Vertical Springs

Vertical Spring problems are more complicated than their horizontal counterparts, but they're really just a more general case. In fact, we can look at springs on any angle of an incline for a continuous range of problems. We begin our analysis with a spring hanging without any mass. The spring length (now unstretched) is $l$. Now, suppose we hang a mass on the spring and then gently let it stretch the spring until equilibrium is achieved, and suppose this stretching to have length $\Delta l$. Then Newton's Second Law says that:

$$\Sigma F_y=-k\Delta l -mg=0,$$

($\Delta l$ is negative) which gives us our expression for the stretch of the spring from the old equilibrium position $\Delta l = -mg/k$. Notice that the new equilibrium position of the system (no force) is different from the old one (no stretch in the spring). The two are the same only when a spring is horizontal.

Now, suppose the spring is further stretched from the new equilibrium position, so that the total stretching is $y$. Then we have

$$\begin{eqnarray*} a &=& F/m\\ \frac{d^2y}{dt^2} &=& -\frac{k}{m}y-g\\ &=& -\om... ...eft(y+\frac{g}{\omega_o^2}\right)\\ &=& -\omega_0^2(y-\Delta l) \end{eqnarray*}$$

where we have set $\omega_0^2=k/m$. Note that we don't know the angular frequency yet - this was simply an algebraic replacement. This last equation suggests that we define a new quantity $\bar y=y-\Delta l$, since derivatives annihilate constants. A quick calculation gives the equation

$$\frac{d^2\bar y}{dt^2}=-\omega_0^2\bar y,$$

which is the usual equation for harmonic motion. Now we can conclude that $\omega_0$ is in fact the angular frequency, and the mass executes oscillations with respect to the new equilibrium position.

Thus, the total effect of having a vertical spring is to shift the equilibrium position down by the amount $mg/k$, so that at equilibrium the spring is under tension and has energy. Make sure you remember this when computing energies.

Finally, note that if we have a spring on an incline, the equilibrium position is determined by

$$\Sigma F_y=-k\Delta l -mg\sin(\theta)=0,$$

where $\theta$ is incline angle. Thus we see that the separation between the new and old equilibrium depends on that angle and goes to zero as the angles does, which is a horizontal spring. Here the two positions are degenerate (the same).