Solves M·x'' + b·x' + c·x = F₀·sin Ωt for a mass on a spring, a mathematical and a physical pendulum and a series RLC circuit. It finds the frequency, period, damping, the amplitude and phase of forced oscillations, resonance, the law x(t) and the amplitude-frequency response.
The equation of oscillation
A linear one-degree-of-freedom system is described by M·x'' + b·x' + c·x = F₀·sin Ωt: M is the inertia (mass, moment of inertia, inductance), b the resistance, c the stiffness (1/C for a circuit), F₀ and Ω the amplitude and frequency of the driving term. Divided by M it becomes x'' + 2n·x' + ω₀²·x = h·sin Ωt with the natural frequency ω₀ = √(c/M), the damping coefficient n = b/(2M) and h = F₀/M.
Free oscillations
Without damping and force x = x₀·cos ω₀t + (v₀/ω₀)·sin ω₀t, the amplitude is A = √(x₀² + (v₀/ω₀)²), the period T = 2π/ω₀. For a spring ω₀ = √(c/m), for a mathematical pendulum ω₀ = √(g/l), for a circuit ω₀ = 1/√(LC).
Damping
For n < ω₀ the motion is damped: x = e^(−nt)(C₁·cos ω_d t + C₂·sin ω_d t), ω_d = √(ω₀² − n²). The logarithmic decrement δ = n·T_d is the logarithm of the ratio of successive peaks. For n = ω₀ (critical damping) and n > ω₀ the motion is aperiodic.
Forced oscillations and resonance
When the transient dies out, oscillations at the frequency of the force remain, x = B·sin(Ωt − ε), with B = h/√((ω₀² − Ω²)² + 4n²Ω²). The maximum occurs at Ω = √(ω₀² − 2n²), close to ω₀ for light damping. Without damping at Ω = ω₀ the amplitude grows linearly, which is resonance.
Amplitude-frequency response
The dynamic factor μ = B·ω₀²/h depends only on Ω/ω₀ and ζ = n/ω₀. For Ω → 0 it equals 1, at Ω = ω₀ it equals 1/(2ζ), and for Ω ≫ ω₀ it tends to zero.
Example
A 2 kg mass on a 50 N/m spring, b = 4 N·s/m, force 10·sin 4t N. This gives ω₀ = 5 rad/s, T₀ = 1.257 s, n = 1 1/s, ζ = 0.2, ω_d = 4.899 rad/s, the forced amplitude B = 0.415 m, the phase shift 41.6°, μ = 2.08. The amplitude peaks at 0.510 m at Ω = 4.80 rad/s.
FAQ
How to find the frequency and period of free oscillations?
The natural frequency is ω₀ = √(c/M) in radians per second, the period T = 2π/ω₀, the frequency f = ω₀/(2π) in hertz. For a mass on a spring ω₀ = √(c/m), for small oscillations of a mathematical pendulum ω₀ = √(g/l), for a circuit ω₀ = 1/√(L·C). The calculator gives all three.
How do damped and aperiodic motions differ?
If n < ω₀ the system oscillates with decreasing amplitude and the conditional period T_d = 2π/√(ω₀² − n²). For n ≥ ω₀ there are no oscillations: the body returns smoothly to equilibrium, and the fastest return without overshoot is the critical damping n = ω₀.
What is the logarithmic decrement?
The natural logarithm of the ratio of two successive peaks: δ = n·T_d. It is used to measure damping experimentally: n = δ/T_d, b = 2M·n. The quality factor Q = ω₀/(2n) shows how many times the amplitude at resonance exceeds the static deflection.
What happens at resonance?
When the frequency of the driving force approaches the natural frequency, the amplitude rises sharply. With damping it is limited to about h/(2n·ω₀) and peaks at Ω = √(ω₀² − 2n²). Without damping the amplitude grows linearly in time, in theory without limit.
How to apply the result to a pendulum and a circuit?
Choose the model: its parameters are reduced to the equivalent M, b, c, F₀. For a mathematical pendulum M = m·l², c = m·g·l; for a circuit M = L, b = R, c = 1/C, and the charge q plays the role of the coordinate. Angles are entered in degrees; the formulas are linear, so they hold for small angles.