recall and use E = 21mω2x02 for the total energy of a system undergoing simple harmonic motion
Simple Harmonic Oscillations – Total Mechanical Energy (Cambridge 9702 – Codes 17.1‑17.3)
1. Syllabus Coverage Overview
| Syllabus Code | What the Syllabus Expects | What These Notes Provide | Comments / Gaps |
|---|---|---|---|
| 17.1 | Definition of SHM, equations of motion, total mechanical energy \(E=\tfrac12 m\omega^{2}x_{0}^{2}\) and its constancy, link to \(k\) and \(\omega\). | Full derivation, SHM equations, energy‑vs‑time graph, table of characteristic points, worked examples, practice questions. | ✓ Covered. |
| 17.2 | Derive and use the energy formula; discuss kinetic ↔ potential energy exchange; explain why total energy is constant. | Derivation, energy‑exchange discussion, “Common mistakes”, conversion reminders. | ✓ Covered. |
| 17.3 | Explain damping (light, critical, heavy), forced oscillations, resonance; sketch displacement‑time graphs for each case. | Brief subsection added (see § 7). | Partial – only introductory description; no quantitative treatment. |
| AO3 – Practical Skills | Plan, conduct and analyse a SHM investigation; evaluate uncertainties; suggest improvements. | Mini‑lab outline with added guidance on uncertainty analysis and systematic‑error evaluation. | ✓ Enhanced. |
| Cross‑topic Links | Relate SHM to pendulums, LC circuits, wave‑particle duality, etc. | Section 8 added with explicit links. | ✓ Added. |
2. Learning Objective (AO1)
Recall and use the formula for the total mechanical energy of a system undergoing simple harmonic motion (SHM):
\$E = \frac{1}{2}\,m\,\omega^{2}\,x_{0}^{2}\$
3. Key Terms (AO1)
| Symbol | Quantity | Definition / Relationship |
|---|---|---|
| \(m\) | Mass | Mass of the oscillating object (kg) |
| \(k\) | Spring constant | Force required per unit extension (N m\(^{-1}\)) |
| \(\omega\) | Angular frequency | \(\displaystyle\omega = \sqrt{\frac{k}{m}} = 2\pi f = \frac{2\pi}{T}\) (rad s\(^{-1}\)) |
| \(f\) | Frequency | Number of cycles per second (Hz); \(f = \omega/2\pi\) |
| \(T\) | Period | Time for one complete cycle; \(T = 2\pi/\omega\) |
| \(x_{0}\) | Amplitude | Maximum displacement from equilibrium (m) |
| \(K\) | Kinetic energy | \(K = \tfrac12 m v^{2}\) (J) |
| \(U\) | Elastic potential energy | \(U = \tfrac12 k x^{2} = \tfrac12 m\omega^{2}x^{2}\) (J) |
4. Simple Harmonic Motion Equations (AO1)
- Restoring force: \(\displaystyle F = -kx\) (directly proportional to displacement).
- Displacement: \(\displaystyle x(t)=x_{0}\cos(\omega t+\phi)\).
- Velocity: \(\displaystyle v(t)= -\omega x_{0}\sin(\omega t+\phi).\)
- Acceleration: \(\displaystyle a(t)= -\omega^{2}x_{0}\cos(\omega t+\phi).\)
- Maximum speed: \(v{\max}= \omega x{0}\).
5. Derivation of the Total‑Energy Formula (AO2)
- Write kinetic and potential energies:
\[
K = \tfrac12 m v^{2}, \qquad
U = \tfrac12 k x^{2}.
\]
- Replace \(k\) using \(\displaystyle k = m\omega^{2}\):
\[
U = \tfrac12 m\omega^{2}x^{2}.
\]
- Insert the SHM expressions for \(x(t)\) and \(v(t)\) (choose \(\phi =0\) for simplicity):
\[
K = \tfrac12 m\omega^{2}x_{0}^{2}\sin^{2}(\omega t),\qquad
U = \tfrac12 m\omega^{2}x_{0}^{2}\cos^{2}(\omega t).
\]
- Sum the two energies:
\[
E = K+U = \tfrac12 m\omega^{2}x_{0}^{2}\bigl[\sin^{2}(\omega t)+\cos^{2}(\omega t)\bigr]
= \boxed{\tfrac12 m\omega^{2}x_{0}^{2}}.
\]
Because \(\sin^{2}\theta+\cos^{2}\theta = 1\), the total energy is independent of time – it is a constant of the motion.
