# Simple Harmonic Motion for Mass Attached to Vertical Spring

So I was looking at the case where we have an object of mass $m$ attached to a spring with spring constant $k$. The spring is attached to the ceiling. I was working to come up with an equation of motion for releasing the mass at the point where the spring is relaxed in the case where no load is attached.

Defining the initial point as $y=0$ and taking upwards as the positive direction I was able to obtain,

$$my''=-ky-mg \implies y''+\omega^2y=-g$$

Here $\omega^2=\frac{k}{m}$. This lead to a solution

$$y(t)=\frac{g}{\omega^2}[cos(\omega t)-1]$$

Assuming this to be true, the object would oscillate with the same amplitude for an indefinite period of time. Intuitively, I would think that the gravity would act as a damping force that causes the oscillations to die out but this does not appear to be the case. Does it make sense that the object in this case will oscillate forever?

• I disagree with your solution for $y(t)$ (I get $y(t)=e^{i\omega^2 t}-\frac{g}{\omega^2}$) – Jordan Abbott Jan 16 '18 at 19:26
• Ok, I will have to go back check over my work. Thank you. – Baugh_Mania Jan 16 '18 at 19:26
• Define the shifted variable $Y=\omega^2 y+ g$ and find the EOM for $Y$... – ZeroTheHero Jan 16 '18 at 19:48
• @Ian B. Is the absence of $g$ in your equation for $y(t)$ just a typo? there should also be a relative sign between the two forces acting ( otherwise an equilibrium length may not exist) – CAF Jan 16 '18 at 19:48
• Yes it is. Thank you for pointing that out. In regards to the sign, I assumed both would have a negative sign. If upwards is positive then $F_g=-mg$ but to make the spring force positive at negative values of $y$ and negative for positive values then I figured it would have to be $F_s = -ky$ with $k>0$. – Baugh_Mania Jan 16 '18 at 20:12

## 2 Answers

In your equation, $y$ is the extension from the relaxed or natural length of the spring. If instead you measure instead the extension of the spring from its equilibrium length $y_e$ (where the net force on the mass is zero), you will find the same form of the equation for a horizontal mass on a spring set up. $$-ky - mg = m\ddot y = -k \left( y + \frac{mg}{k} \right) = -k(y-y_e)$$

Now let $Y = y-y_e = y+mg/k = y+g/\omega^2$ and in this shifted variable you obtain $$\ddot Y = -\omega^2Y.$$

Such a result is to be expected because, as noted in the other answer, gravity is a non-dissipative force, it is conservative and as such does not do any net work on the system.

For time decaying oscillations, an inhomogeneous time dependent driving force for example would schematically look like the $y(t)$ derived in the other answer, so that for large enough $t$, the spring is at rest at its equilibrium length $y_e=-g/\omega^2$.

The solution has no damping terms. A damped oscillation would be of the form

$$y'' +2 \zeta \omega_n y' + \omega_n^2 y + g =0$$

Where $k = m \omega_n^2$ and $c = 2 \zeta m \omega_n$ are the stiffness and damping coefficients respectively.

with solution

\begin{aligned} y(t) & = C + \exp(-\beta t) \left( A \sin (\omega t)+B \sin (\omega t) \right) \\ \beta & = \omega_n \zeta \\ \omega & = \omega_n \sqrt{1-\zeta^2} \\ C & = -\frac{g}{\omega_n^2} \end{aligned}

and the coefficients $A$ and $B$ depending on the initial conditions.