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)

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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.

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2. Learning Objective (AO1)

Recall and use the formula for the total mechanical energy of a system undergoing simple harmonic motion (SHM):

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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)

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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: \(v_{\max}= \omega x_{0}\).

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5. Derivation of the Total‑Energy Formula (AO2)

1. Write kinetic and potential energies: \[ K = \tfrac12 m v^{2}, \qquad U = \tfrac12 k x^{2}. \]
2. Replace \(k\) using \(\displaystyle k = m\omega^{2}\): \[ U = \tfrac12 m\omega^{2}x^{2}. \]
3. Insert the SHM expressions for \(x(t)\) and \(v(t)\) (choose \(\phi =0\) for simplicity): \[ K = \tfrac12 m\omega^{2}x_{0}^{2}\sin^{2}(\omega t),\qquad U = \tfrac12 m\omega^{2}x_{0}^{2}\cos^{2}(\omega t). \]
4. 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.

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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	\(\tfrac12 m\omega^{2}x_{0}^{2}\)
\(x=\pm\frac{x_{0}}{\sqrt2}\)	\(\pm\frac{\omega x_{0}}{\sqrt2}\)	\(\tfrac14 m\omega^{2}x_{0}^{2}\)	\(\tfrac14 m\omega^{2}x_{0}^{2}\)

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7. Graphical Illustration (AO1 – 17.2)

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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.

1. Angular frequency: \[ \omega = \sqrt{\frac{k}{m}} = \sqrt{\frac{100}{0.25}} = 20\ \text{rad s}^{-1}. \]
2. Amplitude: \(x_{0}=0.05\ \text{m}\).
3. Energy: \[ E = \frac12(0.25)(20)^{2}(0.05)^{2}=0.125\ \text{J}. \]

Answer: \(E = 0.125\ \text{J}\) (constant throughout the motion).

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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.

1. Angular frequency: \[ \omega = \frac{2\pi}{T}= \frac{2\pi}{0.50}=12.57\ \text{rad s}^{-1}. \]
2. Spring constant: \[ k = m\omega^{2}=0.40(12.57)^{2}=63.2\ \text{N m}^{-1}. \]
3. Energy: \[ E = \tfrac12 m\omega^{2}x_{0}^{2} = \tfrac12(0.40)(12.57)^{2}(0.06)^{2} \approx 0.11\ \text{J}. \]

Answer: \(k \approx 63\ \text{N m}^{-1}\), \(E \approx 0.11\ \text{J}\).

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10. Conversion Pitfall Reminder (AO2)

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11. Practice Questions (AO2)

1. 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.
2. 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.
3. 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.

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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).

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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.

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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.

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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 \(E=\tfrac12 m\omega^{2}x_{0}^{2}\).

1. 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.
2. 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.
3. 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): \[ \frac{\Delta E}{E}= \sqrt{\left(\frac{\Delta m}{m}\right)^{2} +\left(2\frac{\Delta\omega}{\omega}\right)^{2} +\left(2\frac{\Delta x_{0}}{x_{0}}\right)^{2}}. \]
4. 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).

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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.