AP Physics C Mechanics- 7.4 Energy of Simple Harmonic Oscillators- Study Notes- New Syllabus
AP Physics C Mechanics- 7.4 Energy of Simple Harmonic Oscillators – Study Notes
AP Physics C Mechanics- 7.4 Energy of Simple Harmonic Oscillators – Study Notes – per latest Syllabus.
Key Concepts:
- Total Mechanical Energy in SHM
- Conservation of Energy in SHM
- Kinetic Energy in SHM
- Potential Energy in SHM
Total Mechanical Energy in SHM
In simple harmonic motion, the total mechanical energy of the system is the sum of its kinetic energy and potential energy at any instant:
\( \mathrm{E_{total} = K + U} \)
- \( \mathrm{E_{total}} \): total mechanical energy
- \( \mathrm{K} \): kinetic energy (\( \mathrm{\dfrac{1}{2}mv^2} \))
- \( \mathrm{U} \): potential energy (stored in the spring)
For a spring–mass system, potential energy is given by:
\( \mathrm{U = \dfrac{1}{2}kx^2} \)
Example:
A spring–mass system with spring constant \( \mathrm{k = 50\,N/m} \) and mass \( \mathrm{m = 0.5\,kg} \) oscillates with amplitude \( \mathrm{A = 0.08\,m} \). Show the total energy and express it as the sum of kinetic and potential energy at a general displacement \( \mathrm{x} \).
▶️ Answer / Explanation
Step 1 — total energy (maximum potential at endpoints):
\( \mathrm{E_{total} = \tfrac{1}{2}kA^2 = \tfrac{1}{2}(50)(0.08)^2 = \tfrac{1}{2}(50)(0.0064) = 0.16\,J} \)
Step 2 — potential energy at general position \( \mathrm{x} \):
\( \mathrm{U(x)=\tfrac{1}{2}k x^2} \)
Step 3 — kinetic energy from energy conservation:
\( \mathrm{K(x)=E_{total}-U(x)=0.16-\tfrac{1}{2}k x^2} \)
Example numeric check at \( \mathrm{x=0.04\,m\):
\( \mathrm{U=\tfrac{1}{2}(50)(0.04)^2=0.04\,J;\quad K=0.16-0.04=0.12\,J} \)
Result: The total energy is \( \mathrm{0.16\,J} \) and at any instant \( \mathrm{E_{total}=K+U} \).
Conservation of Energy in SHM
Because only conservative forces (like the spring force or gravity) act on the system, the total mechanical energy remains constant throughout the motion.
\( \mathrm{E_{total} = \dfrac{1}{2}kA^2 = constant} \)
At any displacement \( \mathrm{x} \):
\( \mathrm{\dfrac{1}{2}kA^2 = \dfrac{1}{2}kx^2 + \dfrac{1}{2}mv^2} \)
This equation shows that as the object moves, energy continuously transforms between kinetic and potential forms, but the total remains fixed.
Example :
Same system as above \( (\mathrm{k=50\,N/m},\ A=0.08\,m,\ E_{total}=0.16\,J) \). If at some instant \( \mathrm{x=0.06\,m} \), compute \( \mathrm{K} \) and \( \mathrm{U} \) and verify energy conservation.
▶️ Answer / Explanation
Step 1 — potential energy:
\( \mathrm{U=\tfrac{1}{2}k x^2=\tfrac{1}{2}(50)(0.06)^2=\tfrac{1}{2}(50)(0.0036)=0.09\,J} \)
Step 2 — kinetic energy by conservation:
\( \mathrm{K=E_{total}-U=0.16-0.09=0.07\,J} \)
Step 3 — direct check using velocity: find \( v \) from \( \mathrm{K=\tfrac{1}{2}mv^2}\):
\( \mathrm{v=\sqrt{\dfrac{2K}{m}}=\sqrt{\dfrac{2(0.07)}{0.5}}=\sqrt{0.28}=0.529\,m/s} \)
Result: \( \mathrm{U=0.09\,J,\ K=0.07\,J} \) and \( \mathrm{K+U=0.16\,J} \) — energy conserved.
