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: F_applied = F_o cos(w_at + q_o) .

Our differential equation is now:

-kx + -bv + F_o cos(w_at + q_o) = ma , or

ma + bv + kx = F_o cos(w_at + q_o) .

We recognize this as an inhomogeneous second order linear differential equation. The homogeneous equation has the solution we saw before:

x_h(t) = A e^-gt cos(w₁t + a) where g = -b/2m; w₁ = [w_o² - g²]^1/2 ; w_o = [k/m]^1/2 = frequency of the undamped, unforced mass on a spring; A is the amplitude and a is the initial phase - both of which are dependent on the initial conditions (x_o and v_o).

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) = x_h(t) + x_i(t).

Straightforward method

To find the inhomogeneous solution, x_i(t), we can use our physical intuition:

x_i(t) = h cos(w_at + b) , that is, we might expect the forced motion to oscillate with some amplitude, h, at the same frequency as the applied force, w_a, but with some possible phase difference, b .

To ease the burden of subscripts, I will use w = w_a in the equations below.

To see if our guessed at solution works, and if it does to find h and b , we simply put our solution into the differential equation and see if it works:

ma + bv + kx = F_o cos(wt + q_o) with x_i(t) = h cos(wt + b) gives:

-mw²h cos(wt + b) - bwh sin(wt + b) + kh cos(wt + b) = F_o cos(wt + q_o) .

In this form, it 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(d + e) = cos(d) cos(e) - sin(d) sin(e) and sin(d + e) = sin(d) cos(e) + cos(d) sin(e) .

Using these, our equation becomes:

-mw²h [cos(wt) cos(b) - sin(wt) sin(b)] - bwh [sin(wt) cos(b) + cos(wt) sin(b)] + kh [cos(wt) cos(b) - sin(wt) sin(b)] = F_o [cos(wt) cos(q_o) - sin(wt) sin(q_o)] .

For this equation to be true for all time, we need the coefficient of cos(wt) to be zero AND the coefficient of sin(wt) to be zero. This gives the following two equations:

For cos(wt): -mw²h cos(b) + -bwh sin(b) + kh cos(b) = F_o cos(q_o) and

For sin(w t): +mw²h sin(b) + -bwh cos(b) - kh sin(b) = -F_o sin(q_o) .

This gives two equations for two unknowns: h and b . We can do this, but the algebra is somewhat messy since b is inside both sine and cosine functions.

Exponential Method

Let's try writing the problem this way:

F_o cos(wt + q_o) = Re[F_o e^iqo e^iwt ]

And further let F be a complex number: F = F_o e^iqo , so that

F_o cos(wt + q_o) = Re[F e^iwt ] .

We need to express our guess of x_i(t) = h cos(w_at + b) as x(t) = Re[h e^ib e^iwt ] , and as we did above for force, let x be a complex number: x = h e^ib , so that our guess solution is now:

x(t) = Re[x e^iwt ] .

Substituting these expressions for x(t) and F(t) into our differential equation:

ma + bv + kx = F(t) gives -mw²x + ibw x + k x = F

which gives an express for x 

x = F / [-mw² + ibw + k] ,

or dividing through by m (and recalling that k/m = w_o²):

x = (F/m) / [-w² + ibw /m + w_o²] .

We note that the denominator is a complex number (as are x and F ). Let's call the denominator W e^id 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 e^iq where r = [a²+b²]^1/2 and q = tan^-1(b/a) .)

W = [(w_o²-w²)² + (bw/m)²]^1/2 and d = tan^-1 [(bw/m) / (w_o²-w²)] .

Using this we now have:

x = (F/m) / W e^id = (F_o e^iqo e^-id ) / (mW) = (F_o/mW) e^{i(qo-d )} .

This now gives:

x(t) = Re[x e^iwt] = Re[ (F_o/mW) e^i(qo-d) e^iwt] = Re[(F_o/mW) e^i(wt+qo-d) ]

x(t) = {F_o / m[(w_o²-w²)² + (bw/m)²]^1/2 } cos(wt + q_o - d)

where d = tan^-1[(bw/m) / (w_o²-w²)] .

In terms of g =(b/2m), this can be written as:

x(t) = {F_o / m[(w_o²-w²)² + 4g²w² ]^1/2 } cos(wt + q_o - d)

where d = tan^-1[(2gw ) / (w_o²-w²)] .

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 w approaches w_o, that the amplitude of the resultant oscillation increases. The biggest amplitude results when w = w_o, and this is called being at resonance. The smaller the (b/m) or g , the higher the resonant amplitude.

The phase difference between the applied force and the resulting motion, d , also depends on the frequency. Recall: d = tan^-1[(bw/m) / (w_o²-w²)].

