# Simple Harmonic Oscillator

1. Aug 30, 2010

### Je m'appelle

I need someone to please verify my work.

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

A particle of mass $$m$$ is suspended from the ceiling by a spring of constant $$k$$ and initially relaxed length $$l_0$$. The particle is then let go from rest with the spring initially relaxed. Taking the z-axis as vertically oriented downwards with it's origin in the ceiling, calculate the position 'z' of the particle as a function of time 't'.

2. Relevant equations

SHO equation:

$$m\ddot{x}(t) = -kx(t)$$

3. The attempt at a solution

In this case, the acting forces on the mass are the weight directed downwards and the elastic force of the spring directed upwards, so

$$m\ddot{x}(t) = mg - kx(t)$$

$$\ddot{x}(t) = g - \omega^2 x(t)$$

$$\ddot{x}(t) + \omega^2 x(t) = g$$

It's important to note that I'm assuming 'x' as the displacement of the spring and NOT the displacement of the mass.

$$\ddot{x}(t) + \omega^2 x(t) = g$$

$$x(t) = x_p(t) + x_h(t)$$

The homogeneous solution will be

$$x_h(t) = acos(\omega t) + bsin(\omega t)$$

And the particular solution will be

$$x_p(t) = \frac{g}{\omega^2}$$

By using the initial conditions $$x(0) = 0, \dot{x}(0) = 0$$ we get

$$x(0) = 0 = a + \frac{g}{\omega^2}$$

$$a = - \frac{g}{\omega^2}$$

$$\dot{x}(0) = 0 = b\omega,\ b = 0$$

So by rearranging we get to

$$x(t) = -\frac{g}{\omega^2}cos(\omega t) + \frac{g}{\omega^2}$$

But notice that this is the displacement of the spring and not of the mass, in order to find the displacement of the mass we would have to add the initial length of the relaxed spring, that is, $$l_0$$.

So the final answer for the displacement of the mass in the z-axis will be

$$z(t) = l_0 - \frac{mg}{k} (cos(\sqrt{\frac{k}{m}} t) - 1)$$

Could someone please verify this, as there is a different answer in a solution manual I found, and I'm not so sure which one is correct.

2. Aug 30, 2010

### diazona

Hm, it seems reasonable to me. What's the other answer you found?

3. Aug 30, 2010

### Je m'appelle

Well, I think it is correct after all.

I had two solutions, mine and a friend's which he claimed to have obtained from a solutions manual, but I think that wasn't the case.

I'm sure now that the answer I've provided here is the correct one, as I've analyzed the case when $$t = 0$$, that is when the system is at rest, the displacement of the mass is simply the length of the relaxed spring which is reasonable.

Thanks anyway diazona

4. Aug 30, 2010

### diazona

Ah. Clearly the real lesson of this problem is that you should never trust your friends :tongue2: (just kidding of course)

5. Aug 31, 2010

### Terocamo

Are you sure it is not -l0,which direction are you taking it positive?

Last edited: Aug 31, 2010