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Physics 6B

Physics 6B. Oscillations. Prepared by Vince Zaccone For Campus Learning Assistance Services at UCSB. Definitions of quantities describing periodic motion. Period (T): time required for a motion to go through a complete cycle Frequency (f): number of oscillations per unit time

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Physics 6B

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  1. Physics 6B Oscillations Prepared by Vince Zaccone For Campus Learning Assistance Services at UCSB

  2. Definitions of quantities describing periodic motion • Period (T): time required for a motion to go through a complete cycle • Frequency (f): number of oscillations per unit time • Standard unit for frequency is Hertz (Hz) • 1 Hz = 1 cycle/second Period and frequency are reciprocals • Angular frequency: • The amplitude (A) is the maximum displacement from equilibrium.

  3. Simple Harmonic Motion A spring exerts a restoring force that is proportional to the displacement from equilibrium:

  4. Simple Harmonic Motion • Simple harmonic motion occurs when the restoring force is proportional to the displacement from equilibrium. • Period of a mass on a spring: • Total energy in simple harmonic motion:

  5. Equations for Simple Harmonic Motion • Position as a function of time: • Velocity as a function of time: • Acceleration as a function of time: • Note that v(t) is just the derivative of x(t), and a(t) is the derivative of v(t)

  6. Energy in Simple Harmonic Motion • Potential energy as a function of time: • Kinetic energy as a function of time: ½ kA2 _

  7. Example 1 A 0.98 kg block slides on a frictionless horizontal surface with speed 1.32m/s. The block encounters an unstretched spring with force constant 245N/m, as shown. a) how far is the spring compressed before the block comes to rest? b) how long is the block in contact with the spring before it comes to rest? c) how far is the spring compressed when the kinetic energy of the block is equal to the potential energy stored in the spring? Prepared by Vince Zaccone For Campus Learning Assistance Services at UCSB

  8. Example 1 A 0.98 kg block slides on a frictionless horizontal surface with speed 1.32m/s. The block encounters an unstretched spring with force constant 245N/m, as shown. a) how far is the spring compressed before the block comes to rest? b) how long is the block in contact with the spring before it comes to rest? c) how far is the spring compressed when the kinetic energy of the block is equal to the potential energy stored in the spring? We can use energy conservation for the first part, setting the initial kinetic energy of the block equal to the final potential energy stored in the spring. Block at rest (spring fully compressed) Prepared by Vince Zaccone For Campus Learning Assistance Services at UCSB

  9. Example 1 A 0.98 kg block slides on a frictionless horizontal surface with speed 1.32m/s. The block encounters an unstretched spring with force constant 245N/m, as shown. a) how far is the spring compressed before the block comes to rest? b) how long is the block in contact with the spring before it comes to rest? c) how far is the spring compressed when the kinetic energy of the block is equal to the potential energy stored in the spring? Δx We can use energy conservation for the first part, setting the initial kinetic energy of the block equal to the final potential energy stored in the spring. Block at rest (spring fully compressed) Prepared by Vince Zaccone For Campus Learning Assistance Services at UCSB

  10. Example 1 A 0.98 kg block slides on a frictionless horizontal surface with speed 1.32m/s. The block encounters an unstretched spring with force constant 245N/m, as shown. a) how far is the spring compressed before the block comes to rest? b) how long is the block in contact with the spring before it comes to rest? c) how far is the spring compressed when the kinetic energy of the block is equal to the potential energy stored in the spring? Δx We can use energy conservation for the first part, setting the initial kinetic energy of the block equal to the final potential energy stored in the spring. Block at rest (spring fully compressed) Prepared by Vince Zaccone For Campus Learning Assistance Services at UCSB

