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Reading Taylor 5.1-5.2 (today) and midterm review (Monday). Recall Chapter 5: Oscillations Hooke's Law You learned about Hooke's law for the force exerted by a stretched spring in basic physics, F = -kx . The importance of Hooke's law goes far beyond the potential uses of springs for closing doors. It is the result of the simple observation that for any system near a point of stable equilibrium, the force is approximately a linear restoring force (almost always). Let's see how this happens. For simplicity, begin with a one-dimensional system. Without loss of generality, we can redefine x so that the equilibrium position is at x=0.There is a potential energy function for the system, and we can expand it in a Taylor series about x=0: U(x) ≅ U(x=0) + (∂U(x=0)/∂x) x + ½ (∂²U(x=0)/∂x²) x² + ... = U(0) + U'(0)x + ½U''(0)x² + ... = U(0) + ½U''(0) x² where I've used a prime to represent the derivative with respect to x, in analogy to using a dot to represent a derivative with respect to time. The linear term is zero since ∂U(x=0)/∂x = 0 (the definition of an equilibrium point) and higher order terms are neglected. The force is the negative of the gradient of the potential. For our one dimensional case the gradient is simply the x derivative, and recalling that in the expansion for the potential, the quantities evaluated at x=0 are constant, we find F(x) = -∂U(x)/∂x = -U''(0) x Since this is a point of stable equilibrium, the second derivative of the potential energy evaluated at the equilibrium position is positive. Therefore, the force has the same form as the spring force, where the second derivative of U plays the role of k: U''(0) = k. This argument is true aside from some aberrant cases where the second derivative U''(0) = 0. We conclude that the motion near a point of stable equilibrium is simple harmonic. This is amplified if we draw the potential energy near a point of stable equilibrium. We can apply what we learned about one-dimensional motion to see the turning points, and see that the shape of the PE near the equilibrium is parabolic. Example: The Simple Pendulum A simple pendulum consists of a mass m suspended from a massless string of length l. The string is attached to a support (ceiling) and the system is free to oscillate (in a plane for now). Show that for small oscillations, the mass is subject to a linear restoring force. The potential energy is all gravitational potential energy, given by where θ=0 when the string is vertical. For small oscillations, expand U around θ=0 (up to second order in θ) U ≅ ½mgl&theta² + ... ≅ ½m(g/l)x² where x = lθ is the displacement from equilibrium. The corresponding force is which is a linear restoring force with spring constant k = mg/l. Simple Harmonic Motion Let's now investigate the motion of an object subject to a linear restoring force. When the linear restoring force is the only (relevant) force in the problem, the result is known as simple harmonic oscillation . Another way to state this is that the potential energy is quadratic in the displacement, to within a constant, U = U 0 + ½kx². These are equivalent since F = -∂U/∂x = -kx. There are a number of ways to express the result, and we'll explore four common ways here. The equation of motion for the simple harmonic oscillator is m D t ² x = -kx    or    D t ² x = -(k/m)x = -ω²x This is a linear, homogeneous, second order differential equation. We expect two independent constants in the solution. The equations says that the position is described by a function whose second derivative is proportional to the original function. Functions with this property are sine, cosine, and exponential. The Exponential Solutions Let's begin by trying the exponential function. Make a trial solution of X(t) = Ce αt . By the above, we don't mean that the position is an imaginary number, but rather we will take the real (or imaginary) part of the resulting function as the result. Taking the second derivative and inserting into the differential equation yields: Notice that the exponential factors containing the time dependence are the same, and since the exponential is never always zero, so we can factor it out and get the relation and, since k and m are positive real numbers we know that ω² is a positive real number, therefore α = ±iω. This is two solutions, +ω and -ω, each with an arbitrary constant. Putting these pieces together we can write the solution in exponential form: Again, x is not a complex number. One must take the real or imaginary part of this expression for the value of x, that is: x(t) = Re[X(t)]   or   x(t) = Im[X(t)]. Note that for now we will assume that C 1 and C 2 are real numbers. In general, the can be complex numbers, as discussed below. The Sine and Cosine Solutions One could have started with a sine or cosine trial solution as well. The exponential solution can be transformed into the sine/cosine solution by writing X(t) = ½B 1 (e iωt + e -iωt ) - i½B 2 (e iωt - e -iωt ) x(t) = B 1 cos(ωt) + B 2 sin(ωt) The constants B 1 and B 2 are combinations of C 1 and C 2 . You are likely familiar with this form from earlier physics courses. The Phase-Shifted Cosine Solution Another common way to write the solution is as single cosine (or sine) with a phase. This is entirely equivalent to the earlier solution with a sine and cosine function -- the sum of a sine and a cosine is equivalent to a single sine or cosine function with a phase. The constants A and δ can be related to either B 1 and B 2 or C 1 and C 2 . Solution as the Real Part of a complex Exponential A fourth form that is commonly used in physics is to write the solution as a single complex exponential with either a complex amplitude, or a real amplitude and phase. x(t) = Re( Ce iωt ) = Re( Ae i(ωt - δ) ) Once you are comfortable with this form, it is one of the most convenient to use, being the most compact to write and easy to manipulate. © 2007 Robert Harr