Jumps, Hysteresis & Phase Lags Section 4.5 Consider a damped linear oscillator subject to a sinusoidal external force (as in Sect. 3.6): Note: The text uses different notation in this section than in Sect. 3.6! I follow notation of this section! Student exercise: Verify that these linear oscillator results are the same as in Sect. 3.6! Newton’s 2nd Law equation of motion: m(d2x/dt2) = - r(dx/dt) - kx + F0cos(ωt)

The steady state solution (from Ch. 3, new notation) is: x(t) = A(ω)cos[ωt - φ(ω)] A(ω) = F0/[(k -mω2)2 + (rω)2]½ tan[φ(ω)] = (rω)/(k -mω2) This has resonance, etc., as thoroughly discussed in Ch. 3! Now, consider a new problem! A similar oscillator, but the constant k depends on position x: k = k(x) Newton’s 2nd Law equation of motion is: m(d2x/dt2) = - r(dx/dt) - k(x)x + F0cos(ωt)

m(d2x/dt2) = - r(dx/dt) - k(x)x + F0cos(ωt) For the spring-mass prototype, this means that “stiffness” k depends on how far the mass m is displaced from equilibrium: k = k(x):  F(x) is nonlinear! A particular x dependence for k that is often used is: k(x)  k0 [1 + βx2] (1) Presumably, βx2 << 1, so the nonlinear term is small. The choice in Eq. (1) is equivalent in our general discussion to keeping terms up to x3 the nonlinear restoring force F(x) (terms up to x4 in U(x)).

However A(ω) & φ(ω) are nothing like the linear oscillator results! Newton’s 2nd Law equation of motion is then really, just the same nonlinear equation we already briefly looked at: m(d2x/dt2) = - r(dx/dt) - k0x - k0βx3 + F0cos(ωt) This is known as the Duffing Equation. It is well known & well studied in nonlinear dynamics. It can be shown that the numerical steady state solution is similar in form to the linear oscillator solution: x(t) = A(ω)cos[ωt - φ(ω)] However A(ω) & φ(ω) are nothing like the linear oscillator results!

Qualitative Picture of Numerical Results for the Duffing Equation Both the amplitude A(ω) & the phase angle φ(ω) display hysteresis & “jumps” as a function of ω!

For ω increasing: A(ω) increases to a peak until it reaches ω = ω2 where it suddenly “jumps” discontinuously, decreasing to a small value. For ω decreasing: A(ω) increases until it reaches ω = ω1, where it suddenly “jumps” discontinuously, increasing to a large value. For ω1 < ω < ω2, A(ω) displays hysteresis (its behavior depends on whether ω is increasing or decreasing). φ(ω) shows similar behavior! This is not possible for a linear system, but its not unusual for a nonlinear one!

F(x) = - kx, x  a, F(x) = - kx - c x  a The Duffing Equation is Rather than discuss this case further, the text treats a simpler case, which has many of the same qualitative features. In this model, the spring constant k has 2 values depending on whether x is less than or greater than some constant a: F(x) = - kx, x  a, F(x) = - kx - c x  a The Duffing Equation is like this case with many very closely (infinitesimally) spaced a values so that k(x) is continuous a function of x.

x(t) = A(ω)cos[ωt - φ(ω)] Newton’s 2nd Law equation of motion is now: m(d2x/dt2) = - r(dx/dt) - kx +F0cos(ωt), x  a m(d2x/dt2) = - r(dx/dt) - kx +F0cos(ωt), x  a Assume k > k It can be shown that the numerical steady state solution is again similar in form to the linear oscillator solution: x(t) = A(ω)cos[ωt - φ(ω)] However again A(ω) & φ(ω) are nothing like the linear oscillator results!

Qualitative Picture of Numerical Results for the Model System F(x) = - kx, x  a; = - kx -c, x  a , x(t) = A(ω)cos[ωt - φ(ω)] First, the figure shows the qualitative results for the steady state linear oscillator amplitude A(ω) for force constants k & k For the nonlinear oscillator & very large a (a  ) we expect the result to be the same as for the linear oscillator with force constant k & ω0 = (k/m)½ (peaked function on the left). For very small a (a  0) we expect the result to be the same as for the linear oscillator with force constant k & ω0 = (k/m)½ (peaked function on the right). For intermediate a, we expect qualitatively that the result for A(ω) may jump discontinuously from one curve to another as ω is changed.

Figure shows qualitative results for intermediate a < maximum of A(ω). For increasing ω starting at ω = 0: The amplitude follows the linear oscillator A(ω) result for constant k. A(ω) moves up the curve to point A. When A(ω) = a, the system is now governed by constant k. So, the amplitude “JUMPS” discontinously to the amplitude curve A(ω) for constant k. Between A & B in the figure, it follows the dashed path & “jumps” from A(ω) to A(ω). It follows path B, G, C, D. At D, A(ω) = a; it “jumps” down from A(ω) to A(ω). At F (ω = ω2). As ω increases, the amplitude follows A(ω).

For decreasing ω starting at large ω: The amplitude follows the curve A(ω) for force constant k until ω = ω1 (at E) where A(ω) = a. There, the amplitude “JUMPS” discontinously to the amplitude curve A(ω) for k (E to G in the figure). Between G & B & down, the amplitude follows A(ω) until A(ω) = a. There, the amplitude “JUMPS” discontinously down from A(ω) to A(ω) (at A). As ω decreases, the amplitude follows A(ω).

path is different for increasing ω than for increasing ω. SUMMARY The amplitude of this driven nonlinear oscillator displays hysteresis, because the amplitude vs. ω path is different for increasing ω than for increasing ω. Increasing ω: Path ABGCDF Decreasing ω: Path FEGBA

The results for the phase angle φ(ω) also display JUMPS & HYSTERESIS The results for the phase angle φ(ω) also display JUMPS & HYSTERESIS. Qualitative results are in the figures. Fig a: φ(ω) for linear oscillators with k & k. Fig b: Amplitude paths for increasing & decreasing ω