6. Energy at Characteristic Points (AO1)
| Position \(x\) | Velocity \(v\) | Kinetic Energy \(K\) | Potential Energy \(U\) | Total Energy \(E\) |
|---|---|---|---|---|
| \(x=0\) (equilibrium) | \(v=\pm\omega x_{0}\) | \(\tfrac12 m\omega^{2}x_{0}^{2}\) | 0 | \(\tfrac12 m\omega^{2}x_{0}^{2}\) |
| \(x=\pm x_{0}\) (amplitude) | 0 | 0 | \(\tfrac12 m\omega^{2}x_{0}^{2}\) | |
| \(x=\pm\frac{x_{0}}{\sqrt2}\) | \(\pm\frac{\omega x_{0}}{\sqrt2}\) | \(\tfrac14 m\omega^{2}x_{0}^{2}\) | \(\tfrac14 m\omega^{2}x_{0}^{2}\) |
7. Graphical Illustration (AO1 – 17.2)
8. Worked Example 1 – Using Angular Frequency (AO2)
Problem: A 0.25 kg mass is attached to a spring with \(k = 100\ \text{N m}^{-1}\). It is pulled 5 cm from equilibrium and released. Find the total mechanical energy.
- Angular frequency:
\[
\omega = \sqrt{\frac{k}{m}} = \sqrt{\frac{100}{0.25}} = 20\ \text{rad s}^{-1}.
\]
- Amplitude: \(x_{0}=0.05\ \text{m}\).
- Energy:
\[
E = \frac12(0.25)(20)^{2}(0.05)^{2}=0.125\ \text{J}.
\]
Answer: \(E = 0.125\ \text{J}\) (constant throughout the motion).
9. Worked Example 2 – Starting from the Period (AO2)
Problem: A 0.40 kg block oscillates with period \(T = 0.50\ \text{s}\) and amplitude \(0.06\ \text{m}\). Determine the spring constant \(k\) and the total mechanical energy.
- Angular frequency:
\[
\omega = \frac{2\pi}{T}= \frac{2\pi}{0.50}=12.57\ \text{rad s}^{-1}.
\]
- Spring constant:
\[
k = m\omega^{2}=0.40(12.57)^{2}=63.2\ \text{N m}^{-1}.
\]
- Energy:
\[
E = \tfrac12 m\omega^{2}x_{0}^{2}
= \tfrac12(0.40)(12.57)^{2}(0.06)^{2}
\approx 0.11\ \text{J}.
\]
Answer: \(k \approx 63\ \text{N m}^{-1}\), \(E \approx 0.11\ \text{J}\).
10. Conversion Pitfall Reminder (AO2)
- Always convert linear frequency \(f\) to angular frequency \(\omega\) before using the energy formula: \(\omega = 2\pi f\).
- Do not insert the period \(T\) directly; first compute \(\omega = 2\pi/T\).
- Check units: \(m\) in kg, \(x_{0}\) in metres, \(\omega\) in rad s\(^{-1}\) → \(E\) in joules.
11. Practice Questions (AO2)
- A 0.5 kg block oscillates on a frictionless horizontal spring with \(\omega = 10\ \text{rad s}^{-1}\) and amplitude \(x_{0}=0.08\ \text{m}\). Calculate the total mechanical energy.
- If a simple pendulum (treated as a mass‑spring analogue) has total energy \(2.0\ \text{J}\) and \(\omega = 4\ \text{rad s}^{-1}\), find its amplitude.
- During SHM, the kinetic energy is observed to be \(3.0\ \text{J}\) when the displacement equals half the amplitude. Determine the total energy of the system.
12. Common Mistakes to Avoid (AO1)
- Confusing the amplitude \(x_{0}\) with the instantaneous displacement \(x\). The energy formula uses the maximum displacement only.
- Omitting the factor \(\tfrac12\) in the energy expression.
- Using linear frequency \(f\) instead of angular frequency \(\omega\) without conversion.
- Applying the formula to motions that are not simple harmonic (large‑angle pendulum, non‑linear springs, heavily damped systems).
13. Damped & Forced Oscillations – Brief Overview (17.3)
Although the energy formula above applies only to ideal (undamped) SHM, the syllabus also expects students to recognise the following concepts:
- Light (underdamped) damping: Amplitude decays exponentially, displacement still sinusoidal; energy decreases as \(E(t)=E_{0}e^{-2\beta t}\) where \(\beta\) is the damping coefficient.
- Critical damping: System returns to equilibrium without oscillating; fastest non‑oscillatory return.
- Heavy (over‑damped) damping: Return to equilibrium is slower than critical; no oscillation.
