Oscillations of Mechanical Systems
Math 240
Free oscillation No damping Damping Forced Oscillations of Mechanical Systems oscillation No damping Damping Math 240 — Calculus III
Summer 2015, Session II
Thursday, July 30, 2015 Oscillations of Mechanical Agenda Systems
Math 240
Free oscillation No damping Damping Forced oscillation 1. Free oscillation No damping Damping No damping Damping
2. Forced oscillation No damping Damping Oscillations of Mechanical Introduction Systems
Math 240
Free oscillation No damping Damping Forced oscillation We have now learned how to solve constant-coefficient linear No damping Damping differential equations of the form P (D)y = F for a sizeable class of functions F . We are going to use this knowledge to study the motion of mechanical systems consisting of a mass attached to a spring. Let’s begin by modeling our system using a differential equation. Oscillations of Mechanical The setup Systems
Math 240
Free oscillation No damping Damping Forced oscillation A mass of m kg is attached to No damping Damping the end of a spring with spring constant k N/m whose natural length is l0 m. At equilibrium, the mass hangs without moving at a displacement of L0 m, so that mg = kL0. Spring-mass system in static equilibrium. Oscillations of Mechanical The setup Systems
Math 240
Free To analyze the system in motion, we let y(t) denote the oscillation No damping position of the mass at time t and take y = 0 to coincide with Damping the equilibrium position. The forces that act on the mass are Forced oscillation No damping Damping 1. The force of gravity, Fg = mg. 2. The spring force, Fs = −k(y(t) + L0). 3. A damping force proportional to the velocity of the mass, dy Fd = −c dt . 4. Any external driving A damped spring-mass system. force, F (t). Oscillations of Mechanical Our equation Systems
Math 240
Free oscillation Newton says, the equation governing motion of the mass is No damping Damping d2y Forced m = Fg + Fs + Fd + F (t) oscillation dt2 No damping dy Damping = mg − k(L + y) − c + F (t). 0 dt Rearranging gives us the linear differential equation c k 1 y00 + y0 + y = F (t). m m m We may also have initial conditions 0 y(0) = y0 and y (0) = v0. These indicate that at t = 0 the mass is displaced a distance of y0 m and released with a downward velocity of v0 m/s. Oscillations of Mechanical Free oscillation Systems
Math 240
Free oscillation No damping Damping Forced First consider the case where there are no external forces acting oscillation No damping on the system, that is, set F (t) = 0. Our differential equation Damping reduces to c k y00 + y0 + y = 0. m m We will study the two subcases
I no damping: c = 0 and
I damping: c > 0. Oscillations of Mechanical No damping Systems Math 240 Setting c = 0 in our equation, we have Free 00 2 oscillation y + ω0y = 0, No damping p Damping where ω0 = k/m. This equation has the general solution Forced oscillation y(t) = c1 cos ω0t + c2 sin ω0t. No damping Damping From the constants c1 and c2 we can derive
p 2 2 A I the amplitude, A0 = c + c , 0 1 2 c 2 I the phase, φ = arctan(c2/c1). c 1 Using these new constants, the equation of our motion is
y(t) = A0 cos(ω0t − φ).
This is simple harmonic motion. The constant ω0 is called the circular frequency. Oscillations of Mechanical Simple harmonic motion Systems
Math 240
Free oscillation No damping Damping Forced oscillation No damping Damping
2π p m This function is periodic with a period of T = ω = 2π k . q 0 1 ω0 1 k Its frequency is f = T = 2π = 2π m . Note that these quantities are independent of the initial conditions. They are properties of the system itself. Oscillations of Mechanical Damped motion Systems
Math 240
Free The motion of the system is damped when c > 0. Our oscillation equation is then No damping Damping c k y00 + y0 + . Forced oscillation m m No damping The auxiliary polynomial has roots Damping √ −c ± c2 − 4km r = . 2m The behavior of the system will depend on whether there are distinct real roots, a repeated real root, or complex conjugate roots. This can be determined using the (dimensionless) quantity c2/(4km). We say that the system is 2 I underdamped if c /(4km) < 1 (complex conjugate roots), 2 I critically damped if c /(4km) = 1 (repeated real root), 2 I overdamped if c /(4km) > 1 (distinct real roots). Oscillations of Mechanical Underdamping Systems
Math 240 The two complex roots of the auxiliary polynomial give rise to Free oscillation the general solution No damping −ct/(2m) Damping y(t) = e (c1 cos µt + c2 sin µt), √ Forced 2 oscillation where µ = 4km − c /(2m). Using amplitude and phase, it’s No damping −ct/(2m) Damping y(t) = A0e cos(µt − φ).
