Energy in simple harmonic motion
Section: Oscillations | Syllabus: Cambridge AS 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 = 1/2mω^2 x_0^2 Where m is mass, ω is angular frequency, and x_0 is amplitude. Energy as a Function of Displacement E_k = 1/2mω^2(x_0^2 - x^2) E_p = 1/2mω^2 x^2 Figure 17.3: Energy-Displacement Graph A 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 Energy Question: A 50 g pendulum bob is displaced 12 cm and released.
The period is 2.9 s. Calculate the maximum kinetic energy. Solution 1. ω = 2π / T = 2π / 2.9 = 2.167 rad s^-1. 2. E_max = 1/2mω^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 Damper Question: A 6.6 × 10^5 kg mass oscillates with a period of 6 s and amplitude 0.8 m. Calculate the energy transferred. Solution 1. 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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