A torsion pendulum with natural angular frequency and quality factor obeys the linear…

PHYSICS · Multi-part · Fermi University Physics · Mathematical Methods for Physics · Electronics, Instrumentation & Communication · Signals, systems, control and digital signal processing

Mathematical Methods for Physics, Ch Fourier series and periodic decomposition

Problem

A torsion pendulum with natural angular frequency ω0\omega_0 and quality factor Q=10Q = 10 obeys the linear equation x¨+γx˙+ω02x=f(t)\ddot{x} + \gamma\dot{x} + \omega_0^2 x = f(t), with γ=ω0/Q\gamma = \omega_0/Q. The pendulum is idealised as behaving exactly like the driven damped oscillator, and the drive f(t)f(t) is a square wave of amplitude AA and angular frequency ω=ω0/2\omega = \omega_0/2, f(t)=∑n odd4Anπsin⁡(nωt).f(t) = \sum_{n\,\text{odd}} \frac{4A}{n\pi}\sin(n\omega t). ((a)) Someone tries to find the drive's rate of change f˙(t)\dot{f}(t) by differentiating this series term by term. Decide whether the resulting series converges, and state in one sentence what is actually true about the slope of a square wave at its jumps. ((b)) Find the amplitude of the pendulum's steady-state response at the fundamental (n=1n=1) and at the third harmonic (n=3n=3), each in units of A/ω02A/\omega_0^2, and state which of the two dominates the output. ((c)) Twice the drive frequency is exactly ω0\omega_0. Explain why the output nonetheless shows no resonant peak at ω0\omega_0. Then replace the drive by a sawtooth of the same period, f(t)=∑n≥1(−1)n+12Anπsin⁡(nωt)f(t) = \sum_{n\ge1}(-1)^{n+1}\frac{2A}{n\pi}\sin(n\omega t), and find the amplitude of the steady response at n=2n=2 in units of A/ω02A/\omega_0^2.

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