Chapter 15 Simple Harmonic Motion GOALS When you have mastered the contents of this chapter, you will be able to achieve the following goals: Definitions Define each of the following terms, and use it in an operational definition: period frequency simple harmonic motion restoring force amplitude damping phase angle UCM and SHM Correlate uniform circular motion and simple harmonic motion. SHM Problems Solve problems involving simple harmonic motion. Energy Transformations Analyze the transfer of energy in simple harmonic motion. Superposition Explain the application of the principle of superposition to simple harmonic motion. Natural Frequencies Calculate the natural frequencies of solids from their elastic moduli and density values. PREREQUISITES Before beginning this chapter you should have achieved the goals of Chapter 5, Energy, Chapter 7, Rotational Motion, and Chapter 13, Elastic Properties of Materials. 123
Chapter 15 Simple Harmonic Motion OVERVIEW - Examples of repeated motion are all around us. Examples include a child in a swing and the up and down motion of a fish bobber. Most repeated motions are either uniform, simple harmonic motion, or reduce to nearly this type of motion for small amplitudes. In this chapter you will learn how repeated motion of this type is treated mathematically. SUGGESTED STUDY PROCEDURE - Begin your study by carefully reading the following Chapter Goals: Definitions, UCM and SHM, SHM Problems, and Energy Transformations. For an expanded discussion of the terms listed under the goal of Definitions, see the Definitions section of this Study Guide chapter. Now, read text sections 15.1-15.4 looking carefully at the examples given. The mathematics in Section 15.4 become somewhat difficult, but the use of Figure 15.5 will help you visualize the meaning of the equations. Remember to look in the second section of this for the answers to all questions posed throughout the reading sections. At the end of the chapter, read the Chapter Summary and complete Summary Exercises 1-14. Now do Algorithmic Problems 1-6 and check your answers carefully. Next, complete Exercises and Problems 2, 5, 7, 12, and 21. For additional work on the concepts presented in this chapter, turn to the Examples section of this Study Guide. Now you should be prepared to attempt the Practice Test provided at the end of this Study Guide chapter. If you have difficulties with any of the problems, please refer to that particular part of the text or to the sections of this Study Guide chapter. This study procedure is outlined below. --------------------------------------------------------------------------------------------------------------------- Chapter Goals Suggested Summary Algorithmic Exercises Text Readings Exercises Problems & Problems --------------------------------------------------------------------------------------------------------------------- Definitions 15.1 1-4 UCM and SHM 15.2,15.3 5-14 4 2 SHM Problems 15.2,15.3,15.4 1-3,5,6 5,7,12,21 - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - Superposition 15.6 15,16 9 Natural 15.5,15.6 17,18 8 Frequencies 124
DEFINITIONS PERIOD - The time required for a system to go through one complete cycle of motion and usually measured in seconds. The period of motion of a second hand on a clock is 60 seconds as it takes 60 seconds for a second hand to go completely around the face of a watch. The period of the hour hand is 12 hours, since it makes two complete cycles in 24 hours. SIMPLE HARMONIC MOTION - If the restoring force is proportional to the displacement and oppositely directed, the system will execute SHM when displaced from equilibrium. The motion of a simple pendulum through small angles is a very familiar example of a system that executes simple harmonic motion. AMPLITUDE - Maximum displacement from equlibrium position. The amplitude of motion of a pendulum is the maximum angle from the vertical to which the pendulum swings. PHASE ANGLE - Angular displacement at the starting time. It is common for us to start a SHM system moving by displacing it from its rest position. The initial location determines the value of the phase angle. FREQUENCY - Number of complete oscillations per second. If you have a SHM system that requires 8 seconds to make a complete cycle, then that system goes 1/8th of a cycle in one second. We say its frequency is 1/8 cycles per second or 1/8 hertz. RESTORING FORCE - Force acting to return a displaced body to equilibrium position. DAMPING - A loss of energy during vibration because of friction. It shows its pressure by a decrease in the amplitude of vibration. In all real, isolated oscillating