4.2.1Describe the interchange between kinetic energy and potential energy during SHM Apply the expressions for kinetic, potential, and total energy of a particle in SHM Solve problems, both graphically and by calculation, involving energy changes during SHM. Topic 4: Oscillations and waves 4.2 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 E K and E P occurs. Topic 4: Oscillations and waves 4.2 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 E K and E P occurs.  Note that the sum of E K and E P is constant. Topic 4: Oscillations and waves 4.2 Energy changes during SHM x E K + E P = E T = CONST relation E K and E P FYI  If friction and drag are both zero E T = 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 waves 4.2 Energy changes during SHM E K + E P = E T = CONST relation E K and E P 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 =  (x 0 2 – x 2 ).  Then v 2 =  2 (x 0 2 – x 2 ) (1/2)mv 2 = (1/2)m  2 (x 0 2 – x 2 ) E K = (1/2)m  2 (x 0 2 – x 2 ).  Recall that v max = v 0 =  x 0 so that we have E K,max = (1/2)mv max 2 = (1/2)m  2 x 0 2.  These last two are in the Physics Data Booklet. Topic 4: Oscillations and waves 4.2 Energy changes during SHM E K + E P = E T = CONST relation between E K and E P E K = (1/2)m  2 (x 0 2 – x 2 ) relation between E K and x E K,max = (1/2)m  2 x 0 2 relation E K,max and x 0

Solve problems, both graphically and by calculation, involving energy changes during SHM. Topic 4: Oscillations and waves 4.2 Energy changes during SHM E K + E P = E T = CONST relation between E K and E P E K,max = (1/2)m  2 x 0 2 relation E K,max and x 0 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. x 0 = 4 m.  E K,max = (1/2)m  2 x 0 2 = (1/2)(3)( )(4 2 ) = 26.5 J.  E K = E K,max when x = 0. x

Solve problems, both graphically and by calculation, involving energy changes during SHM. Topic 4: Oscillations and waves 4.2 Energy changes during SHM E K + E P = E T = CONST relation between E K and E P E K,max = (1/2)m  2 x 0 2 relation E K,max and x 0 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:  E K = E K,max when x = 0. Thus E P = 0.  From E K + E P = E T = CONST we have = E T = 26.5 J. x

Solve problems, both graphically and by calculation, involving energy changes during SHM. Topic 4: Oscillations and waves 4.2 Energy changes during SHM E K + E P = E T = CONST relation between E K and E P E K,max = (1/2)m  2 x 0 2 relation E K,max and x 0 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 E T = 26.5 J then  From E K + E P = 26.5 = CONST so we have E P = 26.5 J E P = 11.5 J. x

Apply the expressions for kinetic, potential, and total energy of a particle in SHM.  Since E P = 0 when E K = E K,max we have E K + E P = E T E K,max + 0 = E T  From E K = (1/2)m  2 (x 0 2 – x 2 ) we get E K = (1/2)m  2 x 0 2 – (1/2)m  2 x 2 E K = E T – (1/2)m  2 x 2 E T = E K + (1/2)m  2 x 2 Topic 4: Oscillations and waves 4.2 Energy changes during SHM E K + E P = E T = CONST relation between E K and E P E K = (1/2)m  2 (x 0 2 – x 2 ) relation E K and x E K,max = (1/2)m  2 x 0 2 relation E K,max and x 0 E T = (1/2)m  2 x 0 2 relation between E T and x 0 E P = (1/2)m  2 x 2 potential energy E P

Solve problems, both graphically and by calculation, involving energy changes during SHM. Topic 4: Oscillations and waves 4.2 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. x 0 = 4 m.  E T = (1/2)m  2 x 0 2 = (1/2)(3)( )(4 2 ) = 26.5 J.  E P = (1/2)m  2 x 2 = (1/2)(3)( )(2 2 ) = 6.62 J.  E T = E K + E P so 26.5 = E K or E K = 19.9 J. x E T = (1/2)m  2 x 0 2 relation between E T and x 0 E P = (1/2)m  2 x 2 potential energy E P

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. x 0 = 3 m.  E T = (1/2)m  2 x 0 2 = (1/2)(2)( )(3 2 ) = 116 J. (b) What is the potential energy of this system when x = 2.50 m?  E P = (1/2)m  2 x 2 = (1/2)(2)( )(2.5 2 ) = 80.6 J. Solve problems, both graphically and by calculation, involving energy changes during SHM. Topic 4: Oscillations and waves 4.2 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. x 0 =.004 m.  E T = (1/2)m  2 x 0 2 = (1/2)(2)(21 2 )( ) =.007 J. (b) What is the potential energy at t = s?  From the graph x = m so that  E P = (1/2)m  2 x 2 = (1/2)(2)(21 2 )( ) =.002 J. (c) What is the kinetic energy at t = s?  From E K + E P = E T we get  E K =.007 so that E k =.005 J. Solve problems, both graphically and by calculation, involving energy changes during SHM. Topic 4: Oscillations and waves 4.2 Energy changes during SHM