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Reading

F&C sections 2.4, 4.3, 4.5, 4.6

Recall

Solution to one dimensional motion with a velocity dependent force.

Projectile Motion With Linear Air Resistance

We have earlier looked at projectile motion with no air resistance. We found that the projectile follows a parabolic path. Let's look at how the motion is affected by linear air resistance. (The reason we look at linear air resistance is that we can find a closed form solution to the equations of motion. It is not necessarily true that linear air resistance is the most important case.) For convenience, let's write the air resistance force as -mgv, and I will assume that the initial velocity is in the x direction (not done in the book, but it saves me typing). The equation of motion is then:

mdv/dt = -mgv - mgk 

The m's can be cancelled and the equation can be separated into components as:

a_x = -gv_x

a_z = -gv_z - g 

These equations are separated, that is, the x equation depends only on x and its derivatives and the z equation depends only on z and its derivatives (plus constants). These can be solved individually:

v_x = v_0xe^-gt       v_z = v_0ze^-gt - (1 - e^-gt)g/g 

If you are only learning about differential equations for the first time, this solution may seem to come out of thin air. But it really isn't that difficult. If you were given some polynomial equation and told that x=5 is a solution, you would probably check by substituting 5 into the equation and checking that equality held. The same can be done with differential equations. To check that the above functions are solutions to the differential equations, substitute them, take the appropriate derivatives, and check that the equality holds. For the x equation we get:

-gv_0xe^-gt = -gv_0xe^-gt    which is certainly true.

For the z equation we get:

-g(v_0z + g/g)e^-gt =  -gv_0ze^-gt + (1 - e^-gt)g - g    which is also true.

This solution checks out, but how do we get that solution in the first place (equivalent to finding that x=5 is a solution to a polynomial equation), and how do we know that we have the whole solution (equivalent to knowing when we have found all of the zeroes of a polynomial). There are a number of ways to solve DE's, for instance "guessing" the solution, or the method of operators that I will demonstrate here. Let D stand for the "operation" of taking a derivative (a time derivative for our purposes). Then a_x = Dv_x, and v_z = Dz. The above differential equations become:

(D + g)v_x = 0    and     (D + g)v_z = g

The first equation says that the combination of a time derivative plus gamma times the x velocity is zero. The solution is a constant times  e^-gt . The constant is determined from the initial conditions of the problem. Note also the similarity of this equation to a polynomial equation. This would be a first order polynomial, which we know has one zero. In complete analogy, we are guaranteed that this DE has one general solution, which we have, so we've completely solved the equation.

The second equation is a bit more difficult because the right hand side is not zero but a constant. First, solution that satisfies (D + g)v_z = 0 can be added to a solution, and the result is still a solution to the DE. The DE with the right hand side zero is called a homogenous DE, and the solution is the homogenous solution. We have the homogeneous solution from the x part of the problem. The solution to the DE when the right hand side is not zero is called the specific solution, and the general solution to the problem is the sum of the homogenous and specific solution. The specific solution to this equation is simple, v_z = g/g . Notice that there are no unknown constants in the specific solution. Adding in the homogenous solution, and using the initial conditions to determine the unknown constant gives us the solution given above.

Damped Harmonic Oscillator

Now let's apply this technique to the important problem of the damped harmonic oscillator. We again assume that the damping (resistance) term is linear in the velocity. (This again makes the problem possible to solve, but in this case, it is also an important example with interesting behavior to understand.)

mx" + cx' + kx = 0.

Let g = c/2m, w₀² = k/m, and divide through by m to get:

x" + 2gx' + w₀²x = 0  or using the D notation   (D² + 2gD + w₀²)x = 0.

Now the operator equation looks like a second order polynomial. A second order polynomial can always be factored (allowing for the use of imaginary numbers) and it turns out that we can do the same with our operator expression because none of the terms in parenthesis are effected by the derivative operation (they commute).

[D + g - (g² - w₀²)^1/2] [D + g + (g² - w₀²)^1/2] x = 0

The order of the two terms is arbitrary. Clearly if the result of the first (rightmost) term in brackets is zero, then the equation is satisfied. But we have solved this earlier, and the solution is an exponential function. The same holds here. Let's call the quantity g² - w₀² = q². Then there are two solutions for x:

x = A_± exp[-(g±q)t]

This result may seem a bit strange at first. After all, in the undamped case, the solution was a sine or cosine function not an exponential, and letting g-->0 takes us to the undamped case. What happens when g=0? Then q²=-w₀² leading us to conclude that q is imaginary. And x is the exponential of an imaginary number times t.

Imaginary Exponential

What is the exponential of an imaginary number? Well, as derived in appendix D,

e^iq = cosq + i sinq.

This is how the result derived above squares with the undamped harmonic oscillator solution.

© 31 March 1999 R. Harr