Graphical analysis of a damped harmonic system

[freshcoast]

Homework Statement

Homework Equations

http://en.wikipedia.org/wiki/Harmonic_oscillator#Damped_harmonic_oscillator

The Attempt at a Solution

Part a)
I believe to find the mass I can use the equation
$$ T = 2pi * sqrt( k / m ) $$

with T = 0.6s

Part b)

I am confused on this part, I feel like I am to use one of the equations I posted above but I don't know what to set the equations equal to so I am having a hard time solving for the b.

I believe I won't be able to continue without finding the previous parts.

 

[ehild]

Use the equation 15.42. The amplitude is an exponentially decreasing function of time. Compare the amplitudes in the picture. What is the ratio of the amplitudes at 0.25 s and at 3.5 s?

ehild

 

[freshcoast]

Hmm. I see, Since I know two amplitudes and two times, ill call amplitude at t1 A, and t2 B, which will give me 2 equations

Ae^-bt1/2m = Be^-bt2/2m I just have an algebraic question, When I take the ratio of the two amplitudes and the ratio of both the e functions and use the ln function to get rid of the "e's", the "b's" in the equation, do they cancel each other out? or does it become one variable and I am to solve for that?

 

[ehild]

"Amplitude" means the factor in front of the cosine term, the deviation of the maximum or minimum from the zero value. At t_i,
A=x_me^-bt₁/2m.
At t2,
B=x_me^-bt₂/2m
Take the ratio A/B

A/B=e^{-b/2m (t₁-t₂)}

Take the logarithm of both sides. The b-s do not cancel out.

Plug in A, B, t1, t2: you find b/2m, but the approximate value of m is known already.

ehild

 

[freshcoast]

Oh I see, I didn't know the time was going to equal t2-t1 instead of t2/t1.

Anyways continuing on.

part c) would since I found b, the equation would just be like the one in (15-42) but plug in the "b" that was found?

part d) Since I know the dampening factor now, I can compare two equations using equation (15-42) at two different periods with two different amplitudes?

part e) I think I have solved this one, but I know I am to just take the slope at 3s, which will give me the angular velocity and from there I can just choose another spot where the slope is 0 to find the angular acceleration. I think I may have to clarify to the teacher whether or not he is looking for the angular velocity/acceleration or linear.

 

[ehild]

Oh I see, I didn't know the time was going to equal t2-t1 instead of t2/t1.

The exponents subtract if you divide two exponentials: e^x/e^y=e^x-y.

Anyways continuing on.

part c) would since I found b, the equation would just be like the one in (15-42) but plug in the "b" that was found?

yes, plug in everything you have found.

part d) Since I know the dampening factor now, I can compare two equations using equation (15-42) at two different periods with two different amplitudes?

Compare two subsequent maxima. How do you get the energy of the system, knowing the amplitude?

part e) I think I have solved this one, but I know I am to just take the slope at 3s, which will give me the angular velocity and from there I can just choose another spot where the slope is 0 to find the angular acceleration. I think I may have to clarify to the teacher whether or not he is looking for the angular velocity/acceleration or linear.

The problem clearly asks acceleration and velocity. Look out, at what stage is the vibration at t=3 s?

ehild

 

[Chestermiller]

Plot a graph of the natural log of the maximum and minimum amplitudes of the oscillation as a function of time. Fit a straight line to the data. The slope of this straight line should be -b/(2m). Using this plot will allow you to include much more of the data in determining the value of b/(2m), thus reducing the uncertainty.

Your equation for the period is incorrect. It should be:

T=\frac{2π}{\omega '}=\frac{2π}{\sqrt{\frac{k}{m}-\frac{b^2}{4m}}}

Therefore,

\frac{k}{m}-\frac{b^2}{4m}=(\frac{2π}{T})^2

You now have two equations in two unknowns to solve for b and m.

 

[freshcoast]

Compare two subsequent maxima. How do you get the energy of the system, knowing the amplitude?

I see, I believe I am to take the 1/2kx² of two subsequent maxima.

The problem clearly asks acceleration and velocity. Look out, at what stage is the vibration at t=3 s?

Do you mean the amplitude? at 3s I believe it looks like it equals to 0. I should've mentioned the points I will be looking at is the time a little before and after 3s.

Plot a graph of the natural log of the maximum and minimum amplitudes of the oscillation as a function of time. Fit a straight line to the data. The slope of this straight line should be -b/(2m). Using this plot will allow you to include much more of the data in determining the value of b/(2m), thus reducing the uncertainty.

Hmm, I tried this but I am not given a number for a natural log of a negative number?

Your equation for the period is incorrect. It should be:


T=2π ω ′ =2π k m −b 2 4m − − − − − − − √


Therefore,


k m −b 2 4m =(2π T ) 2


You now have two equations in two unknowns to solve for b and m.

I don't see which one is my second equation?

 

[Chestermiller]

Hmm, I tried this but I am not given a number for a natural log of a negative number?

I don't see which one is my second equation?

Just plot the natural logs of the absolute values. How did the graph look with just the natural logs of the maxima plotted? Was it a straight line? Now include the points for the natural logs of the absolute values of the minima.

The second equation is

-b/(2m) = slope of the plotted line

 

[ehild]

I see, I believe I am to take the 1/2kx² of two subsequent maxima.

Take the ratio of them.

Do you mean the amplitude? at 3s I believe it looks like it equals to 0. I should've mentioned the points I will be looking at is the time a little before and after 3s.

Yes, the acceleration is about zero, as the object goes through its equilibrium point at t=3 s. What is the speed at the equilibrium point of an SHM?

ehild

 

[Chestermiller]

Plot a graph of the natural log of the maximum and minimum amplitudes of the oscillation as a function of time. Fit a straight line to the data. The slope of this straight line should be -b/(2m). Using this plot will allow you to include much more of the data in determining the value of b/(2m), thus reducing the uncertainty.

Your equation for the period is incorrect. It should be:

T=\frac{2π}{\omega '}=\frac{2π}{\sqrt{\frac{k}{m}-(\frac{b}{2m})^2}}

Therefore,

\frac{k}{m}-(\frac{b}{2m})^2=(\frac{2π}{T})^2

You now have two equations in two unknowns to solve for b and m.

I made a typo error in a couple of the equations in the earlier version of the above quote. That error has been corrected here.

 

[Chestermiller]

I plotted up the data as described in my previous postings, and got

slope of plotted line = -0.128

Therefore, b/(2m) = 0.128 s^-1

 

[ehild]

The first question in the problem asks the estimated value of the mass, assuming low damping. So it is safe to use the simple formula k/m=(2∏/T)² to get the mass. Indeed, 2∏/T>>b/2m=0.128.

ehild

 

[Chestermiller]

The first question in the problem asks the estimated value of the mass, assuming low damping. So it is safe to use the simple formula k/m=(2∏/T)² to get the mass. Indeed, 2∏/T>>b/2m=0.128.

ehild

Yes. Thanks. I can see that now. But, at the beginning, it didn't seem obvious to me.

Chet

 

[ehild]

It was suggested: "You may assume the system is lightly damped".

ehild

 

[Chestermiller]

It was suggested: "You may assume the system is lightly damped".

ehild

I guess I was supposed to conclude from this that (b/2m)²<<k/m, but I didn't. In any event, lucky for me, it was not necessary to make this assumption, and the method I suggested gave the correct answer anyway.

Chet