systems the friction forces finally bring the system to rest in its position of lowest energy. ANSWERS TO QUESTIONS FOUND IN THE TEXT SECTION 15.1 Introduction The universality of simple harmonic motion, the simplest kind of periodic motion, is striking. It has been suggested that the pervasive existence of simple harmonic motion is one of the miracles provided by the nature of our universe. A mathematical derivation can be used to show how the universality of SHM arises (see Taylor's Theorem in the enrichment section 15.7). The simplicity of SHM and the completeness with which we can treat SHM, makes it a favorite model for physicists. It is widely used to explain atomic phenomena such as the vibration of atoms in solids and the oscillations of atomic nuclei. EXAMPLES SHM PROBLEMS 1. Of all the possible forces in the universe, there is only one force that results in simple harmonic motion. (a) What is the equation for that force? (b) What is this force called? (c) Sketch the magnitude of this special force F as a function of displacement. (d) Sketch the force F as a function of the displacement. (e) Explain in words the relationship between this force and the magnitude and direction of the acceleration of the body upon which it acts. 125
What Data Are Given? Consider an ideal system which undergoes simple harmonic motion in one dimension. What Data Are Implied? Assume that friction can be neglected and that the amplitude and period have constant, finite values. What Physics Principles Are Involved? The problem only involves the dynamics of SHM as discussed in Section 15.3. What Equations Are to be Used? Newton's Second Law of Motion and Hooke's Law are all that is required for this problem. F = ma (4.1) F (Hooke) = - kx (13.3) Algebraic Solution (a) The equation for the force is Hooke's Law (Equation 13.3) Force = -kx where k is a positive constant and x is the displacement. (b) This force is called a Hooke's Law force on a linear restoring force. (c) F = -kx, so F = kx for x 0 F = -kx for x 0 See Fig. 15.1(c) (d) F = -kx See Fig.15.1(d) (e) Combining Equations (4.1) and (13.3) we obtain the equation of motion for SHM ma = -kx (1) so a = -k/mx where k and m are both positive constants. So the acceleration of the body is opposite in direction to the displacement. It is directly proportional to the displacement with a proportionality constant equal to k/m. 126
2. Draw a graph of the total energy, kinetic energy, and potential energy of a onedimensional system that undergoes simple harmonic motion. What Data Are Given? Consider an ideal system in one dimension, say x, which uses SHM. What Data Are Implied? Neglect friction and consider a finite system. What Physics Principles Are Involved? The energy concepts discussed in Section 15.4 What Equations Are to be Used? Potential Energy = (½) kx 2 (15.12) Kinetic Energy = (½)mv 2 = (½)m(k/m)(A 2 - x 2 ) (15.16) Total energy = KE + PE = (½)kA 2 2 = (½)mv max (15.14 & 15.17) Algebraic Solution 3. A 8 kg monkey grabbed onto an elastic grapevine hanging from a jungle tree. The vine stretched 16 m. (a) What force was applied to the vine by the monkey? (b) What is the effective force constant k of the vine, assuming the applied force did not stretch it beyond its elastic limit? (c) Write Newton's Second Law of Motion equation for the motion of the monkey up and down. (d) What is the frequency in Hertz of the up and down motion? (e) Suppose the monkey wishes to reach down 2 meters from its equilibrium position to catch a banana. What is the total energy of the monkey as it reaches the banana with respect to an assumed total energy of zero at the equilibrium position? (f) What is the velocity of the monkey as it passes through the equilibrium position? (g) Calculate the period of oscillation and sketch the motion of the monkey as a function of time. Show the units on both the vertical and time axes. What Data Are Given? A vertical unstretched vine will stretch down 16 m when an 8 kg mass is hung onto it. That mass is then displaced 2 m and allowed to oscillate up and down. What Data Are Implied? The equlibrium position, 16 m below the starting position of the monkey, is to be taken as the zero energy elevation. Ideal SHM is assumed near the surface of the earth where g = 9.8 m/s 2. What Physics Principles Are Involved? This problem requires the coordination of the kinematics, dynamics, and energy relationships of SHM, Sections 15.2, 15.3, and 15.4 and 15.4. 127
