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Appendix A: Simple Harmonic Motion Appendix A: Simple Harmonic Motion A.1 We Start with Hooke’s Law Harmonic oscillator is depicted here and Hooke’s law defines the following equation: F ¼Àkx: But using Newton’s second law of motion, we can write © Springer International Publishing Switzerland 2017 245 B. Zohuri, Dimensional Analysis Beyond the Pi Theorem, DOI 10.1007/978-3-319-45726-0 246 Appendix A: Simple Harmonic Motion F ¼ ma; where in both equations F is the force, x is displacement of mass m, and k is spring constant as well as the acceleration of the mass. By equating both equations, we obtain the following: ma ¼Àkx or d2x m ¼Àkx dt2 and d2x k þ x ¼ 0: dt2 m ω2 k Define 0 ¼ m then we have 1 d2x ¼ þ x ¼ 0 ω2 2 0 dt or d2x þ ω2x ¼ 0 ðA:1Þ dt2 0 ω ω π 2π where is called angular frequency and can be defined as ¼ 2 f ¼ T where f is frequency and T is period of oscillation. Define dx x_ ¼ dt then we have d2x dx_ dx dx_ dx dx_ ¼ €x ¼ ¼ ¼ x_ : dt2 dt dx dx dt dx Substituting the above result in Eq. A.1, then we have dx_ x_ þ ω2x ¼ 0; dx 0 Appendix A: Simple Harmonic Motion 247 _ _ ω2 : xdx þ 0xdx ¼ 0 Integrating over the differential equation, we have ð ð _ _ ω2 ; xdx þ 0xdx ¼ 0 1 1 x_ 2 þ ω2x2 ¼ cte; 2 2 0 _ 2 ω2 2 ω 2; x þ 0x ¼ 2cte ¼ K ¼ ðÞA 0 _ 2 2ω2 ω2 2; x ¼ A 0 À 0x pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi dx ¼ x_ ¼Æω A2 À x2; dt 0 Separating of variable gives the following results: dx pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi ¼ ω0 dt A2 À x2 or ð ð dx pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi ¼ ω0 dt: ðA:2Þ A2 À x2 Two possible solutions x arcsin ¼ ω t þ ϕ; A 0 x arccos ¼ ω t þ ϕ: A 0 ϕ Integrating constant term Note: To do left-hand side integral, we can do the following steps: Assume x ¼ A sin y ) dx ¼ A cos ydy and pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi qffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi A2 À x2 ¼ A2 À A2 sin 2y ¼ A 1 À sin 2y ¼ A cos y: Therefore, 248 Appendix A: Simple Harmonic Motion ð ð ð dx A cos y pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi ¼ dy ¼ dy ¼ y: A2 À x2 A cos y Since x ¼ A sin y, we can conclude the following: x x sin y ¼ ) y ¼ arcsin : A A Therefore, if we substitute the last step result in Eq. A.2, then we have ð x arcsin ¼ ω dt ¼ ω t þ ϕ: A 0 0 Similarly we can have same results for the second solution: ð x arccos ¼ ω dt ¼ ω t þ ϕ: A 0 0 So the general solution is written as follows: x ¼ A cos ðÞω0t þ ϕ : ω π 2π 1 However, remember we assumed that ¼ 2 f ¼ T where f ¼ T and f is frequency while T is the period. Appendix B: Pendulum Problem B.1 Definition A pendulum is a mass (or bob) on the end of a string of negligible mass that, when initially displaced, will swing back and forth under the influence of gravity over its central (lowest) point. The regular motion of a pendulum can be used for time- keeping; pendulums are used to regulate (Fig. B.1). A simple is an idealization, working on the assumption that: • The rod or cord on which the bob swings is massless, inextensible, and always remains taut. • The motion occurs in a two-dimensional plane, i.e., the bob does not trace an ellipse. • The motion does not lose energy to friction. The differential equation, which represents the motion of the pendulum very similar to simple harmonic motion, is d2θ g þ sin θ ¼ 0 ðB:1Þ dt2 l See Appendix A for Eq. B.1 derivation as well as the following pages. In order to derive the simple pendulum equation and prove the dimensional analysis case about we show the following depiction (Fig. B.2): Note: The path of the pendulum sweeps out an arc of a circle. The angle θ is measured in radians, and this is crucial for this formula. The blue arrow is the gravitational force acting on the bob, and the violet arrows are that same force resolved into components parallel and perpendicular to the bob’s instantaneous motion. The direction of the bob’s instantaneous velocity always points along the red axis, which is considered the tangential axis because its direction is always tangent to the circle. Consider Newton’s second law: © Springer International Publishing Switzerland 2017 249 B. Zohuri, Dimensional Analysis Beyond the Pi Theorem, DOI 10.1007/978-3-319-45726-0 250 Appendix B: Pendulum Problem Fig. B.1 Simple gravity point of suspension pendulum assumes no air resistance and no friction amplitude Massless Rod length q Massive bob Bob’s Trajectory