CIE AS/A Level Physics 17.2 Energy in simple harmonic motion Study Notes- 2025-2027 Syllabus
CIE AS/A Level Physics 17.2 Energy in simple harmonic motion Study Notes – New Syllabus
CIE AS/A Level Physics 17.2 Energy in simple harmonic motion Study Notes at IITian Academy focus on specific topic and type of questions asked in actual exam. Study Notes focus on AS/A Level Physics latest syllabus with Candidates should be able to:
- describe the interchange between kinetic and potential energy during simple harmonic motion
- recall and use \( \mathrm{E = \tfrac{1}{2} m\omega^2 x_0^2} \) for the total energy of a system undergoing simple harmonic motion
Energy Interchange in Simple Harmonic Motion (SHM)
In simple harmonic motion, the total mechanical energy remains constant (when no damping is present). However, the form of the energy continually changes between:
- Kinetic energy (KE) — due to motion of the particle
- Potential energy (PE) — due to displacement from equilibrium (elastic potential energy for springs, gravitational for pendulums)
Key Concepts:
- At the equilibrium position:
- Displacement = 0
- KE is maximum
- PE is minimum (zero)
- At the maximum displacement (±x₀): ‣
- Velocity = 0
- KE = 0
- PE is maximum
- At intermediate positions:
- Both KE and PE are present
- Their sum is constant
Mathematical Expressions:
Kinetic Energy: \( \mathrm{KE = \tfrac{1}{2} m v^2 = \tfrac{1}{2} m \omega^2 (x_0^2 – x^2)} \)
Potential Energy: \( \mathrm{PE = \tfrac{1}{2} m \omega^2 x^2} \)
Total Energy (constant): \( \mathrm{E = \tfrac{1}{2} m \omega^2 x_0^2} \)
Meaning:
Energy continuously transforms from potential → kinetic → potential as the oscillator moves. This smooth interchange is the hallmark of SHM.
Example
A mass on a spring passes through its equilibrium position. Describe the kinetic and potential energies at this instant.
▶️ Answer / Explanation
At equilibrium:
- Displacement = 0 → PE = 0
- Velocity is maximum → KE is maximum
The energy is entirely kinetic at this point.
Example
A particle oscillates with amplitude 0.10 m. At a displacement of 0.06 m, what happens to its kinetic and potential energies qualitatively?
▶️ Answer / Explanation
Potential energy increases with displacement:
\( \mathrm{PE \propto x^2} \rightarrow PE \ \text{is increased at } x = 0.06\ m}
Kinetic energy decreases accordingly because:
\( \mathrm{KE = Total\ Energy – PE} \)
So at x = 0.06 m, PE is significant but not maximum, and KE is reduced but not zero.
Example
A 0.50 kg mass performs SHM with amplitude 0.08 m and angular frequency \( \mathrm{5\ rad\,s^{-1}} \). Calculate the kinetic and potential energies when its displacement is 0.05 m.
▶️ Answer / Explanation
Total energy:
\( \mathrm{E = \tfrac{1}{2} m\omega^2 x_0^2} \) \( \mathrm{= \tfrac{1}{2}(0.50)(5^2)(0.08^2)} \)
\( \mathrm{E = 0.5 \times 25 \times 0.0064 = 0.08\ J} \)
Potential energy at x = 0.05 m:
\( \mathrm{PE = \tfrac{1}{2} m\omega^2 x^2} \) \( \mathrm{= \tfrac{1}{2}(0.50)(25)(0.05^2)} \)
\( \mathrm{PE = 0.5 \times 25 \times 0.0025 = 0.03125\ J} \)
Kinetic energy:
\( \mathrm{KE = E – PE = 0.08 – 0.03125 = 0.04875\ J} \)
Final Answer:
- Potential energy = \( \mathrm{0.031\ J} \)
- Kinetic energy = \( \mathrm{0.049\ J} \)
Total Energy in Simple Harmonic Motion (SHM)
For a system undergoing simple harmonic motion with:
- mass = \( \mathrm{m} \)
- angular frequency = \( \mathrm{\omega} \)
- amplitude = \( \mathrm{x_0} \)
The total mechanical energy is given by:
\( \mathrm{E = \tfrac{1}{2} m\omega^2 x_0^2} \)
Meaning:
- This energy remains constant for undamped SHM.
- Energy is shared between kinetic and potential forms, but total energy stays the same.
- Total energy depends on amplitude , larger amplitude → more energy.
Key Points:
- Total energy ∝ \( \mathrm{x_0^2} \)
- Total energy ∝ mass
- Total energy ∝ \( \mathrm{\omega^2} \) (stiffer systems store more energy)
Example
A 0.40 kg mass oscillates with amplitude 0.10 m and angular frequency \( \mathrm{6\ rad\,s^{-1}} \). Calculate the total energy of the system.
▶️ Answer / Explanation
Use:
\( \mathrm{E = \tfrac{1}{2} m\omega^2 x_0^2} \)
Substitute:
\( \mathrm{E = \tfrac{1}{2}(0.40)(6^2)(0.10^2)} \)
\( \mathrm{E = 0.5 \times 0.40 \times 36 \times 0.01} \)
\( \mathrm{E = 0.072\ J} \)
Total energy = 0.072 J.
Example
If the amplitude of a SHM system is doubled while mass and angular frequency remain constant, how does the total energy change?
▶️ Answer / Explanation
Total energy:
\( \mathrm{E \propto x_0^2} \)
Doubling amplitude:
\( \mathrm{x_0 \rightarrow 2x_0} \)
Therefore:
\( \mathrm{E’ = \tfrac{1}{2} m\omega^2 (2x_0)^2 = 4E} \)
Total energy becomes four times larger.
Example
A 0.25 kg mass oscillates with total energy 0.20 J and angular frequency \( \mathrm{4\ rad\,s^{-1}} \). Find the amplitude of oscillation.
▶️ Answer / Explanation
Use the total energy formula:
\( \mathrm{E = \tfrac{1}{2} m\omega^2 x_0^2} \)
Rearrange for amplitude:
\( \mathrm{x_0^2 = \frac{2E}{m\omega^2}} \)
Substitute values:
\( \mathrm{x_0^2 = \frac{2(0.20)}{0.25 \times 16}} \)
\( \mathrm{x_0^2 = \frac{0.40}{4} = 0.10} \)
Amplitude:
\( \mathrm{x_0 = \sqrt{0.10} = 0.316\ m} \)
Amplitude ≈ 0.32 m.