IB DP Physics: HL复习笔记4.1.4 Energy in SHM

• The total energy of an object oscillating with SHM is the sum of its potential energy (gravitational or elastic) and kinetic energy

E = E_P + E_K

• Where:
  • E = total energy in joules (J)
  • E_P = potential energy in joules (J)
  • E_K = kinetic energy in joules (J)

  

Graph of total energy E, potential energy E_P and kinetic energy E_K of an object oscillating with SHM

• If energy losses due to friction or drag are zero or ignored, the total energy E  of the system is conserved 
• The potential energy store of the object is at a maximum at the point of maximum displacement from the equilibrium position
  • The point of maximum displacement is amplitude x₀
  • Kinetic energy is zero at amplitude
  • Potential energy is equal to the total energy of the system at this point

• Energy is transferred from the object's potential energy store to its kinetic energy store as the object moves from amplitude to the equilibrium position
  • The object has both potential and kinetic energy
  • The sum of the potential and kinetic energy is equal to the total energy of the system
  • The total energy of the system is conserved

• The kinetic energy store of the object is at a maximum at the equilibrium position
  • This is because velocity is at a maximum as the object passes through the equilibrium position
  • Kinetic energy is equal to the total energy of the system at this point

• Energy is transferred from the object's kinetic energy store to its potential energy store as the object moves from the equilibrium position to amplitude
  • The object has both potential and kinetic energy
  • The sum of the potential and kinetic energy is equal to the total energy of the system
  • The total energy of the system is conserved

(i) Determine the total energy of the object by reading the maximum value of the kinetic energy from the graph 

• • From the graph, read the maximum value of the object's kinetic energy (E_K)_MAX = 60 mJ
  • Recall that, at the equilibrium position (x = 0), the total energy E is exactly equal to the maximum value of the kinetic energy (E_K)_MAX
  • Since energy losses can be neglected, the total energy is constant

E = 60 mJ

(ii) Read the amplitude of the object's oscillations from the graph 

• • The maximum displacement positions are the locations on either side of the equilibrium position where the kinetic energy is zero E_K = 0

x₀ = 2.0 cm

(iii)

Step 1: Recall the equation for the kinetic energy E_K of an object in terms of its mass m and velocity v

Step 2: Rearrange the above equation to calculate the velocity v

Step 3: Substitute the numbers into the equation to calculate the maximum velocity of the object 

• • Mass of the object, m = 0.50 kg
  • You must convert the maximum kinetic energy must from millijoules (mJ) into joules (J)

E_K = 60 mJ = 0.06 J

v = 0.49 m s^–1

(iv)

Step 1: Read the value of the kinetic energy E_K of the object when the displacement is x = 1.0 cm

E_K = 50 mJ

Step 2: Write down the relationship between total energy E, kinetic energy E_K and potential energy E_P

E = E_P + E_K

Step 3: Rearrange the above equation to calculate the potential energy E_P

E_P = E –  E_K

Step 4: Substitute the numbers in the above equation 

• • E_K = 50 mJ
  • E = 60 mJ

E_P = 60 mJ – 50 mJ

E_P = 10 mJ