Although the amplitude decays exponentially, this motion has a constant quasiperiod T = 2π = √ 4πm . µ 4km−c2 Oscillations of Mechanical Critical damping Systems
Math 240
Free 2 oscillation Critical damping happens when c /(4km) = 1. Then the No damping Damping equation 2 Forced 00 c 0 c oscillation y + y + 2 y = 0 No damping m 4m Damping has general solution −ct/(2m) y(t) = e (c1 + c2t).
The motion is not oscillatory—it will pass through y = 0 at most once. Oscillations of Mechanical Overdamping Systems
Math 240 When c2/(4km) > 1 we have two real roots of the auxiliary Free oscillation polynomial: No damping √ √ Damping −c + c2 − 4km −c − c2 − 4km Forced r1 = and r2 = . oscillation 2m 2m No damping Thus, our general solution is Damping −ct/(2m) µt −µt y(t) = e (c1e + c2e ), √ where µ = c2 − 4km/(2m).
Overdamped motion is qualitatively similar to critically damped—it is not oscillatory and passes through the equilibrium position at most once. Oscillations of Mechanical Sanity check Systems
Math 240
Free oscillation No damping Damping Forced oscillation No damping Damping Notice that in all three cases of damped motion, the amplitude diminishes to zero as t → ∞. This is certainly what we expect in such a system. Oscillations of Mechanical Forced oscillation Systems
Math 240
Free oscillation No damping Damping Forced Let’s investigate the nonhomogeneous situation when an oscillation No damping external force acts on the spring-mass system. We will focus on Damping periodic applied force, of the form
F (t) = F0 cos ωt,
for constants F0 and ω. Our general equation is now c k F y00 + y0 + y = 0 cos ωt. m m m Oscillations of Mechanical No damping Systems
Math 240
Free oscillation No damping Damping Forced oscillation Setting c = 0, we want to solve No damping F Damping y00 + ω2y = 0 cos ωt, 0 m p where again ω0 = k/m is the circular frequency. We have seen that the complementary function is
yc(t) = A0 cos(ω0t − φ).
Our trial solution will depend on whether ω = ω0. Oscillations of Mechanical Forced harmonic oscillation Systems
Math 240
Free oscillation No damping Damping When ω 6= ω0, we can use the trial solution Forced oscillation yp(t) = A cos ωt + B sin ωt. No damping Damping Then we can find the particular solution F0 yp(t) = 2 2 cos ωt, m(ω0 − ω ) so the general solution is F0 y(t) = A0 cos(ω0t − φ) + 2 2 cos ωt. m(ω0 − ω ) From this we see that the motion will look like a superposition of two simple harmonic oscillations. Oscillations of Mechanical Forced harmonic oscillation Systems
Math 240
Free oscillation If ω/ω0 is a rational number, say, p/q, then the resulting No damping Damping motion will have a period of Forced 2πq 2πp oscillation T = = . No damping ω0 ω Damping Otherwise, it will be oscillatory but not periodic. Oscillations of Mechanical Forced harmonic oscillation Systems
Math 240
Free oscillation No damping Damping Interesting things happen when ω is very close to ω : Forced 0 oscillation No damping Damping Oscillations of Mechanical Resonance Systems
Math 240 If instead we have ω = ω0, then we must use the trial solution Free oscillation yp(t) = t(A cos ω0t + B sin ω0t). No damping Damping This leads to the particular solution Forced F oscillation y (t) = 0 t sin ω t No damping p 0 Damping 2mω0 and general solution F0 y(t) = A0 cos(ω0t − φ) + t sin ω0t. 2mω0 Notice that the amplitude increases without bound as t → ∞. Oscillations of Mechanical Damping Systems
Math 240
Free Finally, we will consider a damped nonhomogeneous equation oscillation No damping 00 c 0 k F0 Damping y + y + y = cos ωt, Forced m m m oscillation No damping with c > 0. The trial solution yp(t) = A cos ωt + B sin ωt Damping yields the particular solution F y (t) = 0 (k − mω2) cos ωt + cω sin ωt . p (k − mω2)2 + c2ω2 Letting q 2 2 2 2 2 2 cω H = m (ω0 − ω ) + c ω and η = arctan 2 2 m(ω0 − ω ) turns it into F y (t) = 0 cos(ωt − η). p H Oscillations of Mechanical Damping Systems Math 240 As before, the system can be underdamped, critically damped, Free or overdamped. Which one will determine the complementary oscillation No damping function. Damping Forced In each case of damped harmonic motion, the amplitude dies oscillation No damping out as t gets large. But the driving force has a constant Damping amplitude and thus it will dominate. We therefore refer to the complementary function as the transient part of the solution and call yp the steady-state solution.