# Dynamics of a Vertical Mass-Spring Simple Harmonic Oscillator with Gravity

I am having some trouble obtaining the elastic potential energy and gravitational potential energy of a simple mass spring system.

In this experiment, masses attached to a spring were dropped from a position in which the spring had not be extended. For example, the 20g mass had an equilibrium position of $y=-34.00cm$ and it reached a maximum vertical displacement of $y=-35.83cm$.

Based on this, I found the amplitude of oscillation to be $17.92cm$

Then, using the spring constant and the mass, I determined the natural frequency: $\omega=21.21 rad{\cdot}s^{-1}$

I was able to create the following inital value problem: $$y(t)=c_{1}\cos({\omega}t) + c_{2}\sin({\omega}t)$$ $$y(0)=0$$ $$A=17.92\times 10^{-3}m$$

I solved it by To begin the solution considering the case $y=0$: $$y(t)=c_{1}cos(({\omega})(0)) + c_{2}sin(({\omega})(0))$$

$$y(0)=c_{1}$$

$$0=c_{1}$$

Now, I used the amplitude to determine that $c_{2}=17.91\times 10^{-3}$

Skipping a few simple steps, I created to the following function:

$$y(t)=17.92\times 10^{-2}\cos(21.21t)-17.92\times 10^{-2}$$

Now, onto the elastic potential energy,

$$E_{e}=\frac{k\times y(t)^2}{2}$$

$$E_{e}=\frac{(9)(17.92\times 10^{-2}\cos(21.21t)-17.92\times 10^{-3})^2}{2}$$

$$E_{e}=0.15-0.29 \cos(21.21 t)+0.14 \cos^2(21.21 t)$$

This function does not at all resemble what it should look like, a simple periodic function.

I have a feeling that my problem is due to my assigned coordinate system.

Any help at all would be immensely appreciated.

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## 2 Answers

I am a little confused as to how you found your amplitude of oscillation. The amplitude is the maximum distance from the equilibrium position. If you say that the equilibrium position is $-34.00$cm, and the maximum vertical displacement is $-35.83$cm, then the amplitude is $1.83$cm.

It appears that you are confusing $y$, the height you measured from some arbitrary point (like the table, or the ground), with the distance from the equilibrium height, $x = y - y_{eq}$. Your working is also suspect. In your derivation you say $0 = c_1$, yet in the next line you say $c_1 = 1.791 \times 10^{-3}$. You should recheck and make sure everything you do makes sense.

Anyway, to answer your question about the potential energy not being a 'simple periodic function', it does not satisfy the same simple harmonic equation $d^2x/dt^2 = - \omega^2 x$that the distance from the equilibrium point does. However, it Is still periodic, but with half the period, or twice the frequency, $2\omega$.

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the coordinate system you use is arbitrary, and is valid as long as you keep clear what quantities you are measuring... – nervxxx Jan 17 at 22:19
no. when you say height, you are measuring it from some reference point. Meaningful quantities only come from differences of height. If you dropped the spring from rest 34cm from the equilibrium position (place where the mass comes to a rest/oscillates about), then the amplitude is 34cm. – nervxxx Jan 17 at 22:35
ah when I meant 'come to a rest' I meant after a long long time when damping has removed all its kinetic energy and it stops moving forever. that's the equilibrium point. I didn't mean the height at which it first comes to rest at the bottom of the oscillation, sorry about that. – nervxxx Jan 17 at 23:06
although... I'm a little confused. So you mean to say that you dropped it from y = 0, and the furthest distance it reached was -35.83cm (presumably on the first descent). Then after letting it bounce up and down for many times and when it finally settles down and stops moving completely, it is at y = -34cm? – nervxxx Jan 17 at 23:09
I suggest you redo the experiment, if you can, because the results you have seem inconclusive. if you're stuck with the derivation/math, I'll be happy to help you, but try to sort it out by yourself first. – nervxxx Jan 17 at 23:51

You should try to express the solution as (check that it is a solution of the harmonic oscillator equation)

$$y(t)=A \cos(\omega t+\phi)$$

where $A$ is the amplitude and $\phi$ is the initial phase. Let $t=0$ so

$$y(0)=A\cos(\phi) \rightarrow \cos(\phi)=\frac{y(0)}{A}$$ can you continue from here?

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According to my coordinate system, the vertical displacement at time zero is equal to zero. So, should the phase $\phi$ not also equal zero? If so, my solution is equal to the one described by you. Or, is my coordinate system inherently flawed? – Richard P Jan 17 at 22:12
Remember that $cos(\phi)=0$ means that $\phi=\frac{\pi}{2},\frac{3\pi}{2},...$ I'm going to check the numbers in your first post, be sure to check @nervxxx answer – Nivalth Jan 17 at 22:14
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