Download Damped Harmonic Oscillator with Applied Force

Damped Harmonic Oscillator with Applied Force
How will the mass on a spring in the presence of damping respond to an applied force?
Let's be specific and apply a force: Fapplied = Fo cos(at + o) .
Our differential equation is now: -kx + -bv + Fo cos(at + o) = ma , or
ma + bv + kx = Fo cos(at + o) . We recognize this as an inhomogeneous second
order linear differential equation. The homogeneous equation has the solution we saw
before:
xh(t) = A e-t cos(1t + ) where  = -b/2m; 1 = [o2 - 2]1/2 ; o = [k/m]1/2 =
frequency of the undamped, unforced mass on a spring; A is the amplitude and  is the
initial phase - both of which are dependent on the initial conditions (xo and vo).
The complete solution to the complete (inhomogeneous) equation is a combination of the
homogeneous equation (which includes constants for the initial conditions) and the
inhomogeneous equation: x(t) = xh(t) + xi(t) .
Straightforward method
To find the inhomogeneous solution, xi(t), we can use our physical intuition:
xi(t) = h cos(at + ) , that is, we might expect the forced motion to oscillate with
some amplitude, h, at the same frequency as the applied force, a, but with some possible
phase difference, .
To ease the burden of subscripts, I will use  = a in the equations below.
To see if our guessed at solution works, and if it does to find h and , we simply put our
solution into the differential equation and see if it works:
ma + bv + kx = Fo cos(t + o) with xi(t) = h cos(t + ) gives:
-m2h cos(t + ) - bh sin(t + ) + kh cos(t + ) = Fo cos(t + o) .
In this form, ti is hard to see if this equation will be satisfied for all values of t. To get it
in a more useful form, we need to apply a couple of trig identities:
cos() = cos() cos() - sin() sin() and sin() = sin() cos() + cos() sin() .
Using these, our equation becomes:
-m2h [cos(t) cos() - sin(t) sin()] - bh [sin(t) cos() + cos(t) sin()]
+ kh [cos(t) cos() - sin(t) sin()] = Fo [cos(t) cos(o) - sin(t) sin(o)] .
For this equation to be true for all time, we need the coefficient of cos(t) to be zero
AND the coefficient of sin(t) to be zero. This gives the following two equations:
For cos(t):
-m2h cos() + -bh sin() + kh cos() = Fo cos(o) and
For sin(t):
+m2h sin() + -bh cos() - kh sin() = -Fo sin(o) .
This gives two equations for two unknowns: h and  . We can do this, but the algebra is
somewhat messy since  is inside both sine and cosine functions.
Exponential Method
Let's try writing the problem this way: Fo cos(t + o) = Re[Fo eio eit ]
And further let F be a complex number: F = Fo eio , so that
Fo cos(t + o) = Re[F eit ]
We want to find x(t) = Re[h ei eit ] , and as we did above for force, let x be a
complex number: x = h ei , so that
x(t) = Re[x eit ] .
Substituting these expressions for x(t) and F(t) into our differential equation:
ma + bv + kx = F(t) gives -m2x - bx + kx = F
which gives an express for x
x = F / [-m2 + ib + k] , or dividing through by m (and recalling that k/m = o2):
x = (F/m) / [-2 + ib/m + o2] .
We note that the denominator is a complex number (as are x and F ). Let's call the
denominator W ei where the magnitude of the complex number (W) is equal to the
square root of the real part squared plus the imaginary part squared:
(Recall that a+ib = r ei where r = [a2+b2]1/2 and  = tan-1(b/a) .)
W = [(o2-2)2 + (b/m)2]1/2 and  = tan-1 [(b/m) / (o2-2)] .
Using this we now have:
x = (F/m) / W ei = (Fo eio e-i) / (mW) = (F/mW) ei(o-) . This now gives:
x(t) = Re[x eit] = Re[ (F/mW) ei(o-) eit] = Re[(F/mW) ei(t+o-) ]
x(t) = {Fo / m[(o2-2)2 + (b/m)2]1/2 } cos(t + o - )
where  = tan-1[(b/m) / (o2-2)] .
In terms of =(b/2m), this can be written as:
x(t) = {Fo / m[(o2-2)2 + 422 ]1/2 } cos(t + o - )
where  = tan-1[(2) / (o2-2)] .
This is what we would have gotten from all the algebra in the first method.
The full solution is this solution added to the homogeneous solution. Note that this
inhomogeneous solution has no arbitrary constants that can be fit to the initial conditions
- the homogeneous solution does.
Analysis
Note that as  approaches o, that the amplitude of the resultant oscillation increases.
The biggest amplitude results when  = o, and this is called being at resonance. The
smaller the (b/m) or , the higher the resonant amplitude.
The phase difference between the applied force and the resulting motion, , also depends
on the frequency. Recall:  = tan-1[(b/m) / (o2-2)].
For  < o,  > 0 but less than 90o, which means the motion lags a little behind the force.
For the case where  is small, the amplitude approaches the constant value of Fo/mo2 =
Fo/k .
For  = o,  = 90o, which means the motion lags behind the force by 90o.
For  > o,  > 90o, which means the motion approaches being out of phase with the
force. For the case where  is large, the amplitude approaches Fo / m2 which
approaches zero.
Homework Problem: Problem #11: Find x(t) and for a particle of mass, m,
subject to a spring force of –kx (but no damping force) and an applied force, F(t)
= Fo sin(t).
