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Topic 4: Oscillations and waves 4.2 Energy changes during SHM

Topic 4: Oscillations and waves 4.2 Energy changes during SHM. 4.2.1 Describe the interchange between kinetic energy and potential energy during SHM. 4.2.2 Apply the expressions for kinetic, potential, and total energy of a particle in SHM.

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Topic 4: Oscillations and waves 4.2 Energy changes during SHM

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  1. Topic 4: Oscillations and waves4.2 Energy changes during SHM 4.2.1 Describe the interchange between kinetic energy and potential energy during SHM. 4.2.2 Apply the expressions for kinetic, potential, and total energy of a particle in SHM. 4.2.3 Solve problems, both graphically and by calculation, involving energy changes during SHM.

  2. Topic 4: Oscillations and waves4.2 Energy changes during SHM Describe the interchange between kinetic energy and potential energy during SHM. Consider the pendulum to the right which is placed in position and held there. Let the green rectangle represent the potential energy of the system. Let the red rectangle represent the kinetic energy of the system. Because there is no motion yet, there is no kinetic energy. But if we release it, the kinetic energy will grow as the potential energy diminishes. A continuous exchange between EK and EP occurs.

  3. x relation EK and EP EK + EP = ET = CONST Topic 4: Oscillations and waves4.2 Energy changes during SHM Describe the interchange between kinetic energy and potential energy during SHM. Consider the mass- spring system shown here. The mass is pulled to the right and held in place. Let the green rectangle represent the potential energy of the system. Let the red rectangle represent the kinetic energy of the system. A continuous exchange between EK and EP occurs. Note that the sum of EK and EP is constant. FYI If friction and drag are both zero ET = CONST.

  4. relation EK and EP EK + EP = ET = CONST x Energy time Topic 4: Oscillations and waves4.2 Energy changes during SHM Describe the interchange between kinetic energy and potential energy during SHM. If we plot both kinetic energy and potential energy vs. time for either system we would get the following graph:

  5. relation between EK and EP EK + EP = ET = CONST relation between EK and x relation EK,max and x0 EK = (1/2)m2(x02 – x2) EK,max = (1/2)m2x02 Topic 4: Oscillations and waves4.2 Energy changes during SHM Apply the expressions for kinetic, potential, and total energy of a particle in SHM. Recall the relation between v and x that we derived in the last section:  v =  (x02 – x2). Then v2 = 2(x02 – x2) (1/2)mv2 = (1/2)m2(x02 – x2) EK = (1/2)m2(x02 – x2). Recall that vmax = v0 = x0 so that we have EK,max = (1/2)mvmax2 = (1/2)m2x02. These last two are in the Physics Data Booklet.

  6. relation between EK and EP EK + EP = ET = CONST x relation EK,max and x0 EK,max = (1/2)m2x02 Topic 4: Oscillations and waves4.2 Energy changes during SHM Solve problems, both graphically and by calculation, involving energy changes during SHM. EXAMPLE: A 3.00-kg mass is undergoing SHM with a period of 6.00 seconds. Its amplitude is 4.00 meters. (a) What is its maximum kinetic energy and what is x when this occurs? SOLUTION:  = 2/T = 2/6 = 1.05 rads-1. x0 = 4 m. EK,max = (1/2)m2x02 = (1/2)(3)(1.052)(42) = 26.5 J. EK = EK,max when x = 0.

  7. relation between EK and EP EK + EP = ET = CONST x relation EK,max and x0 EK,max = (1/2)m2x02 Topic 4: Oscillations and waves4.2 Energy changes during SHM Solve problems, both graphically and by calculation, involving energy changes during SHM. EXAMPLE: A 3.00-kg mass is undergoing SHM with a period of 6.00 seconds. Its amplitude is 4.00 meters. (b) What is its potential energy when the kinetic energy is maximum and what is the total energy of the system? SOLUTION: EK = EK,max when x = 0. Thus EP = 0. From EK + EP = ET = CONST we have 26.5 + 0 = ET= 26.5 J.

  8. relation between EK and EP EK + EP = ET = CONST x relation EK,max and x0 EK,max = (1/2)m2x02 Topic 4: Oscillations and waves4.2 Energy changes during SHM Solve problems, both graphically and by calculation, involving energy changes during SHM. EXAMPLE: A 3.00-kg mass is undergoing SHM with a period of 6.00 seconds. Its amplitude is 4.00 meters. (c) What is its potential energy when the kinetic energy is 15.0 J? SOLUTION: Since ET = 26.5 J then From EK + EP = 26.5 = CONST so we have 15.0 + EP = 26.5 J EP = 11.5 J.

  9. relation between EK and EP EK + EP = ET = CONST relation EK and x relation EK,max and x0 relation between ET and x0 potential energy EP EK = (1/2)m2(x02 – x2) EK,max = (1/2)m2x02 ET = (1/2)m2x02 EP = (1/2)m2x2 Topic 4: Oscillations and waves4.2 Energy changes during SHM Apply the expressions for kinetic, potential, and total energy of a particle in SHM. Since EP= 0 when EK = EK,max we have EK + EP = ET EK,max + 0 = ET From EK = (1/2)m2(x02 – x2) we get EK = (1/2)m2x02 – (1/2)m2x2 EK = ET – (1/2)m2x2 ET = EK + (1/2)m2x2

  10. x relation between ET and x0 potential energy EP ET = (1/2)m2x02 EP = (1/2)m2x2 Topic 4: Oscillations and waves4.2 Energy changes during SHM Solve problems, both graphically and by calculation, involving energy changes during SHM. EXAMPLE: A 3.00-kg mass is undergoing SHM with a period of 6.00 seconds. Its amplitude is 4.00 meters. (a) What is its potential energy when x = 2.00 m. What is its kinetic energy at this instant? SOLUTION:  = 2/T = 2/6 = 1.05 rads-1. x0 = 4 m. ET = (1/2)m2x02 = (1/2)(3)(1.052)(42) = 26.5 J. EP = (1/2)m2x2= (1/2)(3)(1.052)(22) = 6.62 J. ET = EK + EP so 26.5 = EK + 6.62 or EK = 19.9 J.

  11. x Topic 4: Oscillations and waves4.2 Energy changes during SHM Solve problems, both graphically and by calculation, involving energy changes during SHM. • PRACTICE: A 2.00-kg mass is undergoing SHM with a period of 1.75 seconds. • What is the total energy of this system? •  = 2/T = 2/1.75 = 3.59 rads-1. x0 = 3 m. • ET = (1/2)m2x02 = (1/2)(2)(3.592)(32) = 116 J. • (b) What is the potential energy of this system when x = 2.50 m? • EP = (1/2)m2x2 = (1/2)(2)(3.592)(2.52) = 80.6 J. FYI In all of these problems we assume the friction and the drag are both zero. The potential energy formula is not on the Physics Data Booklet.

  12. Topic 4: Oscillations and waves4.2 Energy changes during SHM Solve problems, both graphically and by calculation, involving energy changes during SHM. • PRACTICE: A 2-kg mass is undergoing SHM with a dis- placement vs. time plot shown. • What is the total energy of this system? •  = 2/T = 2/0.3 = 21 rads-1. x0 = .004 m. • ET = (1/2)m2x02 = (1/2)(2)(212)(.0042) = .007 J. • (b) What is the potential energy at t = 0.125 s? • From the graph x = 0.002 m so that • EP = (1/2)m2x2 = (1/2)(2)(212)(.0022) = .002 J. • (c) What is the kinetic energy at t = 0.125 s? • From EK + EP = ETwe get • EK + .002 = .007 so that Ek = .005 J.

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