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CHAPTER 10 Elasticity and Oscillations

CHAPTER 10 Elasticity and Oscillations. Oscillations. Oscillation or vibration = back-and-forth motion. Repeats itself, hence it is periodic. Vibrations common: Vocal cords when singing/speaking String/rubber band, Spring Simple Pendulum – Grandfather clock.

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CHAPTER 10 Elasticity and Oscillations

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  1. CHAPTER 10 Elasticity and Oscillations

  2. Oscillations • Oscillation or vibration = back-and-forth motion. • Repeats itself, hence it is periodic. • Vibrations common: • Vocal cords when singing/speaking • String/rubber band, • Spring • Simple Pendulum – Grandfather clock

  3. An object in stable equilibrium: • If displaced slightly from equilibrium point, will oscillate about the equilibrium point. • It will experience a restoring force. • The restoring force - directly proportional to displacement. • The restoring force is opposite to displacement. F = -kx

  4. Types of Equilibrium

  5. Simple Harmonic Motion F(x) x Simple Harmonic Motion (SHM) Periodic motion in which • Magnitude of the restoring force is proportional to displacement. • Restoring force is opposite to the displacement. • Spring (Elastic) force: F = -kx • So spring motion is SHM.

  6. Springs • Hooke’s Law:The force exerted by a spring is proportional to the distance by which the spring is stretched or compressed from its relaxed position. FX = -k x where xis the displacement from the relaxed position and k is the constant of proportionality. relaxed position (Equilibrium Position) FX = 0 x x=0

  7. Springs FX = -kx relaxed position FX > 0 Compressed x • x < 0 x = 0 relaxed position FX < 0 Stretched x • x > 0 x = 0

  8. x=A F=-kA v=0 a=-kA/m x=0 F=0 v=-vo a=0 x=-A F=kA v=0 a=kA/m x=0 F=0 v=+vo a=0 x=A F=-kA v=0 a=-kA/m

  9. x +A t -A • AMPLITUDE (x = A) = Maximum displacement • from equilibrium. • PERIOD (T) = Time to make one complete • cycle. • FREQUENCY (f) = # of cycles per second.

  10. Total energy of the spring-mass system: E = U(spring) + K(mass) = ½kx2 + ½mv2 x=A F=-kA v=0 a=-kA/m U=½kA2 K=0 x=0 F=0 v=-vo a=0 U=0 K=½mvo2 x=-A F=kA v=0 a=kA/m U =½kA2 K=0 x=0 F=0 v=+vo a=0 U=0 K=½mvo2

  11. Total Energy • At equilibrium point, E1 = U + K = ½mvo2 + 0 • At maximum displacement points, • E2 = U + K = ½kA2 + 0 • At any other position where displacement is x • and the velocity is vx; • E3 = ½kx2 + ½mvx2 • Conservation of energy: E1 = E2 = E3 • ½kA2 = ½mvo2 = ½kx2 + ½mvx2

  12. Total Energy E1 = E2 = E3 ½kA2 = ½mvo2 = ½kx2 + ½mvx2 ½mvo2 = ½kA2 OR vo = (k/m) A

  13. Energy in SHM • A mass is attached to a spring and set to motion. The maximum displacement is x = A • Energy = U + K = constant! = ½ k x2 + ½ m v2 • At maximum displacement x = A, v = 0 Total Energy = ½ k A2 + 0 • At zero displacement x = 0 Total Energy = 0 + ½ mvm2 Conservation of energy: ½ k A2 = ½ m vm2 Hence vo = (k/m) A • Analogy w/ marble in bowl

  14. US x 0 m x x=0 Energy in SHM ½ k A2 = ½ m vm2 vo = (k/m) A

  15. x t a t Sinusoidal Nature of SHM Sinusoidal equation = equation involving sine and cosine. v t

  16. Example #1 A mass on a spring oscillates back & forth with simple harmonic motion of amplitude A. At what points during its oscillation is the magnitude of the acceleration of the block biggest? • When x = +A or -A • When x = 0 (i.e. zero displacement) • The acceleration of the mass is constant.

