Physics GRE Practice

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2 Chapter 3: Harmonic Motion in 1-D Harmonic Motion occurs when the acceleration of a system is proportional to the negative of its displacement. or a x (1) ẍ x (2)

3 Examples of Harmonic Motion musical notes suspension bridge swing pendulum mass on spring planetary motion

4 Hooke s Law and the Simple Harmonic Oscillator Some review... F = mẍ = kx (3) k is a positive number called the force constant. the equilibrium is stable; restoring force We derived V (x) = 1 2 k x 2 for a spring in the last class.

5 Why is Hooke s Law Important? Or, why does it appear so frequently? Suppose there is a system with: a stable equilibrium, which we take to be the origin (x 0 = 0) conservative system, so we can specify V (x)

6 Taylor s Theorem Before proceeding we take a brief detour to review Taylor s theorem. Consider a function f(x) with continuous derivatives within a certain range of x values, near x 0. f (x 0 + h) = f (x 0 ) + h f (x 0 ) + h2 2! f (x 0 ) +... (4) If we set x 0 = 0, we get the more common form of the series (usually called the Maclaurin s series) for the function f (x): f (x) = f (0)+x f (0)+ x 2 2! f (0)+ x 3 3! f (0)+...+ x n n! f (n) (0) (5)

7 Taylor s Theorem - Why Should You Care? The Taylor and Maclaurin series possess two very important properties. 1. They can be integrated/differentiated term by term. 2. The resulting series converge to the integral/derivative of f (x).

8 Taylor s Theorem - Example Find the Taylor series expansion of e x. f (x) = f (0)+x f (0)+ x 2 2! f (0)+ x 3 3! f (0)+...+ x n n! f (n) (0) (6) e x = e 0 + x e 0 + x 2 e x = 1 + x + x 2 2! + x 3 2! e0 + x 3 3! e (7) 3! +... (8)

9 Back to Hooke s Law The Taylor expansion for potential energy about the equilibrium position is: V (x) = V (0) + x V (0) + x 2 2! V (0) (9) If we consider only small values of x, the first 3 terms is a valid approximation. We set V (0) = 0 (zeropoint of potential energy is arbitrary). V (0) = 0. Why? Because we are considering systems in stable equilibrium. (Slope of V (x) = 0.) So Taylor expansion of potential energy becomes: V (x) = x 2 2! V (0) (10)

10 So Taylor expansion of potential energy becomes: V (x) = x 2 2! v (0) = 1 2 x 2 V (0) (11) Renaming V (0) = k, we get V (x) = 1 2 k x 2 (12) And remember that F = gradv = V = V î = k x (13) x So Hooke s Law is always valid for sufficiently small displacements from a stable equilibrium.

11 On to the equation of motion Imagine your favorite harmonic oscillator (mine is a mass on a spring). We can derive the equation of motion from Newton s 2nd Law: F = mẍ (14) kx = mẍ (15) If we let ω 2 = k m, then ẍ + k m x = 0 (16) ẍ + ω 2 x = 0 (17)

12 So how do we solve this differential equation? ẍ + ω 2 x = 0 (18)

13 Review of Linear Homogeneous Differential Equations The general form of linear homogeneous second-order equations: y + a y + b y = 0 (19) homogenous means RHS is equal to zero rather than f (x). linear b/c it contains no higher powers of x or its derivatives than the first power.

14 Review of Linear Homogeneous Differential Equations Important properties: y + a y + b y = 0 (20) 1. If y 1 (x) is a solution, then c 1 y 1 (x) is a solution. 2. If y 1 (x) and y 2 (x) are solutions, then y 1 (x) + y 2 (x) is also a solution (principle of superposition). 3. If y 1 (x) and y 2 (x) are linearly independent solutions, then the general solution to the equation is given by c 1 y 1 (x) + c 2 y 2 (x) (the general solution always contains two arbitrary constants). Linearly independent means ay 1 (x) + by 2 (x) = 0 only when a = b = 0.

15 Review of Linear Homogeneous Differential Equations y + a y + b y = 0 (21) Substitute y = e rx, y = r e rx, y = r 2 e rx. This produces an auxiliary equation: r 2 + ar + b = 0 (22) Solve the quadratic eqn: r = a 2 ± 1 2 a 2 4b (23) Or r 1 = a a 2 4b, r 1 = a a 2 4b (24)

16 Solution to y + a y + b y = 0 The general solution is: y = c 1 e r 1x + c 2 e r 2x, r 1 r 2 (25) For the harmonic oscillator, a = 0 and b = ω 2 in Eqn 21. If we plug into Eqn 24, we get r 1 = ω 2 = iω, and r 2 = ω 2 = iω. Thus, the solutions for the harmonic oscillator are: So that the general solutions is: x(t) = e iωt, x(t) = e iωt (26) x(t) = C 1 e iωt + C 2 e iωt (27) But wait, this doesn t look right, does it?

17 Perform some mathematical gymnastics From last slide: x(t) = C 1 e iωt + C 2 e iωt (28) Using the Euler identity e ix = cos(x) + i sin(x): x(t) = C 1 (cos(ωt) + i sin(ωt)) + C 2 (cos(ωt) i sin(ωt)) (29) x(t) = (C 1 + C 2 ) cos(ωt) + i(c 1 C 2 ) sin(ωt) (30) Let B 1 = C 1 + C 2 and B 2 = i (C 1 C 2 ). Then: x(t) = B 1 cos(ωt) + B 2 sin(ωt) (31)

18 Take a deep breath, almost there... From the last slide: x(t) = B 1 cos(ωt) + B 2 sin(ωt) (32) One last little trick. Let s define A = B B2 2. [ B1 x(t) = A A cos(ωt) + B ] 2 A sin(ωt) Draw the triangle and define δ as the angle between the hypoteneuse A and the leg B 1. Then: (33) x(t) = A [cosδ cos(ωt) + sinδ sin(ωt)] (34)

19 Add One Trig identity... From the previous slide: x(t) = A [cosδ cos(ωt) + sinδ sin(ωt)] (35) How about using Equation C.7 from Appendix C to simplify? cos(a ± B) = cosa cosb sina sinb (36) This gives: x(t) = Acos(ωt δ) (37) Phew! Boundary conditions are required to obtain a particular solution, and these depend on the problem/physical situation.

20 So what does all this mean? Let s apply what we have learned. Obtain some position vs. time data for a simple harmonic oscillator. Make a plot of position vs. time using matlab. Obtain a particular solution for your data analytically. Plot your analytic solution with the data. Repeat the above procedure using a different spring. From your matlab analysis, calculate the spring constant for each spring. Devise another procedure to determine the spring constant. Conduct your procedure and compare your two values of the spring constant. Do this for both springs.

Problem 1: Lagrangians and Conserved Quantities. Consider the following action for a particle of mass m moving in one dimension

105A Practice Final Solutions March 13, 01 William Kelly Problem 1: Lagrangians and Conserved Quantities Consider the following action for a particle of mass m moving in one dimension S = dtl = mc dt 1