1 / 79

Chapter 11 Elasticity And Periodic Motion

Chapter 11 Elasticity And Periodic Motion. Stress characterizes the strength of the force associated with the stretch, squeeze, or twist, usually on a “force per unit area” basis. Strain describes the deformation that occurs.

Télécharger la présentation

Chapter 11 Elasticity And Periodic Motion

An Image/Link below is provided (as is) to download presentation Download Policy: Content on the Website is provided to you AS IS for your information and personal use and may not be sold / licensed / shared on other websites without getting consent from its author. Content is provided to you AS IS for your information and personal use only. Download presentation by click this link. While downloading, if for some reason you are not able to download a presentation, the publisher may have deleted the file from their server. During download, if you can't get a presentation, the file might be deleted by the publisher.


Presentation Transcript

  1. Chapter 11 Elasticity And Periodic Motion

  2. Stress characterizes the strength of the force associated with the stretch, squeeze, or twist, usually on a “force per unit area” basis. Strain describes the deformation that occurs. When the stress and strain are small enough, we often find that the two are directly proportional. The general pattern that emerges can be formulated as Stress/Strain = Constant.

  3. Experiments have shown that, for a sufficiently small tensile or compressive stress, stress and strain are proportional, as stated by Hooke’s law. The corresponding proportionality constant, called Young’s modulus (denoted by Y ), is given by Y = (Tensile stress)/(Tensile strain) or Y = (Compressive stress)/(Compressive strain), Or Y = (F /A) / (Δl / l0)

  4. Strain is a pure number, so the units of Young’s • modulus are the same as those of stress: force per • unit area. • Young’s modulus is a property of a specific material, • rather than of any particular object made of that • material. • A material with a large value of Y is relatively • un-stretchable; a large stress is required by a given • strain.

  5. Modeling an Elastic Material as a Spring Fperpendicular = (YAΔl) / l0 Let k = (YA)/ l0 , Δl = x and Fperpendicular = Fx ; then we have Fx = kx.

  6. The pressure in a fluid, denoted by p, is the force Fperpendicular per unit area A transmitted across any cross section of the fluid, against a wall of its container, or against a surface of an immersed object: p = Fperpendicular/A When a solid object is immersed in a fluid and both are at rest, the forces that the fluid exerts on the surface of the object are always perpendicular to the surface at each point.

  7. When Hooke’s law is obeyed, the volume strain is proportional to the volume stress (change in pressure). The corresponding constant ratio of stress to strain is called the bulk modulus, denoted by B. When the pressure on an object changes by a small amount Δp, from p0 to p0 + Δp, and the resulting volume strain is ΔV/V, Hooke’s law takes the form B = - (Δp)/(ΔV/V0) We include a minus sign in this equation because an increase in pressure always causes a decrease in volume. In other words, when Δp is positive, ΔV is negative.

  8. Shear Stress and Strain

  9. If the forces are small enough so that Hooke’s law is obeyed, the shear strain is proportional to the shear stress. The corresponding proportionality constant (ratio of shear stress to shear strain), is called the shear modulus, denoted by S: S = Shear stress / Shear strain = (Fparallel / A) / (x / h) = (Fparallel / A) / Φ

  10. Periodic Motion

  11. Fx = -kx Using Newton’s second law, max = -kx OR ax = -(k/m)x

  12. Amplitude, A • Cycle • Period, T • Frequency, f • f = 1/T • SI unit : Hertz (Hz) = cycle/s = 1/s • Angular frequency, ω • ω = 2πf = 2π/T

  13. Energy in Simple Harmonic Motion Conservation of Mechanical Energy E = (1/2) mvx2 + (1/2)kx2 = constant

  14. A useful equation When x = ± A, vx = 0. At this point, the energy is entirely potential energy and E = (1/2)kA2 . E = (1/2)kA2 = (1/2) mvx2 + (1/2)kx2 vx = ± k/mA2 – x2 We can use this equation to find the magnitude of the velocity for any given position x.

  15. Equations of Simple Harmonic Motion

  16. Periodic Motion

  17. The relationship between uniform circular motion and simple harmonic motion.

  18. Position of the Shadow as a Function of Time x = A cos θ x = A cos(ωt) x = A cos [(2π/T )t] SI unit: m

  19. Velocity in Simple Harmonic Motion v = -Aω sin(ωt) SI unit: m/s Maximum speed of the mass is vmax = Aω

  20. Period of a Mass on a Spring From equation ax = -Aω2 cos(ωt) ax = -Aω2 (maximum acceleration) (1) Also, we know that ax = -(kx)/m ax = -(kA)/m (maximum acceleration)(2) From equations (1) and (2) -Aω2 = -kx/m ω2 = k/m ω= k/m = 2πf = 2π/T T = 2πm / k SI unit: s

  21. The Simple Pendulum

  22. The restoring force F at each point is the component of force tangent to the circular path at that point: F = -mgsinθ If the angle is small, sinθis very nearly equal toθ(in radians). F = -mgθ = -mgx/L F = -(mg/L)x The restoring force F is then proportional to the coordinate x for small displacements, and the constant mg/L represents the force constant k. K=mg/L

  23. NOTE that the frequency does NOT depend on the mass but on the length of the pendulum

  24. A 0.85-kg mass attached to a vertical spring of force constant 150 N/m oscillates with a maximum speed of 0.35 m / s. Find the following quantities related to the motion of the mass: (a) the period, (b) the amplitude, (c) the maximum magnitude of the acceleration.

  25. A peg on a turntable moves with a constant linear speed of 0.67 m / s in a circle of radius 0.45 m. The peg casts a shadow on a wall. Find the following quantities related to the motion of the shadow: (a) the period, (b) the amplitude, (c) the maximum speed, and the maximum magnitude of acceleration.

  26. Ch. 11, Problem 42. • An object of unknown mass is attached to an ideal spring with force constant 120 N/m and is found to vibrate with frequency of 6.00 Hz. Find • period, • the angular frequency, and • the mass of this object.

  27. Ch. 11, Problem 49 • If a pendulum has period T and you double its • length, what is its new period in terms of T? • (b)If a pendulum has length L and you want to triple its frequency, what should be its length in terms of L? • (c)Suppose a pendulum has a length L and period T on earth. If you take it to a planet where the acceleration of freely falling objects is ten times what it is on earth, what should you do to the length to keep the period the same as on earth?

  28. (d) If you do not change the pendulum’s length in part (c ), what is its period on that planet in terms of T? (e) If a pendulum has a period T and you triple the mass of its bob, what happens to the period (in terms of T)?

  29. CHAPTER 11, PROBLEM 24 Find the period, frequency, and angular frequency of (a) the second hand and (b) the minute hand of a wall clock. CHAPTER 11, Problem 47 A certain simple pendulum has a period on earth of 1.60 s. What is the period on the surface on Mars, where the acceleration due to gravity is 3.71 m/s2 ?

  30. Chapter 11, Problem 33 • A mass is oscillating with amplitude A at the end of a spring. How far (in terms of A) is this mass from equilibrium position of the spring when the elastic potential energy equals the kinetic energy? • Chapter 11, Problem 34 • If a vibrating system has total energy E0, what will its total energy be (in terms of E0) if you double the amplitude of vibration? • If you want to triple the total energy of a vibrating system with amplitude A0, what should its new amplitude be (in terms of A0)?

  31. Chapter 11, Problem 58 An object suspended from a spring vibrates with simple harmonic motion. At an instant when the displacement of the object is equal to one-half the amplitude, what fraction of the total energy of the system is kinetic and what fraction is potential?

More Related