Energy in simple harmonic motion
Section: Oscillations | Syllabus: Cambridge A Level Physics (9702)
Energy Interchange in SHM
During SHM, energy is continually swapped between kinetic energy (E_k) and potential energy (E_p).
- At Equilibrium (x = 0): Speed is maximum, so E_k is maximum. Displacement is zero, so E_p is minimum.
- At Amplitude (x = ± x_0): Speed is zero, so E_k is zero. Displacement is maximum, so E_p is maximum.
In a pendulum, E_p is gravitational. In a mass-spring system, E_p is elastic.
Total Energy in SHM
For an ideal system with no resistive forces, the total energy (E_total) remains constant throughout the oscillation.
E_total = (1/2)mω^2 x_0^2Where m is mass, ω is angular frequency, and x_0 is amplitude.
Energy as a Function of DisplacementE_k = (1/2)mω^2(x_0^2 - x^2)
E_p = (1/2)mω^2 x^2
Figure 17.3: Energy-Displacement GraphA graph of Energy vs. Displacement (x).
- Total Energy: Horizontal line.
- Potential Energy: U-shaped parabola (y = x^2).
- Kinetic Energy: Inverted parabola (y = -x^2).
Worked Examples
Worked Example: Pendulum EnergyQuestion: A 50 g pendulum bob is displaced 12 cm and released. The period is 2.9 s. Calculate the maximum kinetic energy.
Solution1. ω = 2π / T = 2π / 2.9 = 2.167 rad s^-1.
2. E_max = (1/2)mω^2 x_0^2.
3. E_max = 0.5 × 0.050 × (2.167)^2 × (0.12)^2 = 1.69 × 10^-3 J (or 1.69 mJ).
Worked Example: Tuned Mass DamperQuestion: A 6.6 × 10^5 kg mass oscillates with a period of 6 s and amplitude 0.8 m. Calculate the energy transferred.
Solution1. ω = 2π / 6 = 1.047 rad s^-1.
2. E = 0.5 × (6.6 × 10^5) × (1.047)^2 × (0.8)^2 = 2.31 × 10^5 J (or 231 kJ).
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