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

2. 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

3. 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? Block at rest (spring fully compressed) 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. Prepared by Vince Zaccone For Campus Learning Assistance Services at UCSB

4. 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? Block at rest (spring fully compressed) 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. ΔxΔx Prepared by Vince Zaccone For Campus Learning Assistance Services at UCSB

5. 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? Block at rest (spring fully compressed) 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. ΔxΔx Prepared by Vince Zaccone For Campus Learning Assistance Services at UCSB

6. 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? Block at rest (spring fully compressed) 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. ΔxΔx Prepared by Vince Zaccone For Campus Learning Assistance Services at UCSB

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? Block at rest (spring fully compressed) For part b) we can use the formula for the period of oscillation of a mass-on-a-spring: ΔxΔx v=1.32 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? Block at rest (spring fully compressed) For part b) we can use the formula for the period of oscillation of a mass-on-a-spring: ΔxΔx v=1.32 In this case we only want ¼ of the period. 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? Block at rest (spring fully compressed) Part c) is easiest to understand using energy. We know that E total = Kinetic + Potential. ΔxΔx v=1.32 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? Block at rest (spring fully compressed) Part c) is easiest to understand using energy. We know that E total = Kinetic + Potential. We can also calculate the total energy from the given initial speed: ΔxΔx v=1.32 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? Block at rest (spring fully compressed) Part c) is easiest to understand using energy. We know that E total = Kinetic + Potential. We can also calculate the total energy from the given initial speed: ΔxΔx v=1.32 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. 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? Block at rest (spring fully compressed) Part c) is easiest to understand using energy. We know that E total = Kinetic + Potential. We can also calculate the total energy from the given initial speed: ΔxΔx v=1.32 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: 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? Block at rest (spring fully compressed) Part c) is easiest to understand using energy. We know that E total = Kinetic + Potential. We can also calculate the total energy from the given initial speed: ΔxΔx v=1.32 Now we have to realize that since 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: Prepared by Vince Zaccone For Campus Learning Assistance Services at UCSB

14. 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

15. 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

16. 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? ? mg kx 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. Prepared by Vince Zaccone For Campus Learning Assistance Services at UCSB

17. 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? ? mg kx 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: Prepared by Vince Zaccone For Campus Learning Assistance Services at UCSB

18. 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? ? mg kx 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: Plugging in the given values we get Prepared by Vince Zaccone For Campus Learning Assistance Services at UCSB

19. 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? ? mg kx Plugging in the given values we get Now we can use F spring =weight: 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: 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? ? mg kx Plugging in the given values we get Now we can use F spring =weight: 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: Prepared by Vince Zaccone For Campus Learning Assistance Services at UCSB

21. 11.46 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? Prepared by Vince Zaccone For Campus Learning Assistance Services at UCSB

22. 11.46 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? θ Here is a diagram of the pendulum. Prepared by Vince Zaccone For Campus Learning Assistance Services at UCSB

23. 11.46 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? θ Here is a diagram of the pendulum. We have a formula for the period of this pendulum: Prepared by Vince Zaccone For Campus Learning Assistance Services at UCSB

24. 11.46 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? θ Here is a diagram of the pendulum. We have a formula for the period of this pendulum: We can solve this for the length: Prepared by Vince Zaccone For Campus Learning Assistance Services at UCSB

25. 11.46 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? θ Here is a diagram of the pendulum. We have a formula for the period of this pendulum: We can solve this for the length: What value should we use for the period? Prepared by Vince Zaccone For Campus Learning Assistance Services at UCSB

26. 11.46 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? θ Here is a diagram of the pendulum. We have a formula for the period of this pendulum: 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

27. 11.46 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? θ Here is a diagram of the pendulum. We have a formula for the period of this pendulum: 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