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Skeptic
I said damped oscillator!
So far, we have the general equations for the motion of an oscillator:
y = ymax cos[ (2π/T) t ]
,
v = -vmax sin[ (2π/T) t ]
vmax = 2πymax/T
a = -amax cos[ (2π/T) t ] amax = ymax(2π/T)2 = 2πvmax/T
So, here’s the rub. Friction is always present in the macroscopic world, so we
need to see if we can model an oscillator with friction. If we observe a real
physical oscillator for several minutes, what motion parameters change?
(circle)
Period
amplitude or max displacement
spring constant
mass
{Class Discussion}
From the demonstration in class, sketch the position function for an oscillator with
friction over the top of the graph below. This is what we call a damped oscillator.
In actual practice, this happens with any energy drain on an oscillator, whether it
is friction or any other energy loss. In the next few pages, we will explore
exponential decay and how to use that idea to model various decreases that we
will see in the systems we study.
1
Exponential decay:
Suppose we have a radioactive isotope. The equation for the number of nuclei at
any time is then
N = No e-(t/τ)
A plot of this with No = 106 looks like:
Exponential growth
number
1000000
500000
0
0
5
10
15
20
25
time
At t=5s, N = No/2 so 1/2 = exp-(5/τ)
2 = exp(5/τ)
ln(2) = ln[exp(5/τ)] = 5/τ
τ ln(2) = 5
τ = 5/ln(2) = 7.213 where the 5 is the half life (N = No/2)
The time constant, τ, in the exponential is equal to the halflife, t1/2, divided by
ln(2).
τ = t1/2/ ln(2)
To check, use the 5 second halflife from above in the time constant:
N = No/4 = No exp(t/(5/ln(2)))
1/4 = exp-(t/(5/ln(2)))
ln(1/4)=ln[exp-(t/(5/ln(2)))] = -t/(5/ln(2))
-ln(4)= -t ln(2)/5
2ln(2)= t ln(2)/5
{ ln(ab) = b ln(a)}
2 = t/5
t = 10s , as expected.
Note that the value of N halves every 5 seconds in the plot above, regardless of
the value. If the number is 20, 5 seconds later it will be 10. If the number is 5
million, 5 seconds later it will be 2.5 million.
2
This is an important property of exponential curves, the value halves (or doubles
for exponential growth) after a fixed time, called the half life (or doubling time).
This is true no matter what the current value of the function is.
Another important property is that the slope is proportional to the current value.
This is often the way an exponential dependence of a value is identified, that its
rate of change is proportional to its value.
If you draw a smooth curve through the crests of the position function, does it
seem to have a halflife? If so, it is probably an exponential function. Below is a
data set for an oscillator. See if you can determine the halflife and write the
exponential function with that halflife.
Exponential = A e-(t/τ) τ= t1/2/ln(2)
A = _______ t1/2 = ________ A e-(t/τ) = _______________________
This becomes the new amplitude for your cosine function. To get the rest,
determine the period and then use our expression for the cosine, cos[ (2π/T) t ] .
T = _______s
Cosine term =
cos[
t]
Putting it all together, A e-(t/τ) cos[ (2π/T) t ] =
3
Energy
Consider the spring and mass oscillator. Your equations and tools are:
y = ymaxcos[(2π/T)t]
ω = 2π/T = √[k/m]
v =- ω ymaxsin[(2π/T)t]
vmax = ω ymax = √[k/m] ymax
a = -(k/m)y = -
ω2
y
= -(k/m) ymaxcos[(2π/T)t] amax = ω2 ymax = (k/m) ymax
T = 2π/ω = 2π√[m/k] note the 2π here!
Write the expressions for kinetic energy and for potential energy in a spring.
What is the maximum potential energy in a spring and mass oscillator with
maximum displacement ymax ?
Let the maximum kinetic energy in such an oscillator be KE = ½ m vmax2.
In an ideal oscillator with no friction, etc., how does this value compare with the
maximum potential energy you wrote down?
equal
or
= ________ times as much
Use this relationship and the two expressions to solve for the maximum velocity
in terms of k,m and ymax. Compare with the results above; is everything
consistent?
Finally, on the graph of position of the mass below, indicate by shading the bars
the amount of potential energy and the amount of kinetic energy at the times
where they are located. Maximum is the entire bar shaded.
4