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.

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.

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.

Topic 4: Oscillations and waves4.2 Energy changes during SHM

x

EK + EP = ET = CONST relation EK and EP

FYI

If friction and drag are both zero ET = CONST.

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:

Topic 4: Oscillations and waves4.2 Energy changes during SHM

EK + EP = ET = CONST relation EK and EP

Energy

time x

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 = (x0

2 – x2).Then v2 = 2(x0

2 – x2) (1/2)mv2 = (1/2)m2(x0

2 – x2) EK = (1/2)m2(x0

2 – x2).

Recall that vmax = v0 = x0 so that we have EK,max = (1/2)mvmax

2 = (1/2)m2x02.

These last two are in the Physics Data Booklet.

Topic 4: Oscillations and waves4.2 Energy changes during SHM

EK + EP = ET = CONST relation between EK and EP

EK = (1/2)m2(x02 – x2) relation between EK and x

EK,max = (1/2)m2x02 relation EK,max and x0

Solve problems, both graphically and by calculation, involving energy changes during SHM.

Topic 4: Oscillations and waves4.2 Energy changes during SHM

EK + EP = ET = CONST relation between EK and EP

EK,max = (1/2)m2x02 relation EK,max and x0

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 rad s-1. x0 = 4 m.

EK,max = (1/2)m2x02

= (1/2)(3)(1.052)(42) = 26.5 J.

EK = EK,max when x = 0.

x

Solve problems, both graphically and by calculation, involving energy changes during SHM.

Topic 4: Oscillations and waves4.2 Energy changes during SHM

EK + EP = ET = CONST relation between EK and EP

EK,max = (1/2)m2x02 relation EK,max and x0

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.

x

Solve problems, both graphically and by calculation, involving energy changes during SHM.

Topic 4: Oscillations and waves4.2 Energy changes during SHM

EK + EP = ET = CONST relation between EK and EP

EK,max = (1/2)m2x02 relation EK,max and x0

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.

x

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)m2(x02 – x2) we get

EK = (1/2)m2x02 – (1/2)m2x2

EK = ET – (1/2)m2x2

ET = EK + (1/2)m2x2

Topic 4: Oscillations and waves4.2 Energy changes during SHM

EK + EP = ET = CONST relation between EK and EP

EK = (1/2)m2(x02 – x2) relation EK and x

EK,max = (1/2)m2x02 relation EK,max and x0

ET = (1/2)m2x02 relation between ET and x0

EP = (1/2)m2x2 potential energy EP

Solve problems, both graphically and by calculation, involving energy changes during SHM.

Topic 4: Oscillations and waves4.2 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 rad s-1. x0 = 4 m.

ET = (1/2)m2x02 = (1/2)(3)(1.052)(42) = 26.5 J.

EP = (1/2)m2x2 = (1/2)(3)(1.052)(22) = 6.62 J.

ET = EK + EP so 26.5 = EK + 6.62 or EK = 19.9 J.

x

ET = (1/2)m2x02 relation between ET and x0

EP = (1/2)m2x2 potential energy EP

PRACTICE: A 2.00-kg mass is undergoing SHM with a period of 1.75 seconds. (a) What is the total energy of this system? = 2/T = 2/1.75 = 3.59 rad s-1. x0 = 3 m.ET = (1/2)m2x0

2 = (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)m2x2 = (1/2)(2)(3.592)(2.52) = 80.6 J.

Solve problems, both graphically and by calculation, involving energy changes during SHM.

Topic 4: Oscillations and waves4.2 Energy changes during SHM

x

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.

PRACTICE: A 2-kg mass is undergoing SHM with a dis- placement vs. time plot shown. (a) What is the total energy of this system? = 2/T = 2/0.3 = 21 rad s-1. x0 = .004 m. ET = (1/2)m2x0

2 = (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)m2x2 = (1/2)(2)(212)(.0022) = .002 J.(c) What is the kinetic energy at t = 0.125 s? From EK + EP = ET we get EK + .002 = .007 so that Ek = .005 J.

Solve problems, both graphically and by calculation, involving energy changes during SHM.

Topic 4: Oscillations and waves4.2 Energy changes during SHM