ẍ + kẋ + μx = ƒ cos (σ1t + ε).
(34)
The solution is
| x = | ƒ | cos (σ1t + ε − ε1), |
| R |
(35)
provided
| R = { (μ − σ12)2 + k2σ12}1/2, tan ε1 = | kσ1 | . |
| μ − σ12 |
(36)
Hence the phase of the vibration lags behind that of the force by the amount ε1, which lies between 0 and 1⁄2π or between 1⁄2π and π, according as σ12 ≶ μ. If the friction be comparatively slight the amplitude is greatest when the imposed period coincides with the free period, being then equal to ƒ/kσ1, and therefore very great compared with that due to a slowly varying force of the same average intensity. We have here, in principle, the explanation of the phenomenon of “resonance” in acoustics. The abnormal amplitude is greater, and is restricted to a narrower range of frequency, the smaller the friction. For a complete solution of (34) we must of course superpose the free vibration (30); but owing to the factor e−t/τ the influence of the initial conditions gradually disappears.
For purposes of mathematical treatment a force which produces a finite change of velocity in a time too short to be appreciated is regarded as infinitely great, and the time of action as infinitely short. The whole effect is summed up in the value of the instantaneous impulse, which is the time-integral of the force. Thus if an instantaneous impulse ξ changes the velocity of a mass m from u to u′ we have