# Determining the damping constant from the ratio of amplitudes

1. Nov 11, 2013

### richyw

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

I have measured the amplitude between the first and tenth oscillations for a damped harmonic oscillator (spring with a mass attached). I also have measured the period. First question. Is an (underdamped) harmonic oscillator periodic? Like I know that the solution to the differential equation has a sine/cosine argument in it and it does oscillated with a period. But as time goes on the exponential decreases its amplitude. So can we really call this motion "periodic"?

Anyways my main question is how do I find the damping constant? In the equations section i'll put what my lab manual gives

2. Relevant equations

$$\frac{b}{2M}T=\ln\left(\frac{x_1}{x_2}\right)$$ where b is the damping constant, M is the mass, T is the period and $x_1$ and $x_2$ are the amplitude of the first and second oscillations respectively

3. The attempt at a solution

This makes sense to me, except for one thing. I do not understand what to do if I measured the 1st and 10th amplitude instead of the first and second. What is the proper way to handle this. The two things I can think of would be to divide the ratio I have by 9, and then just plug it in. I also thought I could use the ratio I have and then multiply the period of one oscillation (which I measured independently) by 9. Neither really makes sense to me though TBH.

2. Nov 11, 2013

### nasu

[STRIKE]They both make sense.[/STRIKE]

$x(t)=A\cos (\omega t) e^{-\frac{b}{2m} t}$
and take t=0 and t=10 T.
Then calculate the the ratio between x(0) and x(10T).
The cos part is 1 for both values of time.

Last edited: Nov 11, 2013
3. Nov 11, 2013

### richyw

I can't exactly read what you wrote. The LaTeX is broken I think. If they both make sense then why do they give me different answers? In fact if I do it the first way, I get a negative damping constant...

4. Nov 11, 2013

### nasu

You are right. I did not think. Sorry for the confusion.
Only the one with 10 T makes sense. As it results from the method I described in the previous post.

The Latex looks OK for me.
But without, the equation is
x(t)=Acos(omega*t)*exp[(-b/2m)*t]

5. Nov 11, 2013

### richyw

ah, thanks. This makes sense now.