# Tricky spring on a surface 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)

• Looks like a straight homework question, so I've tagged it as such. Please feel free to help, but don't answer the question fully/directly. Jan 16, 2011 at 15:52
• You say there is no mass attached to the spring. If that is so then the mass is distributed along the length of the spring. So rather than a simple differential equation, you have a partial differential equation for the position and velocity at all points along the spring as a function of time. You want to solve for the lowest eigenvalue mode. Assume time dependence is exp(iomegat), solve for the x dependence of displacement and velocity. Only certain discrete frequencies, (eigenvalues) will satisfy the boundary conditions. Jan 16, 2011 at 15:59
• @Noldorin It is no homework question. and I would like you to restore it as such. if you want proof, I don't mind it being answered later, but I'd prefer to know it sooner as I can't stop thinking about it. I don't mind if it's just help, but I don't like inappropriate tagging. Jan 16, 2011 at 16:50
• @Noldorin I prefer answers that explain things, my solution is 0.1secs but I want to fully understand it. @Omega Centauri I have no clue what a eigenvalue is? Jan 16, 2011 at 17:25
• The force on the bottom of the spring until it takes off is 0=(Normal force)-mg+kx, when it lifts you have (Normal force)=0, that gives you x, you get acc for center of mass by F=ma, solve for t by standard motion formula t=f(x,a). Jan 16, 2011 at 17:32

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.

• I'm still gradually decomposing each step. Jan 19, 2011 at 9:58
• so the period of oscillation at the start where at length deviation of 0.05m, x1 = 0.05*2/3, so angular velocity at the start is: w1 = sqrt(2g/x1) = sqrt(600), so the period of the wave at the start is T=2Pi/w1, T=2Pi/sqrt(600), and I need 1/2T, so the answer is t = Pi/sqrt(600), which is 0.128s I'm probably missing the added acceleration over this period of time... but otherwise I'm just plain stuck... Jan 19, 2011 at 12:08
• @Yuri: There were some errors in the solution. It should be correct now. What exactly do you not understand? Jan 20, 2011 at 6:21
• w0 is the natural frequency, right? I found that w0 = (1/2Pi)*sqrt(mass/k), so where did you get the ω0^2=3k/m from? Jan 20, 2011 at 8:17
• @Yuri: $\omega_0^2=\frac{3k}{m}$ is angular frequency($\frac{rad}{s}$). It follows from the equation of the motion of the spring. You are talking about temporal frequency(herz). Jan 20, 2011 at 9:00

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.