For w < w_o, 0^o < d < 90^o, which means the motion lags a little behind the force. For the case where w is small, the amplitude approaches the constant value of F_o/mw_o² = F_o/k .

For w = w_o, d = 90^o, which means the motion lags behind the force by 90^o.

For w > w_o, d > 90^o, which means the motion approaches being out of phase with the force. For the case where w is large, the amplitude approaches F_o / mw² which approaches zero.

Homework Problem: Problem #12: 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) = F_o sin(wt).

Power

Power is defined to be Work per time, and Work is force thru a distance:

P = dW/dt = d [ò 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) = E_ocos(wt) so F(t) = -eE_ocos(wt) . Hence, our F_o = -eE_o .

We can use the previous results to see the x position of the electron in our model will be:

x(t) = A e^-gt cos(w₁t + a) + (-eE_o/m) cos(wt - d) / [(w²-w_o²)² + 4g²w²]^1/2

where d = tan^-1[(2gw) / (w_o²-w²)] .

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^{-g 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, 

v_steady-state (t) = dx_i(t)/dt = (+eE_o/m)w sin(wt - d) / [(w²-w_o²)² + 4g²w²]^1/2 .

The power now is Fv: 

Power = (-eE_o) cos(wt) * (+eE_o/m)w sin(wt - d) / [(w²-w_o²)² + 4g²w²]^1/2

= {-e²E_o²w / m[(w²-w_o²)² + 4g²w²]^1/2 } cos(wt) sin(wt-d) . 

To find the Average Power, <P>, we need to find the average of cos(w t) sin(wt-d). To do this, we can use a trig identity for the sine of the difference of two angles: 

sin(wt-d) = sin(wt)cos(d) - cos(wt)sin(d)

so that cos(wt) sin(wt-d) = cos(wt) sin(wt) cos(d) - cos²(wt) sin(d) . 

But the time average of cos(wt) sin(wt) is zero, and the time average of cos²(wt) is ½.

Using these time averages gives: 

<Power> = {-e²E_o²w / m[(w²-w_o²)² + 4g²w²]^1/2 }*{0 - ½ sin(d)}

<Power> = +½ e²E_o²w sin(d) / m[(w²-w_o²)² + 4g²w²]^1/2 .

Recall that tan(q ) = y/x and sin(q ) = y/[x²+y²]^1/2 . Using d = tan^-1[(2gw) / (w_o²-w²)], we can express sin(d ) as sin(d) = (2gw) / [(w²-w_o²)² + 4g²w²]^1/2 . Thus we finally get:

<Power> = +e² E_o² w² g / m[(w²-w_o²)² + 4g²w²] for each electron. 

We see that for maximum power absorption, w =w_o ; we also see that the power absorption goes to zero for both w going to zero (due to w² in the numerator) and for w going to infinity (due to w⁴ 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/[em]^1/2 , so n = c/v = [em / e_o m_o ]^1/2 . For most materials, m = m_o , so the expression for n becomes: n = [e/e_o]^1/2 . Recall from Physics II that in materials, e =Ke_o where K was the dielectric constant that depended on the stretchability of the charges. In terms of field values: D = electric displacement = eE = 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 e gives: e = e_o + (-Nex)/E which in tern gives:

n = [e/e_o]^1/2 = [1 + (-Ne/e_oE)x]^1/2 .

If we now substitute our inhomogeneous solution for x (since the homogeneous part will die out due to the e^{-g t} term) due to the oscillating applied Electric Field (the radiation), we get for n:

n = [ 1 + {-Ne/[e_oE_ocos(wt)]}*{-eE_o/m[(w²-w_o²)² + 4g²w²]^1/2} cos(wt-d) ]^1/2 .

If we use the trig identity: cos(wt-d) = cos(wt)cos(d) + sin(wt)sin(d) we get for n:

n = [1 + {Ne² / e_om[(w²-w_o²)² + 4g²w²]^1/2}cos(d) + {Ne² / e_om[(w²-w_o²)² + 4g ²w²]^1/2}sin(d) tan(wt) ]^1/2 .

Recall that tan(q) = y/x and cos(q) = x/[x²+y²]^1/2 , and recall that tan(d) = (2gw )/(w_o²-w²) so that

cos(d) = (w_o²-w²)/ [(w²-w_o²)² + 4g²w²]^1/2 .

Also realize that the term with tan(wt) 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(w_o²-w²) / e_om[(w²-w_o²)² + 4g²w²]}]^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 w <w _o, that the index of refraction is greater than one and increases as the frequency gets bigger. This is hard to see by just looking at the function since w’s appear in both the numerator and denominator. To get a better picture of the behavior of this n(w), see the following graph. 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.

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