Power
Power is defined to be Work per time, and Work is force thru a distance:
P = dW/dt = d [F dx] dt = Fv .
For electrons bound to atoms, we can consider the electrons to be in a potential well, and
hence the binding force acts like a spring force, -kx. We can consider that any energy the
electrons receive might be radiated away or transferred to the atom as a whole and this
would act like a resistive force, -bv. What happens to the atom in this model if we apply
an oscillating electric force by applying an oscillating electric field (electromagnetic
radiation)?
E(t) = Eocos(t) so F(t) = -eEocos(t) . Hence, our Fo = -eEo .
We can use the previous results to see the x position of the electron in our model will be:
x(t) = A e-t cos(1t + ) + (-eEo/m) cos(t - ) / [(2-o2)2 + 422]1/2
where  = tan-1[(2) / (o2-2)] .
To get the power absorbed by the atom, we need to calculate the velocity of the electron.
This is simply dx(t)/dt . We can save ourselves a little time if we recognize that the
homogenous term of x(t) has a dying exponential in it (e-t), and any derivative of this
homogenous term will still have that dying exponential in it. This term depends on the
initial conditions of the atom that should not be important. Thus the only steady-state
velocity will be that due to the inhomogeneous term. Therefore,
vsteady-state (t) = dxi(t)/dt = (+eEo/m)  sin(t - ) / [(2-o2)2 + 422]1/2 .
The power now is Fv:
Power = (-eEo) cos(t) * (+eEo/m)  sin(t - ) / [(2-o2)2 + 422]1/2
= {-e2Eo2 / m[(2-o2)2 + 422]1/2 } cos(t) sin(t-) .
To find the Average Power, <P>, we need to find the average of cos(t) sin(t-). To do
this, we can use a trig identity for the sine of the difference of two angles:
sin(t-) = sin(t)cos() - cos(t)sin()
so that cos(t) sin(t-) = cos(t) sin(t) cos() - cos2(t) sin() .
But the time average of cos(t) sin(t) is zero, and the time average of cos2(t) is ½.
Using these time averages gives:
<Power> = {-e2Eo2 / m[(2-o2)2 + 422]1/2 }*{0 - ½ sin()}
<Power> = +½ e2Eo2 sin() / m[(2-o2)2 + 422]1/2 .
Recall that tan() = y/x and sin() = y/[x2+y2]1/2 . Using  = tan-1[(2) / (o2-2)], we
can express sin() as sin() = (2) / [(2-o2)2 + 422]1/2 . Thus we finally get:
<Power> = +e2 Eo2 2  / m[(2-o2)2 + 422] for each electron.
We see that for maximum power absorption, =o ; we also see that the power
absorption goes to zero for both  going to zero (due to 2 in the numerator) and for 
going to infinity (due to 4 in the denominator).
Dispersion of light
A glass prism and rain can break light into its component colors. This is due to the index
of refraction changing slightly with frequency. Why is that?
Light is bent on going from one medium to another due to the differences in the speed of
light between the mediums. We use the index of refraction to express the speed (n=c/v)
and we use this in Snell's Law to show how the light is refracted (bent). The speed of
light depends on the medium: v = 1/[]1/2 , so n = c/v = []1/2 . For most
materials, , so the expression for n becomes: n = []1/2 . Recall from Physics II
that in materials, =Ko where K was the dielectric constant that depended on the
stretchability of the charges. In terms of field values: D = electric displacement = E =
oE + P, where P is the dipole moment per volume, and the dipole moment is defined to
be (for the electron) -ex, where x is the amount stretched from the neutral position.
Hence P = -Nex where N is the number of electrons per volume. Putting this altogether
for  gives:  = o + (-Nex)/E which in term gives: n = [/o]1/2 = [1 + (-Ne/oE)x]1/2 .
If we now substitute our inhomogeneous solution for x (since the homogeneous part will
die out due to the e-t term) due to the oscillating applied Electric Field (the radiation), we
get for n:
n = [ 1 + {-Ne/oEocos(t)}*{-eEo/m[(2-o2)2 + 422]1/2}cos(t-) ]1/2 .
If we use the trig identity: cos(t-) = cos(t)cos() + sin(t)sin() we get for n:
n = [1 + {Ne2 / om[(2-o2)2 + 422]1/2}cos() +
{Ne2 / om[(2-o2)2 + 422]1/2}sin() tan(t) ]1/2 .
Recall that tan() = y/x and cos() = x/[x2+y2]1/2 , and recall that tan() = (2)/(o2-2)
so that cos() = (o2-2)/ [(2-o2)2 + 422]1/2 . Also realize that the term with tan(t)
in the expression for n will oscillate about zero. If we neglect the oscillating term, we get
for the index of refraction:
n = [1 + {Ne2(o2-2) / om[(2-o2)2 + 422]}]1/2 .
From this we see that the index of refraction depends on the frequency of the light. This
leads to the fact that light of different frequencies will travel at different speeds through
the material. This is called dispersion. (Note that dispersion does not take place in
vacuum, only in materials!) This also leads to the fact that light of different frequencies
(colors) will be bent slightly differently in going from one medium into another! Also,
note that for <o, that the index of refraction is greater than one and increases as the
frequency gets bigger. Also note that when the frequency of the incident radiation gets
close to the resonant frequency, the power absorbed increases making the material
essentially opaque to the light. It is only when the incident radiation has a frequency well
below the resonant frequency that the material will be transparent for Snell's Law to
really be applicable.