NATS 1720 Lecture Notes - Lecture 5: Simple Harmonic Motion, Natural Frequency, Normal Mode
30 Nov 2015
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= 0. 5kg x -9. 8m/s^2 (negative because the force is downwards/negative) = 2. 45m: m x a = -k x y (displacement away from equilibrium) F = 1/2 k/m k = spring constant (elasticity) m= mass (inertia) = 1/2 (2 n/m) / (0. 5kg) = 1 = 0. 318 hz. If mass increases, frequency decreases: suppose a new spring is introduced, and the mass now oscillates at 3hz (natural frequency). First, write f^2 = 1/(2 )^2 (k/m) so k= (2 )^2 f^2 m. = (2 )^2 (3)^2 (0. 5) = 177 n/m. Resonance: if a sho (simple harmonic oscillator) is pushed periodically at some frequency (driving frequency) then its behaviour depends on driving frequency. i. ii. iii. iv. At low frequencies, oscillator vibrates at driver frequency and amplitude. As driver frequency increases , so does oscillator amplitude. Amplitude peaks when fdriver = natural frequency (when frequency of the driver matches the frequency of the oscillator)
