Damped and forced oscillations, resonance
Section: Oscillations | Syllabus: Cambridge AS Level Physics 9702
Damping Damping is the process by which the amplitude of an oscillation decreases over time due to resistive forces (like friction or air resistance) dissipating energy from the system. Types of Damping Light Damping: The system oscillates many times with gradually decreasing amplitude.
Critical Damping: The system returns to equilibrium in the shortest possible time without oscillating. (Used in car shock absorbers). Heavy (Over) Damping: The system returns to equilibrium very slowly without oscillating.
(Used in door closers). Figure 17.4: Damping Displacement-Time Graphs A graph showing three curves for displacement vs time: 1. Under-damped: Decaying sine wave. 2. Critically damped: Fast return to zero with no overshoot.
3. Heavily damped: Very slow return to zero. Forced Oscillations and Resonance A system allowed to oscillate on its own moves at its natural frequency (f_0) . If an external periodic force is applied, the system undergoes forced oscillations at the driving frequency (f) .
Resonance occurs when the driving frequency is equal to the natural frequency (f = f_0). At resonance, the transfer of energy from the driver to the oscillator is most efficient, resulting in a maximum amplitude of oscillation.
Figure 17.5: Resonance Graphs with Damping A graph of Amplitude vs. Driving Frequency. - Sharp Peak: Low damping (high amplitude at f_0). - Broad Peak: High damping (lower amplitude, peak shifts slightly left).
Worked Examples Worked Example: Unbalanced Car Wheel Question: A car of mass 850 kg is supported by four identical springs (k = 37 kN m^-1 each). Calculate the natural frequency and the speed at which a 60 cm diameter wheel will resonate.
Solution 1. Mass per wheel m = 850 / 4 = 212.5 kg. 2. T = 2π√m/k = 2π√212.5 / 37000 0.476 s. 3. Natural frequency f_0 = 1 / T = 2.1 Hz. 4. Circumference C = π d = π × 0.60 = 1.88 m. 5. Resonance Speed v = f_0 × C = 2.1 × 1.88 = 3.95 m s^-1.
Worked Example: Barton's Pendulum Question: A driver pendulum has f_0 = 0.87 Hz. Which length of oscillator pendulum will resonate? (Use T = 2π√L/g). Solution 1. T = 1 / 0.87 = 1.15 s. 2. L = g(T/2π)^2 = 9.81 × (1.15 / 6.28)^2 = 0.33 m (or 33 cm).
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