Kinetic Energy in SHM
The kinetic energy (K) of a system in SHM is at its maximum when the system passes through its equilibrium position, where speed is greatest:
\( \mathrm{K = \dfrac{1}{2}m v^2 = \dfrac{1}{2}k(A^2 – x^2)} \)
- At equilibrium (\( \mathrm{x = 0} \)): \( \mathrm{K = \dfrac{1}{2}kA^2} \) (maximum)
- At extreme position (\( \mathrm{x = \pm A} \)): \( \mathrm{K = 0} \) (minimum)
Key Idea: Kinetic energy increases as the system approaches equilibrium and decreases toward the turning points.
Example :
A spring–mass system with spring constant \( \mathrm{k = 50\,N/m} \) and mass \( \mathrm{m = 0.5\,kg} \) oscillates with amplitude \( \mathrm{A = 0.08\,m} \). For the same spring–mass system, find the maximum kinetic energy and the position where it occurs.
▶️ Answer / Explanation
Step 1 — total energy (from earlier):
\( \mathrm{E_{total}=0.16\,J} \)
Step 2 — kinetic energy is maximal when potential energy is minimal. Potential energy is minimal at equilibrium \( \mathrm{x=0} \) where \( \mathrm{U=0} \). Therefore:
\( \mathrm{K_{max}=E_{total}=\tfrac{1}{2}kA^2=0.16\,J} \)
Step 3 — corresponding maximum speed:
\( \mathrm{v_{max}=\sqrt{\dfrac{2K_{max}}{m}}=\sqrt{\dfrac{2(0.16)}{0.5}}=\sqrt{0.64}=0.80\,m/s} \)
Result: Maximum kinetic energy \( \mathrm{0.16\,J} \) occurs at \( \mathrm{x=0} \); \( \mathrm{v_{max}=0.80\,m/s} \).
Potential Energy in SHM
The potential energy (U) of a system in SHM is stored in the restoring force (like a spring). It varies with displacement from equilibrium:
\( \mathrm{U = \dfrac{1}{2}k x^2} \)
- At maximum displacement (\( \mathrm{x = \pm A} \)): \( \mathrm{U = \dfrac{1}{2}kA^2} \) (maximum)
- At equilibrium (\( \mathrm{x = 0} \)): \( \mathrm{U = 0} \)
Key Relationship: As potential energy increases, kinetic energy decreases — their sum is constant.
The minimum kinetic energy of a system in SHM is zero, and the potential energy equals the total energy at that instant.
Dependence of Energy on Amplitude
Changing the amplitude of oscillation changes the total mechanical energy of the system because total energy depends on \( \mathrm{A^2} \):
\( \mathrm{E_{total} = \dfrac{1}{2}kA^2} \)
Thus:
- Doubling the amplitude quadruples the total energy.
- Energy is directly proportional to the square of amplitude (\( \mathrm{E \propto A^2} \)).
Relevant Equation for a Spring–Mass System:
\( \mathrm{E = \dfrac{1}{2}kA^2 = \dfrac{1}{2}mv_{max}^2} \)
Example
A \( \mathrm{0.25\,kg} \) block attached to a spring with constant \( \mathrm{k = 100\,N/m} \) oscillates with amplitude \( \mathrm{0.10\,m} \). Find: (a) the total energy of the system, (b) the kinetic and potential energies when the block is \( \mathrm{0.06\,m} \) from equilibrium.
▶️ Answer / Explanation
Step 1: Write known quantities:
\( \mathrm{m = 0.25\,kg,\ k = 100\,N/m,\ A = 0.10\,m,\ x = 0.06\,m.} \)
Step 2: Total mechanical energy:
\( \mathrm{E = \dfrac{1}{2}kA^2 = \dfrac{1}{2}(100)(0.10)^2 = 0.5\,J} \)
Step 3: Potential energy at \( \mathrm{x = 0.06\,m} \):
\( \mathrm{U = \dfrac{1}{2}k x^2 = \dfrac{1}{2}(100)(0.06)^2 = 0.18\,J} \)
Step 4: Kinetic energy (by conservation):
\( \mathrm{K = E – U = 0.50 – 0.18 = 0.32\,J} \)
Step 5: Maximum kinetic energy:
\( \mathrm{K_{max} = \dfrac{1}{2}kA^2 = 0.5\,J} \)