Tricky spring on a surface question

Question

I have this relative simple-looking question that I haven't been able to solve for hours now, it's one of those questions that just drive you nuts if you don't know how to do it. This is the scenario:

I have a spring that is on a flat surface, the springs details are like this:

spring constant = 100N/m

height = 0.1m

mass = 0.5kg

g = 10m/s^2

there is nothing attached to the spring. The initial force exerted on the surface is 5N. I compress the spring halfway until the force exerted on the surface is double, now 10N and then let it go.

The (simple) oscillation starts, and at one point the force exerted on the surface will be 0N (weightless).

I need to find out how much time has passed after letting it go, and reaching weightlessness.

as in: 10(N)---time--->0(N)

p.s. not homework, read comments.

Answer 1

I assume for simplicity that the spring constant has a quite a high value so that the settling down of the spring under its own weight is insignificant.

Designations:

$x$-vertical displacement of the center of mass of the spring from its equilibrium position.
$l$-vertical displacement of the top of the spring from its equilibrium position.
$m$-the mass of the spring.
$k$-the spring constant.
$g$-gravitational acceleration.

First of all let's highlight the following relation: $$x=\frac{2}{3}l$$ Its derivation is elementary but too long to present here.

The next step is write down the equation of the conservation of energy:
$$m\frac{\dot{x}^2}{2}+\frac{3}{2}kx^2+mg(x_0-x)= \frac{3}{2}kx_0^2=const$$ $x_0=x(0)$ is an initial displacement of mass center of the spring from its equilibrium position. After differentiating with respect to $t$ we get the equation of the motion of the center of mass of the spring:

$$\ddot{x}+\frac{3k}{m}x-g=0$$ According to initial conditions $x(0)=x_0= \frac{2}{3}l_0$ and $\dot{x}(0)=0$ the solution of this equation:
$$x(t)=\frac{g}{\omega_0^2}+\left(x_0-\frac{g}{\omega_0^2}\right)cos(\omega_0t);\omega_0^2=\frac{3k}{m}$$ At the moment of the departure from the ground the following holds:

$$-mg=kl=\frac{3}{2}kx$$or $$x=-\frac{2g}{\omega_0^2}$$ Minus sign indicates that a vertical coordinate is above the equilibrium. Thus, the time we are looking for is:

$$t=\frac{1}{\omega_0}arccos\left(-\frac{3g}{-g+x_0\omega_0^2}\right)= \frac{1}{\omega_0}\left(\frac{\pi}{2}+arcsin\frac{3g}{x_0\omega_0^2-g}\right)$$ The formula has a meaning if
$$x_0>\frac{4g}{\omega_0^2}$$ I would point out the assumption at the top of the post! For the given data this is probably not a good assumption. But as a first approximation maybe it fits.

Answer 2

Basically at a point on the spring where y is displacement from the equilibrium condition, you'll get a differential equation $d^2/dt^2 (Y \times density) = -d^2Y/dZ^2 \times k$ the spring constant. (sorry I can't use Latex)! If we postulate that solutions look like $e^{ikZ +i\omega t}$, $ 2\pi\omega$ will be the frequency. Plug in $\omega$ or $k$ and you can solve for the other one.

Then at $Z$=height of zero the boundary condition is displacement =0, this will imply that the spatial part of the solution looks like $\sin{(kx)}$. At the top of the spring $dY/dZ$ must be zero, else there would be an unbalanced force, that means that $kZ$ must be an odd multiple of $\pi/2$. The $\omega$ values that satisfy these conditions are your eigenvalues.

Next you need to discover the amplitudes for the infinately number of modes excited. i.E. your initial displacement as a function of Z is proportional to Z, and you must find Ai such that $$Z=\sum_{i= odd N} Ai\cos{(ikZ)} (k*.1 = \pi/2)$$. You may need to lookup Fourier analysis to do this, but you should get a simple formula for all the Ai. The solution at any future time will be the sum of these (note each term as its own frequency dependence). You should discover that when time is $\pi$ times the lowest eigenvalue, you've reversed the value of F everywhere, so that will be your answer.

Undergrad physics should teach solving the wave equation, and show you how to apply these methods.