A child on a swing is slowest at the top and fastest at the bottom, yet, ignoring friction, the swing never gains or loses energy overall. The energy simply moves between two forms. Sections 14.6 and 14.7 of the book show this with graphs and with formulas.
Chapter 14 of the PECTAA Physics 12 book is Simple Harmonic Motion. After SHM and circular motion and phase (section 14.5), section 14.6 gives the graphs of SHM and section 14.7 the conservation of energy in SHM. This note is for FSc Part 2 (2nd year, Class 12) students in Lahore, Faisalabad, Sargodha, Rawalpindi, Multan, Gujranwala and the other Punjab boards. Gujrat district comes under the Gujranwala board. PECTAA writes the books. Your own board sets your paper.
What to know
Graphs of SHM (section 14.6, Fig. 14.9). For x = x₀ cos ωt:
- Displacement (Eq. 14.19): x = x₀ cos ωt. Greatest (x₀) at the extreme positions.
- Velocity (Eq. 14.20): magnitude x₀ω sin ωt. Greatest (x₀ω) at the mean position, zero at the extremes.
- Acceleration (Eq. 14.21): a = −ω²x = −x₀ω² cos ωt. Greatest (x₀ω²) at the extremes, zero at the mean position.
All three are cosine or sine curves of the same period. Velocity is π/2 (a quarter cycle) out of step with displacement. Acceleration is π out of step with displacement, because the minus sign in a = −ω²x makes it point the opposite way at every instant.
Energy in SHM (section 14.7). An oscillating body has potential energy because of its displacement and kinetic energy because of its velocity. The two vary, but their sum stays constant when there is no friction.
Potential energy (Eq. 14.22). For a mass on a spring, the restoring force grows from 0 at the mean position to kx₀ at the extreme, so its average over that stretch is ½kx₀. Work done = average force × distance = ½kx₀ × x₀ = ½kx₀². This is stored as elastic potential energy. At any displacement x,
PE = ½kx².
PE is zero at x = 0 and greatest at x = ±x₀, where the total energy is all potential.
Kinetic energy. With v = ω√(x₀² − x²) and ω² = k/m,
KE = ½mv² = ½k(x₀² − x²).
KE is greatest, ½kx₀², at the mean position, where PE is zero.
Total energy. TE = KE + PE = ½kx₀² = ½mω²x₀². It is constant and proportional to the square of the amplitude. This holds for every body in SHM. The book's Fig. 14.11 shows KE and PE trading places against displacement while TE stays flat.
Key points
- Mean position: speed greatest, acceleration zero, energy all kinetic.
- Extremes: speed zero, acceleration greatest, energy all potential.
- TE = ½kx₀²: doubling the amplitude makes the energy four times larger.
- Displacement and acceleration are π apart; velocity sits π/2 from each.
Worked examples
Example 1 (book example). A 0.010 kg body executes SHM with period 6 s. One second after passing its mean position, its velocity is 0.06 m s⁻¹. Find KE, PE and TE.
- ω = 2π/T = π/3 ≈ 1.047 rad s⁻¹.
- KE = ½mv² = ½ × 0.010 × (0.06)² = 1.8 × 10⁻⁵ J.
- Time is counted from the mean position, so x = x₀ sin ωt and v = x₀ω cos ωt. At t = 1 s, cos(π/3) = 0.5, so 0.06 = x₀ × 1.047 × 0.5 and x₀ ≈ 0.115 m.
- TE = ½mω²x₀² = ½ × 0.010 × 1.097 × (0.115)² ≈ 7.2 × 10⁻⁵ J.
- PE = TE − KE ≈ 7.2 × 10⁻⁵ − 1.8 × 10⁻⁵ = 5.4 × 10⁻⁵ J.
- The book uses v = x₀ω sin ωt here, which counts time from an extreme position, not from the mean position stated in the question. That gives x₀ ≈ 0.066 m, TE ≈ 2.4 × 10⁻⁵ J and PE ≈ 0.6 × 10⁻⁵ J. KE is the same either way.
Example 2 (energy check). A spring with k = 200 N m⁻¹ oscillates with amplitude 0.1 m. Find TE, and KE at x = 0.06 m.
- TE = ½ × 200 × 0.01 = 1 J.
- PE at 0.06 m = ½ × 200 × 0.0036 = 0.36 J.
- KE = 1 − 0.36 = 0.64 J.
Common mistakes in the exam
- Putting greatest speed at the extremes. Speed is zero there.
- Saying acceleration and displacement are π/2 apart. They are π apart.
- Using full force × distance for spring work. Use the average force, ½kx₀.
- Thinking TE varies with x. Only the split between KE and PE varies.
- Ignoring where time starts. From the mean position, x = x₀ sin ωt; from an extreme, x = x₀ cos ωt.
Exam technique
- Sketch all three graphs on one time axis to show the phase gaps.
- In energy derivations, show the average-force step clearly.
- Check numerical answers with KE + PE = TE.
- Check your board's notice for the sections in your paper.
Practice questions
- Where is the potential energy of an oscillator greatest? Answer: At the extreme positions.
- A spring has k = 50 N m⁻¹ and amplitude 0.2 m. Find TE. Answer: ½ × 50 × 0.04 = 1 J.
- If the amplitude is tripled, what happens to TE? Answer: It becomes 9 times larger.
- With TE = 1 J, what is KE where PE = 0.25 J? Answer: 0.75 J.
- What is the acceleration at the mean position? Answer: Zero.
Ready to practise this? Continue in 2nd Year Physics.