equilibrium position Fig. B.2 Force diagram of a simple gravity pendulum F ¼ ma; where F is the sum of forces on the object, m is the mass, and a is the instantaneous acceleration. Because we are only concerned with changes in speed, and because the bob is forced to stay in a circular path, we apply Newton’s equation to the Appendix B: Pendulum Problem 251 tangential axis only. The short violet arrow represents the component of the gravitational force in the tangential axis, and trigonometry can be used to determine its magnitude. Thus, F ¼Àmg sin θ ¼ ma a ¼Àg sin θ; where g is the acceleration due to gravity near the surface of the earth. The negative sign on the right-hand side implies that θ and a always point in opposite directions. This makes sense because when a pendulum swings further to the left, we would expect it to accelerate back toward the right. This linear acceleration a along the red axis can be related to the change in angle θ by the arc length formulas; s is arc length: s ¼ lθ; ds dθ υ ¼ ¼ l ; dt dt d2s d2θ a ¼ ¼ l : dt2 dt2 Thus, d2θ l ¼Àg sin θ dt2 or d2θ g þ sin ¼ 0; ðB:2Þ dt2 l This is the differential equation which, when solved for θ(t), will yield the motion of the pendulum. It can also be obtained via the conservation of mechanical energy principle: any given object, which fell a vertical distance h, would have acquired kinetic energy equal to that which it lost to the fall. In other words, gravitational potential energy is converted into kinetic energy. Change in potential energy is given by ΔU ¼ mgh change in kinetic energy (body started from rest) is given by 1 ΔK ¼ mυ2: 2 Since no energy is lost, those two must be equal: 252 Appendix B: Pendulum Problem Fig. B.3 Trigonometry of a simple gravity pendulum 1 mυ2 ¼ mgh; 2 pffiffiffiffiffiffiffiffi υ ¼ 2gh: dθ Using the arc length formula above, this equation can be rewritten in favor of dt dθ 1 pffiffiffiffiffiffiffiffi ¼ 2gh; dt l where h is the vertical distance the pendulum fell. Consider Fig. B.3. If the pendulum starts its swing from some initial angle θ0, then y0, the vertical distance from the screw, is given by θ y0 ¼ l cos 0 similarly, for y1, we have θ y1 ¼ l cos then h is the difference of the two h ¼ lðÞcos θ À cos θ0 dθ substituting this into the equation for dt gives Appendix B: Pendulum Problem 253 rffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi dθ 2g ¼ ðÞcos θ À cos θ : ðB:3Þ dt l 0 This equation is known as the first integral of motion; it gives the velocity in terms of the location and includes an integration constant related to the initial displacement (θ0). We can differentiate, by applying the chain rule, with respect to time to get the acceleration: rffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi d dθ d 2g ¼ ðÞcos θ À cos θ dt dt dt l 0 d2θ 1 ÀðÞ2g=l sin θ dθ ¼ pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi 2 = θ θ dt 2 ðÞ2g l ðÞcos À cos 0 rdtffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi 1 ÀðÞ2g=l sin θ 2g g ¼ pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi ðÞcos θ À cos θ0 ¼À sin θ 2 ðÞ2g=l ðÞcos θ À cos θ0 l l d2θ g ¼À sin θ; dt2 l which is the same result as obtained through force and dimensional analysis. Appendix C: Similarity Solution Methods for Partial Differential Equations (PDEs) Here we discuss briefly how to handle and solve a partial differential equation of high order by reducing to and ordinary differential equation using self-similar methods given by George W. Bluman and J. D. Cole. C.1 Self-Similar Solutions by Dimensional Analysis Consider the diffusion problem from the last section, with point-wise release (reference: Similarity Methods for Differential Equations (Applied Mathematical Sciences, Vol. 13)—Paperback (Dec. 2, 1974) by George W. Bluman and J.D. Cole (Sect. 2.3): 8 < ∂c ∂2c ¼ D þ Q δðÞx δðÞt : ∂t ∂x2 0 cxðÞ¼; 0 0, cðÞ¼Æ1, t 0: Initial release within infinitely narrow neighborhood of x ¼ 0, such that ΠðÞx =d ¼ δðÞx and L=d !1. Note Q0 has different dimension as the previous Q because of the cross-sectional area S and time contained in δ(t). 1. Dimensional analysis À3 2 À1 À2 fgc ¼ ML , fgD ¼ L T , fgQ0 ¼ ML (mass release per unit cross-sectional area) fgx ¼ L, fgt ¼ T. Thus, we expect 2Pi groups: pffiffiffiffiffi Dt x Π1 ¼ c, Π2 ¼ pffiffiffiffiffi Q0 Dt and the solution to the PDE problem must be of the form Π1 ¼ f ðÞΠ2 or © Springer International Publishing Switzerland 2017 255 B.
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