  17. Example #2 A mass on a spring oscillates back & forth with simple harmonic motion of amplitude A. At what points during its oscillation is the magnitude of the potential energy (U) of the spring biggest? • When x = +A or -A • When x = 0 (i.e. zero displacement) • Potential energy of the spring is constant.

  18. x = 0 x = A Example #3 • A mass on a spring oscillates back & forth with simple harmonic motion of amplitude A. If the mass takes 2 seconds to move from x = 0 to x = A, what is the period of the oscillation? (A) 0.5s (B) 1s (C) 2s (D) 4s (E) 8s

  19. x +A t -A Example #4 A mass on a spring oscillates back & forth with simple harmonic motion of amplitude A. A plot of displacement (x) versus time (t) is shown below. At what points during its oscillation is the total energy (K+U) of the mass and spring a maximum? (Ignore gravity). 1. When x = +A or -A (i.e. max displacement) 2. When x = 0 (i.e. zero displacement) 3. The energy of the system is constant.

  20. x +A t -A Example #5 A mass on a spring oscillates back & forth with simple harmonic motion of amplitude A. A plot of displacement (x) versus time (t) is shown below. At what points during its oscillation is the speed of the block biggest? 1. When x = +A or -A (i.e. max displacement) 2. When x = 0 (i.e. zero displacement) 3. The speed of the mass is constant

  21. EndPoint Equilibrium EndPoint EndPoint Equilibrium EndPoint EndPoint Equilibrium EndPoint Example #6 A spring oscillates back and forth on a frictionless horizontal surface. A camera takes pictures of the position every 1/10th of a second. Which plot best shows the positions of the mass. (A) (B) (C)

  22. What does moving in a circlehave to do with moving back & forth in a straight line ??

  23. vo m A y  x m x = A x = 0 x = - A • Mass m moving in uniform circular motion, with constant speed vo [Angular velocity  = vo/A]. • x-component of position of the mass is same as SHM of the spring-mass system.

  24. vo m A y  x m x = A x = 0 x = - A • Circular motion can be used to describe SHM of the spring-mass system. • Just pay attention to the x-component of the circular motion. • x =Acos = Acos t

  25. Total energy: • E = = ½mvo2 = ½kA2 OR vo = (k/m) A vo = A  = 2A/T = 2Af vo = A  = (k/m) A OR  = (k/m) vo m A y  x m x = A x = 0 x = - A • Also, vo = 2A/T = (k/m) A • OR T = 2 (m/k) When amplitude (A) is doubled, maximum velocity (vo) will …… In the final analysis, period (T) depends only on m and k. When amplitude (A) is doubled, period (T) will ……

  26. x +A t -A ¼T ¾T ½T T Displacement x = Acos t = Acos (2/T)t Velocity v = -[A]sint Acceleration a = -[A2]cost

  27. y vo    x v A 0 T t ¾T m ¼T ½T -A x = A x = 0 x = - A vy = v0cos vx = -v0sin vo = A vx = -v0sin = (-A)sin = (-A)sint

  28. x=A F=-kA v=0 a=-kA/m x=0 F=0 v=-vo a=0 x=-A F=kA v=0 a=kA/m x=0 F=0 v=+vo a=0 x=A F=-kA v=0 a=-kA/m

  29. x +A t A2 t Displacement x = [A]cos t v Velocity v = -[A]sint A t Acceleration a = -[A2]cost a

  30. Simple Harmonic Motion: x(t) = [A]cos(t) v(t) = -[A]sin(t) a(t) = -[A2]cos(t) xmax = A vmax = vo = A amax = A2

  31. P10.25/28 The period of oscillation of an object in an ideal spring-and-mass system is 0.50 s and the amplitude is 5.0 cm. What is the speed at the equilibrium point? T = 0.5 s, A = 5.0 cm, vo = ? vo = A,  = 2/T, vo = (2/T)A

