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Fowles and Cassidy 3.1 to 3.3
One-dimensional problems with a force that depends on time only can be solved by integration.
One-dimensional problems with a force that depends on position only can be turned into an integrable problem by a trick: dv/dt = 1/2 d/dx(v2) . This leads to a one-dimensional work-energy relation, and conservation of (mechanical) energy for conservative forces. A potential energy function is defined as V(x) = -[Int] F(x) dx .
A very important example of a force that depends on position only is the spring force, F(x) = -kx, also known as Hooke's law. This form implies that k is a positive constant, therefore this force is always pulling our object back towards x=0. At x=0, F=0. This position is called the equilibrium position. If the object is at rest at the equilibrium position, then it feels no net force, and will remain in that position, at rest, indefinitely.
If our object is at a location x>0, then the force points in the negative direction, pulling the object back to the origin. The magnitude of the force grows linearly as the object moves further in the positive direction, so eventually, the force will overcome the inertia of the object and pull it back toward the origin.
But then, the object will reach the origin with some non-zero momentum (velocity) in the negative direction. At the origin the force is zero, so the object will not stop, but will continue past the origin to negative x. In the region of negative x, the force points in the positive direction, again pulling the object back to the origin. The magnitude of the force grows linearly as the object moves further in the negative direction, so eventually, the force will overcome the inertia of the object and pull it back toward the origin.
But then, the object will reach the origin with some non-zero momentum (velocity) in the positive direction, and as you might guess, the whole process repeats itself.
This repetitive motion is the signature of an oscillator. An oscillator that moves under the influence of a linear force (a Hooke's law force) is called a simple harmonic oscillator (SHO). The classic example of a SHO is the mass on a spring, neglecting friction. There are many other systems that are approximately SHO's: a pendulum under going small oscillations, the vibration of atoms in a molecule, the vibrations of certain crystals, ... And there are many more systems that behave analogously to the SHO: electronic oscillations, vibrations of strings and membranes, EM waves, and virtually any system of bound particles. The plethora of analagous systems makes the SHO possibly the most important problem in Physics!
Applying what we learned about one-dimensional forces that depend only on position, we can determine a potential energy function:
V(x) = -[Int] F(x) dx = k [Int] x dx = 1/2 kx2
We can of course add an arbitrary constant, but for simplicity let it be zero such that V=0 for x=0. [graph of potential energy]
The equation for energy conservation reads:
T + V = 1/2 mv2 + 1/2 kx2 = E = constant
Draw a line at E=constant on the potential energy graph. Since T + V = E, and T > 0, the object can only be in the region where V < E. At the points where V = E, the kinetic energy is zero, and so is the velocity. If an object moves towards this point, the velocity decreases to zero, then the object reverses direction, and moves away from this point. These points are called turning points of the motion. In this case, the turning points occur at x = ±Sqrt{2E/k}. This is the amplitude of the motion.
Now, we want to find the solutions for the position and velocity as functions of time. We can directly integrate from the energy equation, and this is done in the text. It is possible, but a bit messy, so let's take a different route. Let's go back to the equation of motion:
m[xddot] + kx = 0 or [xddot] + k/m x = 0.
This is a differential equation; in particular it's a second order ordinary differential equation with constant coefficients. This is alot of words. Basically, we are looking for a function (remember, x represents a function) whose second derivative is proportional to itself, with an overall minus sign. Not many functions have this behavior: the second derivative of a polynomial is a lower order polynomial; the second derivative of an exponential is proportional to itself, but without the minus sign; however the sine and cosine functions do have this behavior.
Let's try x(t) = A sin(wt) . We see that this satisfies the differential equation if w = w0 = Sqrt{k/m} . w0 is called the undamped or natural frequency of the oscillator (actually angular frequency, but usually people omit the word angular so get used to it). The same holds true for the cosine function. Our differential equation is second order -- this means that we need two "constants of integration". The A in A sin(wt) is one constant of integration; there are two ways in which to get two constants. We can either shift the sine function by a phase or use the sum of a sine and cosine function:
x(t) = A sin(w0t + f) (1) or
x(t) = A sin(w0t) + B cos(w0t) (2)
These two forms are equivalent, connected by the trigonometric relation for the sine of a sum of two angles. Checking this is left to the reader. For the present, we will use form (1). You may be wondering if this is a complete solution to the problem, or if maybe something is missing. After all, we "guessed" a solution, and found something that it satisfies the differential equation, but we haven't shown anything in a formal sense. Well, this is a complete solution. It can be verified by solving the energy equation, or, as we'll do later in the course, by solving the equation of motion with more sophisticated techniques. For now, we'll just use the solution and not worry about how to get it.
The period of the oscillation is the time it takes to go through one complete cycle. In one cycle, the argument of the sine function must change by 2p. This occurs when the time changes by an amount T = 2p/w0.
From energy conservation, we determined that the amlitude is related to the energy by A = Sqrt{2E/k}= Sqrt{2E/w02m}. Question: How can we relate the amplitude to the maximum velocity of the object undergoing SHO?
If we know the position and velocity at t=0, how do we determine the constants A and f? We begin by writing the relations for the position and velocity at t=0:
x(0) = A sin(f) = x0
v(0) = w0 A cos(f) = v0
We can now use these equations to find A and f in terms of the initial position and velocity. Remember, w0 is a constant. Dividing x(0) by v(0) yields tan(f) = w0x0/v0. Squaring both equations, and adding yields A2 = x02 + v02/w02 .