  11. Example 1 A 0.98 kg block slides on a frictionless horizontal surface with speed 1.32m/s. The block encounters an unstretched spring with force constant 245N/m, as shown. a) how far is the spring compressed before the block comes to rest? b) how long is the block in contact with the spring before it comes to rest? c) how far is the spring compressed when the kinetic energy of the block is equal to the potential energy stored in the spring? Δx We can use energy conservation for the first part, setting the initial kinetic energy of the block equal to the final potential energy stored in the spring. Block at rest (spring fully compressed) Prepared by Vince Zaccone For Campus Learning Assistance Services at UCSB

  12. Example 1 A 0.98 kg block slides on a frictionless horizontal surface with speed 1.32m/s. The block encounters an unstretched spring with force constant 245N/m, as shown. a) how far is the spring compressed before the block comes to rest? b) how long is the block in contact with the spring before it comes to rest? c) how far is the spring compressed when the kinetic energy of the block is equal to the potential energy stored in the spring? Δx v=1.32 For part b) we can use the formula for the period of oscillation of a mass-on-a-spring: Block at rest (spring fully compressed) v=1.32 Prepared by Vince Zaccone For Campus Learning Assistance Services at UCSB

  13. Example 1 A 0.98 kg block slides on a frictionless horizontal surface with speed 1.32m/s. The block encounters an unstretched spring with force constant 245N/m, as shown. a) how far is the spring compressed before the block comes to rest? b) how long is the block in contact with the spring before it comes to rest? c) how far is the spring compressed when the kinetic energy of the block is equal to the potential energy stored in the spring? Δx v=1.32 For part b) we can use the formula for the period of oscillation of a mass-on-a-spring: In this case we only want ¼ of the period. Block at rest (spring fully compressed) v=1.32 Prepared by Vince Zaccone For Campus Learning Assistance Services at UCSB

  14. Example 1 A 0.98 kg block slides on a frictionless horizontal surface with speed 1.32m/s. The block encounters an unstretched spring with force constant 245N/m, as shown. a) how far is the spring compressed before the block comes to rest? b) how long is the block in contact with the spring before it comes to rest? c) how far is the spring compressed when the kinetic energy of the block is equal to the potential energy stored in the spring? Δx v=1.32 Part c) is easiest to understand using energy. We know that Etotal = Kinetic + Potential. Block at rest (spring fully compressed) v=1.32 Prepared by Vince Zaccone For Campus Learning Assistance Services at UCSB

  15. Example 1 A 0.98 kg block slides on a frictionless horizontal surface with speed 1.32m/s. The block encounters an unstretched spring with force constant 245N/m, as shown. a) how far is the spring compressed before the block comes to rest? b) how long is the block in contact with the spring before it comes to rest? c) how far is the spring compressed when the kinetic energy of the block is equal to the potential energy stored in the spring? Δx v=1.32 Part c) is easiest to understand using energy. We know that Etotal = Kinetic + Potential. We can also calculate the total energy from the given initial speed: Block at rest (spring fully compressed) v=1.32 Prepared by Vince Zaccone For Campus Learning Assistance Services at UCSB

  16. Example 1 A 0.98 kg block slides on a frictionless horizontal surface with speed 1.32m/s. The block encounters an unstretched spring with force constant 245N/m, as shown. a) how far is the spring compressed before the block comes to rest? b) how long is the block in contact with the spring before it comes to rest? c) how far is the spring compressed when the kinetic energy of the block is equal to the potential energy stored in the spring? Δx v=1.32 Part c) is easiest to understand using energy. We know that Etotal = Kinetic + Potential. We can also calculate the total energy from the given initial speed: Now we have to realize that when the kinetic and potential energies are equal, they are also each equal to half of the total energy. Block at rest (spring fully compressed) v=1.32 Prepared by Vince Zaccone For Campus Learning Assistance Services at UCSB

