graph x(t).
Thus v(t) = -(g√(m/k))sin(u)
x'(t) = -(mg/k)sin(u)du
Since v(t) = x'(t), I set the two equal.
-(g√(m/k))sin(u) = -(mg/k)sin(u)du
g√(m/k) = (mg/k)du
√(m/k) * k/m = du
du = √(k/m)
And if this is true, then u = t√(k/m)

Therefore my graph is (mg/k)cos(t√(k/m)).
This is also supported by the fact that the acceleration oscillates between g and -g, and x''(t) = a(t) = -gsin(t√(k/m)), which oscillates between g and -g.

But my uncertainty about my answer is that the period is vastly different if the mode is degree or radians, so how do I know which is correct? Of course radians produces very realistic answers, but how would I know that they are the real ones? Is the period of this 2$\pi$√(m/k)? I feel like that seems really familiar, but I'm uncertain. I guess I just don't see the connection, how radians can make all the times right.

2. Oct 12, 2011

### Spinnor

You wrote,

"The lab was if you have a spring hanging from a hook with a spring constant k, place an object of mass m on the spring so it just barely touches it and exerts no force on it, then drop the spring, what is the maximum stretch of the spring? "

Do you drop the spring or the mass?

3. Oct 12, 2011

### Villyer

The mass, sorry.