  32. P10.42/49 A body is suspended vertically from an ideal spring of spring constant 2.5 N/m. The spring is initially in its relaxed position. The body is then released and oscillates about its equilibrium position. The motion is described by y = (4.0 cm)sin[(0.70 rad/s)t] What is the maximum kinetic energy of the body? Kmax = ½ mvo2 y =Asin t and  = (k/m) A = 4.0 cm,  = 0.7 rad/s, m = k/2 vo = A,  = 2/T, vo = (2/T)A

  33. Simple Harmonic Oscillator • w = 2 p f = 2 p / T • x(t) = [A] cos(wt) • v(t) = -[Aw] sin(wt) • a(t) = -[Aw2] cos(wt) • For a Spring F = kx • amax = (k/m) A • Aw2 = (k/m) A w = sqrt(k/m)

  34. Vertical Mass and Spring • If we include gravity, there are two forces acting on mass. With mass, new equilibrium position has spring stretched d • SF = kd – mg = 0 d = mg/k Let this point be y=0 • SF = k(d-y) – mg = -k y • Same as horizontal! SHO • New equilibrium position y = d

  35. FT x m s mg mgsin  The Simple Pendulum • Simple pendulum – small object (pendulum bob) attached to the end of a light inextensible string of length L and swung through small displacements. • Motion of the bob back and forth along arc length (s) is a simple harmonic motion. • Restoring force bringing bob back to eqlbm position F = -mgsin m

  36. Types of Equilibrium

  37. FT L x m s mg mgsin  The Simple Pendulum • Restoring force bringing bob back to eqlbm position = F = -mgsin • For small angles, sin   = s/L  x/L • F = -mg  = -mgx/L, i.e. F  x • Compare with F = -kx, means k  mg/L • OR k/m = g/L . But k/m = 2 • So 2= g/L • Since T = 2(m/k) = 2(L/g) • Period of simple pendulum T = 2(L/g)

  38. The Simple Pendulum • Period of simple pendulum T = 2(L/g) • What should the length of a pendulum clock be for it to measure time accurately? • You need period T = 1s = 2(L/g). [L ~ 25 cm] • What will happen • If L is increased? [ie if L > 25 cm] • If L is decreased? [ie if L < 25 cm] • If g is decreased? (eg clock taken to the moon).

  39. #1 A grandfather clock is running too fast. To fix it, should the pendulum be lengthened or shortened? (A) lengthened (B) shortened (C) need to know the period T = 2(L/g)

  40. #2 A simple pendulum has a period of oscillation of one second here on Earth. On Mars (where g < 9.8 m/s2), the pendulum will have a period of oscillation   (A) greater than one second. (B) equal to one second. (C) less than one second. T = 2(L/g)

  41. Two simple pendulums A and B • Same lengths, Mass of A is twice the mass of B. • Equal vibrational amplitudes. • Periods TA and TB • Energies EA and EB • Choose the correct statement: • TA = TB and EA > EB • TA > TB and EA > EB • TA > TB and EA < EB • TA = TB and EA < EB

  42. A mass attached to a spring oscillates with a simple harmonic motion. If it takes 28 seconds to make 40 cycles, what is the frequency of the oscillation? • 7.0 Hz • 1,120 Hz • 0.70 Hz • 2.80 Hz • 1.43 Hz

  43. A mass on a spring oscillates with SHM. It takes 0.82 sec to move from x = 0 to x =A. What is the period of the oscillation? • 1.22 s • 0.21 s • 0.41 s • 1.64 s • 3.28 s

  44. A mass on a spring oscillates with simple harmonic motion. If the amplitude (A) is increased by a factor 3, the maximum kinetic energy of the mass will increase by a factor • 1 • 1/3 • 1/6 • 1/9 • 3 • 6 • 9

  45. A spring-mass system is in simple harmonic motion and the velocity of the mass is plotted against time as shown below. When is the potential energy of the spring at its minimum? • 2s, 6s • 0s, 4s, 8s • 0s, 2s, 4s, 6s, 8s • 1s, 2s, 3s • The potential energy is constant throughout

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