  17. Example 1 A 0.98 kg block slides on a frictionless horizontal surface with speed 1.32m/s. The block encounters an unstretched spring with force constant 245N/m, as shown. a) how far is the spring compressed before the block comes to rest? b) how long is the block in contact with the spring before it comes to rest? c) how far is the spring compressed when the kinetic energy of the block is equal to the potential energy stored in the spring? Δx v=1.32 Part c) is easiest to understand using energy. We know that Etotal = Kinetic + Potential. We can also calculate the total energy from the given initial speed: Now we have to realize that when the kinetic and potential energies are equal, they are also each equal to half of the total energy. Since we want to find the compression distance, we should use the formula involving potential energy: Block at rest (spring fully compressed) v=1.32 Prepared by Vince Zaccone For Campus Learning Assistance Services at UCSB

  18. Example 1 A 0.98 kg block slides on a frictionless horizontal surface with speed 1.32m/s. The block encounters an unstretched spring with force constant 245N/m, as shown. a) how far is the spring compressed before the block comes to rest? b) how long is the block in contact with the spring before it comes to rest? c) how far is the spring compressed when the kinetic energy of the block is equal to the potential energy stored in the spring? Δx v=1.32 Part c) is easiest to understand using energy. We know that Etotal = Kinetic + Potential. We can also calculate the total energy from the given initial speed: Now we can realize that when the kinetic and potential energies are equal, they are also each equal to half of the total energy. Since we want to find the compression distance, we should use the formula involving potential energy: Block at rest (spring fully compressed) v=1.32 Prepared by Vince Zaccone For Campus Learning Assistance Services at UCSB

  19. Mass-Spring Example Stop the Block Prepared by Vince Zaccone For Campus Learning Assistance Services at UCSB

  20. Example 2 A 0.260kg mass is attached to a vertical spring. When the mass is put into motion, its period is 1.12sec. How much does the mass stretch the spring when it is at rest in its equilibrium position? Prepared by Vince Zaccone For Campus Learning Assistance Services at UCSB

  21. Example 2 A 0.260kg mass is attached to a vertical spring. When the mass is put into motion, its period is 1.12sec. How much does the mass stretch the spring when it is at rest in its equilibrium position? ? Prepared by Vince Zaccone For Campus Learning Assistance Services at UCSB

  22. Example 2 A 0.260kg mass is attached to a vertical spring. When the mass is put into motion, its period is 1.12sec. How much does the mass stretch the spring when it is at rest in its equilibrium position? If we consider all the forces acting on the mass when it is hanging at rest we see that the weight must cancel the spring force. So if we can find the spring constant k, we can solve for x. ? kx mg Prepared by Vince Zaccone For Campus Learning Assistance Services at UCSB

  23. Example 2 A 0.260kg mass is attached to a vertical spring. When the mass is put into motion, its period is 1.12sec. How much does the mass stretch the spring when it is at rest in its equilibrium position? If we consider all the forces acting on the mass when it is hanging at rest we see that the weight must cancel the spring force. So if we can find the spring constant k, we can solve for x. Use the formula for the period of a mass-spring system: ? kx mg Prepared by Vince Zaccone For Campus Learning Assistance Services at UCSB

  24. Example 2 A 0.260kg mass is attached to a vertical spring. When the mass is put into motion, its period is 1.12sec. How much does the mass stretch the spring when it is at rest in its equilibrium position? If we consider all the forces acting on the mass when it is hanging at rest we see that the weight must cancel the spring force. So if we can find the spring constant k, we can solve for x. Use the formula for the period of a mass-spring system: ? kx Plugging in the given values we get mg Prepared by Vince Zaccone For Campus Learning Assistance Services at UCSB

  25. Example 2 A 0.260kg mass is attached to a vertical spring. When the mass is put into motion, its period is 1.12sec. How much does the mass stretch the spring when it is at rest in its equilibrium position? If we consider all the forces acting on the mass when it is hanging at rest we see that the weight must cancel the spring force. So if we can find the spring constant k, we can solve for x. Use the formula for the period of a mass-spring system: ? kx Plugging in the given values we get mg Now we can use Fspring=weight: Prepared by Vince Zaccone For Campus Learning Assistance Services at UCSB

  26. Example 2 A 0.260kg mass is attached to a vertical spring. When the mass is put into motion, its period is 1.12sec. How much does the mass stretch the spring when it is at rest in its equilibrium position? If we consider all the forces acting on the mass when it is hanging at rest we see that the weight must cancel the spring force. So if we can find the spring constant k, we can solve for x. Use the formula for the period of a mass-spring system: ? kx Plugging in the given values we get mg Now we can use Fspring=weight: Prepared by Vince Zaccone For Campus Learning Assistance Services at UCSB

  27. The Simple Pendulum Looking at the forces on the pendulum bob, we see that the restoring force is proportional to sin θ,whereas the restoring force for a spring is proportional to the displacement (which is θ in this case).

