Fill in the missing algebraic steps to prove the expression for the steady state response of an oscillator in Eq. 2.33.
2.5 Impulse Response and Applications
As already pointed out, the steady state motion in Eq. 2.33 was produced by a harmonic driving force F (t) = |F | cos(ωt) which is an idealization since it has no beginning and no end. We now turn to the response of the oscillator to a more general and realistic driving force.
We start by considering the motion of a damped mass-spring oscillator after it is set in motion by an impulse I at time t = t. We let the impulse have unit strength.
Since the impulse is instantaneous, the displacement immediately after the impulse will be ξ = 0 and the velocity, ˙ξ = 1/M. In the subsequent motion, the oscillator is free from external forces but influenced by a spring force and a resistive force−Ru proportional to the velocity u. Then, for an underdamped oscillator, the displacement will be of the form given in Eq. 2.23, i.e.,
ξ(t )= A e−γ t cos(ω0φ), (2.51) where γ = R/2M, ω0 =
ω20− γ2, and ω0=√
K/M. As before, K is the spring constant. The amplitude A and the phase angle φ are determined by the displacement ξ = 0 and the velocity ˙ξ = 1/M at t = t.
In order to make the displacement zero at t = twe must have φ = π/2 which means that the displacement must be of the form
ξ = A exp[−γ (t − t)] sin[(ω0(t− t)].
The corresponding velocity is
˙ξ = A exp[−γ (t − t)][ω0cos[ω0(t− t)] − γ sin[ω0(t− t)].
To make this velocity equal to 1/M at t = t requires that A = 1/(Mω0). In other words, the impulse response function, sometimes called the Green’s function for displacement is
Impulse response function
h(t, t)= (1/ω0M) e−γ (t−t)sin[ω0(t− t)] for t > t h(t, t)= 0 for t < t
(2.52)
[γ = R/2M · ω0=
ω20− γ2· ω20= K/M].
The dependence on t and tis expressed through the combination (t−t)only (i.e., the time difference between the ‘cause’ and the ‘effect’). Since we accept the causality principle that the effect cannot occur before the cause, we have added h(t, t)= 0 for t < tin the definition of the impulse response function h(t, t).
2.5.1 General Forced Motion of an Oscillator
The reason for the particular importance of the impulse response function is that the response to an arbitrary driving force can be easily constructed from it.
To prove this, we consider first the displacement resulting from two unit impulses delivered at t and t. Although not necessary, we assume for simplicity that the displacement and the velocity of the oscillator are zero when the first impulse is delivered at t = t. Then, by definition, the displacement that results at time t is the impulse response function h(t, t).
At the later time twhen the second impulse is delivered, the displacement and the velocity are both different from zero. It should be realized, however, that the change in the displacement produced by the second impulse does not depend on the state of motion when the impulse is delivered (because the system is linear) and the total displacement at time t will be ξ(t)= h(t, t)+ h(t, t). If the impulses at tand t
have the values Iand I, the displacement at t will be Ih(t, t)+ Ih(t, t). We can now proceed to the displacement produced by a general driving force F (t ). The effect of this force is the same as that of a succession of impulses of magnitude F (t)tover the entire time of action of the force up to the time t. The displacement, at time t, produced by one of these impulses is h(t, t)F (t)t with an analogous expression for any other impulse; this follows from the discussion of the impulse response function h(t, t). The sum of the contributions from all the elementary impulses can be expressed as the integral
Response to a driving force F (t) ξ(t )=t
−∞ h(t, t)F (t)dt=∞
0 F (t− τ)h(τ)dτ (2.53) [h(t, t): See Eq. 2.52. τ = t − t].
The range of integration for t is from−∞ to t to cover all past contributions.
Frequently, it is convenient to introduce a new variable, τ = t − t, in which case the range of integration is from τ = 0 to infinity, as indicated in Eq. 2.53.
The validity of this result relies on the linearity of the system so that the incremental change of the displacement will be the same for a given impulse independent of the state of motion when the impulse is delivered.
2.5.2 Transition to Steady State
As an example of the use of Eq. 2.53, we consider a driving force which is turned on at t = 0 and defined by
F (t )= F0cos(ωt) t > 0
F (t )= 0 t < 0. (2.54)
Since the driving force is zero for t < 0, the lower bound of the range of integration in Eq. 2.53 can be set equal to 0. Using the expression for the impulse-response function in Eq. 2.52, we then get
ξ(t )= ξ0e−γ t
t
0
eγ tsin[ω0(t− t)] cos(ωt)dt, (2.55)
where ξ0= F0/(ω0M)= A(ω20/ω0)and A= F0/K. The integration is elementary and it is left as a problem to show that
ξ /A= 1
(1− 2)2+ (D)2[cos(ωt − φ) − (ω0/ω0)e−γ tcos(ω0t− β)], (2.56) where = ω/ω0, tan φ = D/(1 − 2), tan β = γ (1 + 2)/ω0(1− 2), D = R/(ω0M), and A= F0/K.
