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Oscillations

Oscillations. Spring time again…. Let’s revisit Hooke’s law X>0 F x <0 Spring pulls back the mass toward its equilibrium position X=0 F x =0 Equilibrium position, no force X<0 F x >0 Spring pushes the mass back toward its equilibrium position. Restoring force and harmonic motion.

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Oscillations

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  1. Oscillations PHY231

  2. Spring time again… • Let’s revisit Hooke’s law • X>0 Fx<0 • Spring pulls back the mass toward its equilibrium position • X=0 Fx=0 • Equilibrium position, no force • X<0 Fx>0 • Spring pushes the mass back toward its equilibrium position PHY231

  3. Restoring force and harmonic motion • A restoring force always pushes or pulls the object toward its equilibrium position • Hooke’s law is an example of restoring force • If the net force occurring on an object is along the direction of motion and is similar to Hooke’s law then the object follows a simple harmonic motion PHY231

  4. Period of the motion • The period T is the time it takes for the mass to come back to the exact same location with the exact same velocity. For the horizontal spring system • Mass comes back to same x and v periodically • Max displacement • t1=0 s t2=6.3 s • T= t2-t1 =6.3 s • Intermediate pos. • t1=2.5 s t2=8.8 s • T= t2-t1 =6.3 s PHY231

  5. Frequency, angular frequency • The frequency f is the inverse of the period T. It corresponds to the number of cycles completed each second. For a mass on a spring SI unit Hz=1/s • The angular frequency w is defined as 2p times the frequency SI unit rad/s PHY231

  6. k =spring-constant , m=mass , w=ang. freq w = (k/m)0.5 Visualization • Equilibrium position • Zero F • Zero accel • Zero SPE • Maximum KE = 0.5*kA2 • Maximum velocity = Aw • Maximum displacement A • Maximum F = kA • Maximum accel = Aw2 • Maximum SPE = 0.5*kA2 • Zero KE • Zero velocity A A

  7. Time evolution T • Simple harmonic motion is described by the following equations • Maximum values T Aw Aw2

  8. Vertical mass-spring system • Exactly the same as horizontal mass-spring system but for the fact that gravitational potential energy is now also varying when the mass moves up and down • If x refers to the displacement with respect to the equilibrium position then all the previous formulas are still valid

  9. At equilibrium position • When an object is moving in simple harmonic motion, which of the following is at a minimum when the displacement from equilibrium is zero? • A) the magnitude of the velocity • B) the magnitude of the acceleration • C) the kinetic energy • D) the total mechanical energy PHY231

  10. At equilibrium position • A) the magnitude of the velocity • B) the magnitude of the acceleration • C) the kinetic energy • D) the total mechanical energy PHY231

  11. Period T • Consider a mass-spring system on a horizontal frictionless surface set in simple harmonic motion with amplitude A. How does the period of the system change if the mass is quadrupled? • A) It doubles • B) It quadruples • C) It is half • D) It doesn’t change PHY231

  12. Period T Consider a mass-spring system on a horizontal frictionless surface set in simple harmonic motion with amplitude A. How does the period of the system change if the mass is quadrupled? • A) It doubles • B) It quadruples • C) It is half • D) It doesn’t change PHY231

  13. Cosine function • A mass on a spring vibrates in simple harmonic motion at a frequency of 4.0 Hz and an amplitude of 8.0 cm. If a timer is started when its displacement is a maximum (hence x = 8 cm when t = 0), what is the displacement of the mass when t = 3.7 s? • A) zero • B) 0.025 m • C) 0.036 m • D) 0.080 m PHY231

  14. Cosine function • A mass on a spring vibrates in simple harmonic motion at a frequency of 4.0 Hz and an amplitude of 8.0 cm. If a timer is started when its displacement is a maximum (hence x = 8 cm when t = 0), what is the displacement of the mass when t = 3.7 s? • A) zero • B) 0.025 m • C) 0.036 m • D) 0.080 m PHY231

  15. Spring combo • A mass of 4.0 kg, resting on a horizontal frictionless surface, is attached on the right to a horizontal spring with spring constant 20 N/m and on the left to a horizontal spring with spring constant 50 N/m. • If this system is moved from equilibrium, what is the effective spring constant? • A) 30 N/m • B) -30 N/m • C) 70 N/m • D) 14 N/m PHY231

  16. Spring combo • A) 30 N/m • B) -30 N/m • C) 70 N/m • D) 14 N/m Fright-spring Fleft-spring Equilibrium position PHY231

  17. Pendulum • The pendulum is a mass swinging at the end of a light string • If the amplitude of the motion is small, then the motion of the pendulum is approximately a simple harmonic motion • For the mass-spring system acceleration = - constant * displacement We want to find an equivalent equation for the pendulum

  18. Pendulum motion PHY231

  19. Pendulum period • We have found for the pendulum • We see that it is similar in form to the spring mass equation a = - k/m*x • acceleration = - constant * displacement • We can use the results of the spring mass system if we replace: • x by q, a by a, and k/m by g/L • Thus, the period for the pendulum is given by PHY231

  20. Motion: pendulum / spring-mass • We saw that both systems have similar equations • We can visualize the similarities in their motions PHY231

  21. Demo: Pendulum physlet • Dependence of T with the length of the pendulum • We try two different lengths for the pendulum • L1 = 0.50 m and L2=2.0 m • On earth g = 9.8 m/s2 • T1=2*π*√(0.5/9.8 )= 1.4 s • T2=2*π*√(2.0/9.8 )= 2.8 s PHY231

  22. Result with physlet L = 0.50 m T = 1.4 s L = 2.0 m T = 2.8 s

  23. Different planet • What should be the period of a pendulum on the moon if its period on earth is 2.0 s ? • gearth = 5.8*gmoon • A) 0.34 s • B) 0.83 s • C) 4.8 s • D) 12 s PHY231

  24. Different planet • What should be the period of a pendulum on the moon if its period on earth is 2.0 s ? • gearth = 5.8*gmoon • A) 0.34 s • B) 0.83 s • C) 4.8 s • D) 12 s PHY231

  25. Result with physlet Earth T = 2.0 s Moon T = 4.9 s

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