  28. Period of a Pendulum • A simple pendulum with small amplitude exhibits simple harmonic motion • Period of a simple pendulum: • Period of a physical pendulum:

  29. You want to construct a simple pendulum with a 3.25kg bob that will take 1.13sec to swing from a maximum angle of 9.5° to its lowest point after being released from rest. How long should this pendulum be? Here is a diagram of the pendulum. θ Prepared by Vince Zaccone For Campus Learning Assistance Services at UCSB

  30. You want to construct a simple pendulum with a 3.25kg bob that will take 1.13sec to swing from a maximum angle of 9.5° to its lowest point after being released from rest. How long should this pendulum be? Here is a diagram of the pendulum. We have a formula for the period of this pendulum: θ Prepared by Vince Zaccone For Campus Learning Assistance Services at UCSB

  31. You want to construct a simple pendulum with a 3.25kg bob that will take 1.13sec to swing from a maximum angle of 9.5° to its lowest point after being released from rest. How long should this pendulum be? Here is a diagram of the pendulum. We have a formula for the period of this pendulum: θ We can solve this for the length: Prepared by Vince Zaccone For Campus Learning Assistance Services at UCSB

  32. You want to construct a simple pendulum with a 3.25kg bob that will take 1.13sec to swing from a maximum angle of 9.5° to its lowest point after being released from rest. How long should this pendulum be? Here is a diagram of the pendulum. We have a formula for the period of this pendulum: θ We can solve this for the length: What value should we use for the period? We are given a time of 1.13s to go from max angle to the lowest point. Prepared by Vince Zaccone For Campus Learning Assistance Services at UCSB

  33. You want to construct a simple pendulum with a 3.25kg bob that will take 1.13sec to swing from a maximum angle of 9.5° to its lowest point after being released from rest. How long should this pendulum be? Here is a diagram of the pendulum. We have a formula for the period of this pendulum: θ We can solve this for the length: What value should we use for the period? We are given a time of 1.13s to go from max angle to the lowest point. This is only ¼ of a full cycle. So we multiply by 4: T = 4.52s Prepared by Vince Zaccone For Campus Learning Assistance Services at UCSB

  34. You want to construct a simple pendulum with a 3.25kg bob that will take 1.13sec to swing from a maximum angle of 9.5° to its lowest point after being released from rest. How long should this pendulum be? Here is a diagram of the pendulum. We have a formula for the period of this pendulum: θ We can solve this for the length: What value should we use for the period? We are given a time of 1.13s to go from max angle to the lowest point. This is only ¼ of a full cycle. So we multiply by 4: T = 4.52s Now we can plug in to get our answer: Prepared by Vince Zaccone For Campus Learning Assistance Services at UCSB

  35. Damped Oscillations • Oscillations where there is a nonconservative (i.e. friction) force are called damped. • Underdamped: the amplitude decreases exponentially with time: • The frequency of oscillation is also affected. • Critically damped: no oscillations; system relaxes back to equilibrium in minimum time • Overdamped: also no oscillations, but slower than critical damping

  36. Damped Oscillations

  37. Driven Oscillations • An oscillating system may be driven by an external force • This force may replace energy lost to friction, or may cause the amplitude to increase greatly at resonance • Resonance occurs when the driving frequency is equal to the natural frequency of the system

  38. Driven Oscillations • The Amplitude of the oscillation depends on the driving frequency.

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