- Forced (driven) oscillations: An external periodic force \(F{\text{ext}}=F{0}\cos(\omega{\text{d}}t)\) produces a steady‑state amplitude that depends on the driving frequency \(\omega{\text{d}}\). Resonance occurs when \(\omega_{\text{d}}\) ≈ natural \(\omega\), giving the largest amplitude (limited by damping).
- Typical displacement‑time sketches:
- Undamped SHM – sinusoidal with constant amplitude.
- Light damping – sinusoid with decreasing envelope.
- Critical damping – exponential curve that just touches the equilibrium line.
- Forced resonance – large steady‑state amplitude at \(\omega_{\text{d}}\approx\omega\).
These ideas are introduced qualitatively; quantitative treatment (e.g., logarithmic decrement) is beyond the scope of the current notes but can be added in a later module.
14. Cross‑Topic Links (Cambridge 9702)
- Simple pendulum: For small angles, the restoring torque leads to \(\omega = \sqrt{g/L}\); the energy expression becomes \(E=\tfrac12 m g L \theta_{0}^{2}\).
- LC circuit: Electrical analogue of SHM with \(\omega = 1/\sqrt{LC}\); total energy stored is \(\tfrac12 L I{0}^{2} = \tfrac12 C V{0}^{2}\).
- Motion in a circle: Uniform circular motion of radius \(x{0}\) and angular speed \(\omega\) projects onto a SHM displacement \(x = x{0}\cos(\omega t)\). The kinetic energy of the circular motion equals the total SHM energy.
- Quantum harmonic oscillator: The classical expression \(E=\tfrac12 m\omega^{2}x{0}^{2}\) mirrors the quantum result \(E{n} = \left(n+\tfrac12\right)\hbar\omega\), reinforcing the link between classical and quantum physics.
15. Mini‑Lab Outline – Investigating SHM (AO3)
Objective: Determine the angular frequency \(\omega\) of a mass‑spring system experimentally and verify that the measured total mechanical energy matches \(E=\tfrac12 m\omega^{2}x_{0}^{2}\).
- Equipment
- Air‑track with low friction carriage.
- Spring of known or unknown constant (attach to fixed end).
- Set of calibrated masses.
- Motion sensor or photogate + data‑acquisition software.
- Ruler or digital caliper for measuring amplitude.
- Procedure
- Attach the spring to the fixed end of the air‑track and place a mass \(m\) on the carriage.
- Pull the carriage a measured distance \(x_{0}\) from equilibrium (record with ±0.5 mm uncertainty).
- Release gently (no extra push) and record \(x(t)\) for at least 10 complete cycles.
- Fit the data to a sinusoid \(x(t)=x_{0}\cos(\omega t+\phi)\) to obtain \(\omega\) and its standard error.
- Calculate the theoretical total energy using the measured \(\omega\) and \(x{0}\). Optionally, compute kinetic energy at the equilibrium position from the measured maximum speed \(v{\max}= \omega x_{0}\) and compare with the theoretical value.
- Uncertainty Analysis
- Mass \(m\): use the balance’s least count (e.g., ±0.001 kg).
- Amplitude \(x_{0}\): combine ruler uncertainty with the fitting uncertainty from the sinusoidal fit.
- Angular frequency \(\omega\): propagate the standard error from the fit; if using period measurement, combine timing uncertainty (stop‑watch or photogate) with the number of periods measured.
- Energy uncertainty (propagation):
\[
\frac{\Delta E}{E}= \sqrt{\left(\frac{\Delta m}{m}\right)^{2}
+\left(2\frac{\Delta\omega}{\omega}\right)^{2}
+\left(2\frac{\Delta x{0}}{x{0}}\right)^{2}}.
\]
- Evaluation
- Compare experimental \(E{\text{exp}}\) with theoretical \(E{\text{theo}}\) using the percentage difference.
- Discuss systematic errors (air‑track tilt, spring non‑linearity, friction) and suggest improvements (use a longer spring, increase mass to reduce relative friction, repeat with different amplitudes).
16. Summary (AO1)
- The total mechanical energy of an ideal SHM system is constant and given by \(E=\tfrac12 m\omega^{2}x_{0}^{2}\).
- Energy is continuously exchanged between kinetic and elastic potential forms; at any instant \(E=K+U\).
- Accurate use of the formula requires the angular frequency \(\omega\); always convert from \(f\) or \(T\) first.
- Understanding energy in SHM provides a foundation for later topics such as damping, resonance, and analogues in electrical and quantum systems.