What Equations Are to be Used? F = w = mg (4.5) F = -kx (13.3) F = 1/(2π SQR RT(k/m)) (15.8) E spring = 1/2 ka 2 = 1/2 mv 2 + 1/2 kx 2 (15.14) PE gravity = mgh (5.5) V max = ±ωa (15.17) T = 2π SQR RT(m/k) = 1/f (15.9) x = A cos (ωt + φ) (15.2) Algebraic Solution. (a) Force on the vine = mg (b) Force constant = mg/(elongation of the vine) (c) ma = -ky where y is the change in elevation from the elongated position of the vine (d) f = 1/2π SQR RT)(k/m) Hz (e) Total Energy = 1/2 ka 2 - mga at the bottom (f) V = ±ωa (g) T = 1/f sec. Numerical Solutions (a) Force on the vine = (8 kg)(9.8 m/s 2 ) = 7.8 x 10 1 N (b) Force constant = (7.8 x 10 1 N)(1.6 x 10 1 m) = 4.9 N/m (c) acceleration = ((-4.9 N/m)/8 kg)y = (0.61 N/kg m)y (d) f = 0.12 Hz (e) E = (½) (4.9 N/m)(2 m) 2 - (8)(9.8)(2) = -150 Joules (f) V max = ±wa = 2πfA = ± 2(p)(.12) (2 m) = 0.75 m/s (g) T = 1/f = 8.0 seconds Thinking About the Answer Notice how the need to also consider the gravitational potential energy changes this problem. In other kinds of SHM problems negative energies would not be possible. Consider shifting the zero in gravitational potential energy to the bottom of the motion - what is the energy at the equilibrium position then? 128
NATURAL FREQUENCIES 4. A bell manufacturer constructed a copper bell with a resonance frequency of 440 Hz. He found that if he made it of lead (Y lead = 1.7 x 10 10 N/m 2 ) it would have the same resonance frequency only near liquid nitrogen temperature (~73 ø K). Estimate the temperature coefficient for Young's modulus for lead and the room temperature resonance frequency for the lead bell. What Data Are Given? Y lead = 1.7 x 10 10 N/m 2 at normal temperatures. Y copper = 12 x 10 10 N/m 2 at normal temperatures (from Table 13.1), copper normal temperature 20 ø C 293 ø K; LN 2 T 73 ø K What Data Are Implied? The densities of the materials are needed to determine natural frequencies. From Table 8.1 density of lead = 11.3 x 10 3 kg/m 3 ; density of copper = 8.9 x 10 3 kg/m 3. What Physics Principles Are Involved? The dependence of the natural frequency of a solid is given by Equation 15.18. In this problem let us consider only the compressional waves in the materials so that the natural frequency depends upon the square root of the ratio of Young's modulus to the density. The temperature coefficient can be found by calculating the Young's modulus value at 73 ø K; then find the ratio temperature coefficient = change in Y's modulus/change in temperature What Equations Are to be Used? f 0 = (1/2π) SQR RT(Young's Modulus /((density)(area))) (15.19) temperature coefficient = change in Young's Modulus / change in temperature (2) Algebraic Solutions Since all the dimensions of the two bells are the same, the ratio of the natural frequencies will only depend upon the ratio of the square roots of Young's modulus to density. f Pb /f Cu = SQR RT[(Y Pb /ρ Pb )/(Y Cu /ρ Cu )] = (SQR RT)[(Y Pb ρ Cu )/ (ρ Pb Y Cu )] (3) temperature coefficient = (Y R.T. - Y LN2T )/(R. T. - LN 2 T) (4) Numerical Solutions f Pb = {SQR RT[((1.7 x 10 10 ) x (8.9 x 10 3 ))/ ((11.3 x 10 3 ) x (12 x 10 10 ))]} (440 Hz) =150 Hz To have the same frequency the ratio of Young's modulus to density must be the same, so Young's Modulus Pb at 73 ø K/(11.3 x 10 3 kg/m 3 ) = (12 x 10 10 N/m 2 )/(8.9 x 10 3 kg/m 3 ) Young's modulus Pb at 73 ø K = 15.2 x 10 10 N/m 2 temperature coefficient = (1.7 x 10 10-15.2 x 10 10 ) / (293 ø K - 73 ø K) = (-13.5 x 10 10 N/m 2 ) / 220 ø K temperature coefficient = -6.3 x 10 8 N/m 2 / ø K Thinking About the Answer Notice that the temperature coefficient is negative. As the temperature increases, the Young's modulus decreases. This is typical of many materials. They become less elastic as they are heated. Use the molecular theory of matter to explain this phenomenon. 129
PRACTICE TEST 1. A patient's heart of effective mass 0.05 kg has a maximum displacement given by Z = 0.02 sin3πt where Z is in meters and t is in seconds. a. Sketch a graph of the motion of this heart on the axis provide below. b. Find the position of the heart at.75 seconds. c. Give the frequency of oscillation of the heart. d. What is the total energy of the heart? e. What is the maximum displacement of the heart? f. Calculate the maximum velocity of the heart at Z = 0. g. The projection of uniform circular motion is simple harmonic motion. On the axis below, specify the requirements of a mass undergoing uniform circular motion so that the projection of the motion on the Z - axis will be exactly the same as the heart above. 130
2. In another monitoring session, the patient above is again observed. The position of the heart is again a periodic function and is illustrated below. a. Write the equation of the relation found in this graph. b. What is the maximum energy of the heart in this situation 131