# Homework Help: How do you find the Damping Constant? (SHM)

1. Jan 29, 2005

### $id Hi guys, I am a bit stuck at the moment on this experiment I am doing. I am trying to model an oscillating spring mass which is being damped using air resistance and a circle piece of polystyrene. The equation of this will be in the form d^2y/dt^2 + Rdy/dx + ky/x = 0 I know the mass of the oscillating object, the length of the spring, and the spring constant. Does anybody know how to proceed from here. I know that you have to plot a log graph of some experimental properties such as time needed for the amplitude to halve. I also have the auxiliary equation for this. I really need some clear guidance on this plz Thanks a lot sid 2. Jan 29, 2005 ### da_willem I'm not sure I understand the differentia equation you wrote down, what is x for example: The differential equation for damped harmonic motion is: $$m\frac{d^2 x(t)}{dt^2}=-kx(t)-b\frac{dx(t)}{dt}$$ So the friction force is proportional to the velocity and is always opposite to the direction of the motion. The general solution is: $$x(t)=Ae^{-\frac{b}{2m} t} sin(\omega t + \phi)$$ With $$\omega=\sqrt{\frac{k}{m}-\frac{b^2}{4m^2}$$. So there is an oscillation but but it's amplitude decays exponentially. If you would like to know the damping constant b for example you have to observe with what timescale the amplitude decays, or use the relation for the angular frequency. 3. Jan 29, 2005 ###$id

Im not familiar with the Tex codes used here so please bear with me

x =Ae^(-b/2m)t

How can i use that equation to work out B ( I presume logging it would help) The sin(wt + O) will be like a constant i guess.

Also for the equation relating it to the angular frequency. Could you show some working as to where you got it from?

thanks a lot

4. Jan 29, 2005

### dextercioby

He solved the characteristic equation which is a quadratic.

Daniel.

5. Jan 29, 2005

### da_willem

Well it's not unreasonable to presume the solution will exhibit oscillatory motion while exponentially decaying beacause of damping. So using $$x(t)=Asin(\omega t +\phi)e^{-\alpha t}$$ as a test solution, you can find the frequency and alpha by filling this test solution into the original equation. It's best to do this yourself.

Now the damping constant. Suppose you observe after t seconds the amplitude is only a fraction f of it's orginal value. Then you solve f=exp(-bt/2m) to find b (you know how to do this right?). For a more accurate measurement you measure f at defferent times and make a logaritmic plot.

6. Jan 29, 2005

### $id thanks a lot for that equation I can make different measurements of F coz my data logger is continuously following the motion of the spring mass. 7. Jan 29, 2005 ### Parth Dave Does your data logger measure the motion of the spring system? If it does than when you graph it you should see the periodic motion with the damping involved. It will look sort of like a funnel. If you have good programming skills, you could take the data set that you get from the data logger and than write a program to isolate all the maximum points. Than when you graph them, you will get an exponential decay graph. You can then just regress this to find your exponential function. And using the knowledge you got from your differential equation, you can get a very accurate reading of the damping constant. And even if you aren't that great at programming, you can still take about 20 maxima that are evenly spread out and do the same regression. 8. Jan 29, 2005 ### saltydog LaTex Dude, you need to get into LaTeX. Check out the site: https://www.physicsforums.com/showthread.php?t=8997 9. Jan 29, 2005 ###$id

I already have that curve you are describing could you please describe in detail how i work out the damping constant from the regression curve, I presume the regression curve includes the damping constant right?

10. Jan 29, 2005

### da_willem

$$ln(x(t))=ln(Ae^{-\frac{b}{2m} t} sin(\omega t + \phi))=\frac{-bt}{2m} + ln(A)+ln(sin(\omega t + \phi)))$$

The ln(A) will be just an offset, and the ln of a sine will wiggle around the line -bt/2m. Find the slope of this line and you've got b.

11. Jan 29, 2005

### \$id

Thanks

but do i plot ln(X) against t/2m giving a gradient of -b yeah?

12. Jan 29, 2005

### da_willem

That's one possibility. Or just plot x(t) vs t yielding a slope of -b/2m...