Title: InkPhysics / Periodic motion
On every note of this deck, unless a note says otherwise:
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Front: DEFINITION
Define: Natural frequency.
Back: The frequency at which a system oscillates freely after a single displacement, with no driving force.

Front: DEFINITION
Define: Resonance.
Back: The large-amplitude response of a system driven at its natural frequency, when energy transfer from the driver is most efficient.

Front: DEFINITION
Define: Damping.
Back: The loss of energy from an oscillating system to resistive forces, reducing the amplitude over time.

Front: CHECK YOURSELF
A point oscillates in SHM with amplitude 0.040 m and frequency 1.5 Hz. Find its maximum speed and maximum acceleration, and state where each occurs.
Show a hint
Build ω first; both maxima follow in one line each.
Back: ω = 2πf = 2π × 1.5 = 9.4 rad s^−1.
Maximum speed: ωA = 9.4 × 0.040 = 0.38 m s^−1, at the centre, where all the energy is kinetic.
Maximum acceleration: ω²A = 9.42^2 × 0.040 = 3.6 m s^−2, at the extremes, where the displacement and the restoring pull are greatest.

Front: CHECK YOURSELF
What length of simple pendulum has a period of 2.0 s? Take g = 9.81 m s^−2. Would the answer change on the Moon?
Show a hint
Rearrange for l, and remember what sits inside the square root.
Back: Rearranging: l = g(T/2π)^2 = 9.81 × (2.0/2π)^2 = 0.99 m, the metre-long “seconds pendulum” of old clocks.
On the Moon, yes: g is about a sixth, so the same 2.0 s period needs l = 1.62 × (0.318)^2 ≈ 0.16 m. The pendulum reads its local gravity.
The bob's mass, note, was never asked for and never needed.

Front: DEFINITION
Define: Simple harmonic motion.
Back: Oscillation in which the acceleration is proportional to the displacement from equilibrium and directed towards it: a = −ω^2x.

Front: CHECK YOURSELF
A washing machine vibrates violently at one particular spin speed and is calmer both above and below it. Explain the violence, and one design change that would reduce it.
Show a hint
One special frequency, and one knob that reshapes the response curve.
Back: At that spin speed the drum's driving frequency matches the machine's natural frequency: resonance. Each cycle of the imbalance feeds energy in exactly in step, so the amplitude builds to a maximum.
Above and below, the pushes fall out of step with the motion and partly cancel, so the response is far smaller: the machine is off the peak of its resonance curve.
Adding damping, stiffer mounts or shock absorbers, lowers and broadens the peak, so even at the unlucky speed the amplitude stays modest.

Front: CHECK YOURSELF
A 900 kg car takes a bend of radius 50 m at a steady 15 m s^−1. Find the centripetal force required, state what provides it, and explain what happens if the road cannot supply that much.
Show a hint
One formula, then think about which real force is doing the job on a flat road.
Back: F = mv^2/r = 900 × 15^2 / 50 = 4100 N (2 s.f.), directed towards the bend's centre.
On a flat road the provider is friction between the tyres and the surface.
If friction cannot reach 4100 N, on ice say, the car cannot follow that circle: it continues on a straighter path, drifting wide along a tangent, which is a skid described politely.

Front: DEFINITION
Define: Angular speed.
Back: The angle swept out per unit time: ω = θ/t = 2πf, in rad s^−1.

Front: DEFINITION
Define: The radian.
Back: The angle subtended at the centre of a circle by an arc equal in length to the radius.