ENGS 22 Oscillations PDF

ENGS 22 Oscillations 1. Driven harmonic oscillation: ẍ + ω 2 x = f m with initial conditions x(0) = x0 and ẋ(0) = v0 . Solution: Use the Laplace transform to solve for the free and forced responses. (a) Free response: Set f = 0. Apply the Laplace transform to the homogeneous ODE ẍ + ω 2 x =...

ENGS 22 Oscillations 1. Driven harmonic oscillation: ẍ + ω 2 x = f m with initial conditions x(0) = x0 and ẋ(0) = v0 . Solution: Use the Laplace transform to solve for the free and forced responses. (a) Free response: Set f = 0. Apply the Laplace transform to the homogeneous ODE ẍ + ω 2 x = 0 with prescribed initial conditions: x(0) = x0 and ẋ(0) = v0 . In the transform domain, the system can be written as s2 X(s) − sx(0) − ẋ(0) + ω 2 X(s) = 0, which yields X(s) = s 1 sx0 + v0 = x0 2 + v0 2 . s2 + ω 2 s + ω2 s + ω2 Use lines 8 and 9 in Table 2.2.1 to invert X(s). The free response solution back in the time domain reads v0 x(t) = x0 cos(ωt) + sin(ωt), ω which is precisely equation (4.2.3) on page 184. (b) Forced response: Apply zero initial conditions x(0) = 0 and ẋ(0) = 0, and set f (t) = us (t), the unit step function. The Laplace transform of the inhomogeneous ODE ẍ + ω 2 x = 1 us (t) m becomes s2 X(s) + ω 2 X(s) = 1 , ms from which X(s) can be solved: X(s) = 1 1 ω2 = . 2 2 2 +ω ) mω s(s + ω 2 ) ms(s2 Use line 25 in Table 2.2.1 to invert X(s), then the forced response solution back in the time domain becomes  1  1  1 − cos(ωt) = 1 − cos(ωt) , x(t) = mω 2 k which is precisely equation (4.2.7) on page 184. Note that the last step of simplification makes k use of the definition of natural frequency ω 2 = m . 2. Terminal velocity: mv̇ + cv = mg sin θ with initial condition v(0) = v0 . Solution: Apply the Laplace transform to the inhomogeneous ODE with the prescribed initial condition: mg sin θ . m(sV (s) − v(0)) + cV (s) = s Then V (s) mv0 mg sin θ + = ms + c s(ms + c) 1 1  = v0 c + g sin θ s+ m s s+ = c m . 1 ENGS 22 Oscillations Use lines 6 and 12 in Table 2.2.1 to invert V (s), then the solution in the time domain reads v(t)  c c mg sin θ  1 − e− m t = v0 e − m t + c  mg sin θ  − c t mg sin θ e m + = v0 − , c c which is precisely the solution in Example 4.4.1. c Note that the first term is the transient solution as it involves e− m t , in the limit lim , this term t→+∞ should drop out. The second term is interpreted as the steady state solution beacuse as the transient term dies off, the velocity v(t) approaches the terminal velocity mg csin θ , which is a constant. In short, mg sin θ lim v(t) = . t→+∞ c 2