The solution is the sum of two parts. The first, the steady state solution, has the same frequency ω as the driving force and its amplitude remains constant. The second part, often referred to as the transient, decays exponentially with time and can be ignored when γ t >> 1; it has a frequency ω0, i.e., the frequency of free oscillations.
For small values of D, the maximum amplitude occurs very close to = ω/ω0= 1 but as the damping increases, the maximum shifts toward lower frequencies.
2.5.3 Secular Growth
There are several aspects of the solution in Eq. 2.56 that deserve special notice. One concerns the response of an undamped oscillator when the frequency of the driving force equals the resonance frequency, i.e., ω= ω0. Our previous analysis dealt only with the steady state response from the very start and, in the absence of damping, yielded nothing but an infinite amplitude at resonance. The present approach shows how the amplitude grows with time toward the infinite value at t = ∞. With γ = 0, i.e., D= 0 and ω0= ω0, the integral in Eq. 2.55 can be evaluated in a straight-forward manner and we obtain, with F0/(ω0M)= F0ω0/K= Aω0,
ξ(t )= A
t
0 sin[ω0(t− t)] cos(ω0t)dt= A(ω0t /2) sin(ω0t ). (2.57) The amplitude of this motion grows linearly with time, secular growth, toward the steady state value of infinity.6
2.5.4 Beats Between Steady State and Transient Motions
If the frequency ω of the driving force in Eq. 2.54 is not equal to the resonance frequency of the oscillator and if the damping is sufficiently small, the transition to steady state exhibits ‘beats’, i.e., variations in the amplitude. The beats result from the interference between the steady state motion with the frequency ω of the driv-ing force and the transient motion with the frequency ω0of the free motion of the oscillator (see Eq. 2.56). Both of these motions are present during the transition to steady state. The curves shown refer to ω = 1.1ω0and the values of the damping factor D= R/(ω0M)= 0.01 and 0.04 corresponding to the Q-values of 100 and 25.
The interference between the two motions periodically goes from destructive to con-structive as the phase difference (ω− ω0)t ≈ (ω − ω0)t between the two motions
6The general solution in Eq. 2.56 reduces to Eq. 2.57 for γ = 0 and ω = ω0(i.e., = 1). Actually, for these values the expression becomes of the form 0/0 and we have to determine the limit value as γ goes to zero, using exp(−γ t) ≈ 1 − γ t. Then, the steady state term is canceled out, and, since φ = β = π/2 and D= 2γ /ω0, the remaining term indeed reduces to Eq. 2.57.
increases with time. With ω = ω − ω0 = 0.1 ω0, it is increased by 2π in a time interval t given by ωt= 2π which, in the present case, yields 0.1 ω0t = 2π or
t /T0≈ 10, where T0≈ T0is the period of free motion. Thus, t is the time interval between two successive maxima or minima in the resulting displacement function, which is consistent with the result shown in the figure.
At a driving frequency below the resonance frequency, a similar result is obtained.
For example, with ω = 0.9 ω0, the curves are much like those in the figure except that they start out in the positive rather than the negative direction.
If the driving frequency is brought sufficiently close to the resonance frequency, the time interval between beats will be so large that the amplitude of the transient will be damped so much that the beats will be less pronounced. For a more detailed discussion of this question we refer to Example 13 in Ch.11.
2.5.5 Pulse Excitation of an Acoustic Resonator
A simple and instructive demonstration of beats involves an acoustic cavity resonator exposed to repeated wave trains (pulse modulated) of sound.
A microphone in the cavity of the resonator measures the sound pressure and the corresponding signal from the microphone can be displayed on one channel of a dual beam oscilloscope. On the other channel can be shown the input voltage to the loudspeaker which produces the incident sound. The amplitude of this sound is constant during the duration of each pulse train.
The time dependence of the sound pressure in the resonator is quite different from that of the incident wave. It starts to grow toward a steady state value, but before this value has been reached, the incident pulse is terminated and the sound pressure in the cavity starts to decay. This process is repeated for each pulse.
If the ‘carrier’ frequency of the incident sound is equal to the resonance frequency, the growth and (exponential) decay of the sound pressure in the resonator are mono-tonic. During the decay, the resonator re-radiates sound which can be heard as a
‘reverberation’ after the incident sound has been shut off.7
If the frequency of the incident sound is somewhat lower than the resonance frequency, beats resulting from the interference of the free and forced oscillations occur. A similar result is obtained if the carrier frequency is somewhat higher than the resonance frequency. The sound pressure in the resonator is now much smaller than that obtained at the resonance frequency.