Jan 06, 2016
Sect. 6.6: Damped, Driven Pendulum• Consider a plane pendulum subject to an an applied torque N & subject
to damping by the viscosity η of the medium (say, air) in which it swings. Mass m. See figure:
The angle of swing is allowed in the entire range: -π π
The usual assumption of a massless rod. Length = R
• First, a simple pendulum,
no applied torque. Newton’s
2nd Law or Lagrange’s equations
give the equation of motion:
Ngrav = Iα = I(d2/dt2), I = mR2
Ngrav = - mgR sin
mR2(d2/dt2) + mgRsin = 0 (1)NOTE: The motion is clearly NOT simple harmonic!
• For small angles sin & (1) becomes: mR2(d2/dt2) + mgR = 0
Or: (d2/dt2) + (g/R) = 0 (2)
(2) is clearly simple harmonic with frequency ω0 = (g/R)½
• Now, add an applied torque = N
• No damping for now. • We still have gravitational torque
Ngrav = - mgRsin • Newton’s 2nd Law or
Lagrange’s equations give the equation of motion:
N + Ngrav = mR2(d2/dt2) (3)
• There is a critical angle c at which the applied torque & the gravitational torque exactly balance (for which the pendulum is at static equilibrium!). This is given by
N = - Ngrav = mgR sinc (4)
At = c have static equilib. Both angular velocity & angular acceleration vanish: (d/dt) = 0 & (d2/dt2) =0
• At critical angle c for static
equilib the applied torque is
N = mgRsinc (4)
(4) Obviously, the greater the
applied torque N, the larger the critical angle c.
(4) There is a critical torque Nc for which the critical angle
c = (½)π & the pendulum is horizontal; c = (½)π is obviously
the maximum critical angle for static equilibrium. From (4), this is:
Nc = mgR (5)If the applied torque N > Nc, then N > mgRsin for all angles The
pendulum can rotate “over the top” & goes from 0 to 2π. The motion
then is rotation, not oscillation. The rotational motion still satisfies the
equation of motion: N+ Ngrav= mR2(d2/dt2) (3)
Solution to (3) will give time dependent angular velocity ω = (d/dt).
• It’s useful to remember all this
when now considering the motion
of the driven, damped pendulum. • Assume, as usual, that the torque due to the frictional
or damping force is angular velocity ω = (d/dt), with damping constant equal to the viscosity η: Nfr - ηω
• Newton’s 2nd Law gives the equation of motion:
N + Nfr + Ngrav = mR2(d2/dt2) (3)
Putting in the forms for Nfr & Ngrav & putting them on the other side of the equation gives:
N = mR2(d2/dt2) + η(d/dt) + mgRsin (6)
N = mR2(d2/dt2) + η(d/dt) + mgRsin (6) • Define a critical angular speed the angular speed at which the
damping torque Nfr = ηω exactly equals the critical torque for static equilibrium Nc = mgR:
ωc Ncη-1 = mgRη-1 (7)
• Write (6) in dimensionless form: Divide by Nc & use (7):
(N/Nc) = (ω0)-2(d2/dt2) + (ωc)-1(d/dt) + sin (8)
Recall: ω0 = (g/R)½ = harmonic oscillation frequency for the small angle problem.
• (8) is the equation we will now work with in several special cases. There is rich, varied, complex behavior which can be obtained from it! This includes the possibility of chaotic behavior! For discussion of conditions under which (8) can lead to chaos, see the text by Marion, Ch. 4.
(N/Nc) = (ω0)-2(d2/dt2) + (ωc)-1(d/dt) + sin (8)
where: Nc = mgR, ω0 = (g/R)½ , ωc = mgRη-1
(8): A nonlinear, inhomogeneous, 2nd order differential
equation! No general theory of nonlinear differential equations exists! We can get the solution in special cases or else we need numerical solutions.
• Solution to the inhomogeneous equation = solution to homogeneous equation + a particular solution to the inhomogeneous equation. Due to damping, the 2nd term on right side (8), the solution to the homogeneous equation will exponentially decay with time (transient solution!):
exp(-t/ωc) = exp[- ηt(mgR)-1] We look at solutions to (8) at long times (ηt) >> (mgR) so the
transient solution has decayed away & only the particular solution to
The inhomogeneous equation remains Dynamic Steady State
(N/Nc) = (ω0)-2(d2/dt2) + (ωc)-1(d/dt) + sin (8) where: Nc = mgR, ω0 = (g/R)½ , ωc = mgRη-1
• Now, a qualitative discussion of several special cases of solutions to (8) under Dynamic Steady State conditions:
1. Low applied torques: N Nc (the applied torque is less than
the critical torque for static equilibrium at c = (½)π): In this case, after the transient dies away, the pendulum will eventually go to a Static Steady State, at which applied torque & gravitational torque exactly balance (so the pendulum is at static equilibrium!). We had this earlier. It is given by
N = - Ngrav = mgRsinc = Ncsinc
For constant N, after long enough times, the pendulum will stop at angle c !
(N/Nc) = (ω0)-2(d2/dt2) + (ωc)-1(d/dt) + sin (8) Special cases of solutions to (8) in the Dynamic Steady State:
2. Undamped Motion (η = 0) & constant applied torque N: It’s best in this case to go back to Newton’s 2nd Law:
Nnet = N + Ngrav = N - Ncsin = mR2(d2/dt2)
a. The total torque, Nnet is dependent! It takes on
special values at 4 angles:
= 0 Nnet = N
= (½)π Nnet = N - Nc
= π Nnet = N
= 3(½)π, Nnet = N + Nc .
(N/Nc) = (ω0)-2(d2/dt2) + (ωc)-1(d/dt) + sin (8)
Special cases of solutions to (8) in Dynamic Steady State:
2. Undamped Motion (η = 0) & constant applied torque N: Nnet = N + Ngrav = N - Ncsin = mR2(d2/dt2)
b. If N > Nc, the motion is
continuously accelerated rotation!
The pendulum gains energy as
time goes on. The angular speed
increases with time, but with
periodic fluctuations every cycle.
The time averaged angular speed:
<ω> = <d/dt> increases
continually with time. See figure (which is with damping).
(N/Nc) = (ω0)-2(d2/dt2) + (ωc)-1(d/dt) + sin (8)
Special cases of solutions to (8) in Dynamic Steady State:
3. Damping is present, small ωc << ω0. Also, N > Nc
The angular speed ω increases until,
at some time t, the damping term
approaches the applied torque:
η(d/dt) N. At this time, average
angular speed approaches the limiting
value: <ω> <ω>L as in the figure. After this time, the
angular acceleration fluctuates about a zero average value:
<d2/dt2> = 0. In this state, the pendulum undergoes quasi-static
motion: It rotates with angular speed ω which has periodic
fluctuations about its average <ω>L, but remains close to it.
(N/Nc) = (ω0)-2(d2/dt2) + (ωc)-1(d/dt) + sin (8)
Special cases of solutions to (8) in Dynamic Steady State:
3. Damping is present, small ωc << ω0. Also, N > Nc
More on quasi-static motion: Rotates with ω near <ω>L,
fluctuating periodically about it. In this state, the angular
acceleration fluctuates about a 0 average value: <d2/dt2> = 0.
Approximately:
(N/Nc) (ωc)-1(d/dt) + sin (8´)
(8´): Solve analytically! Student
exercise. Then, time average & find:
<ω> = 0, N < Nc
<ω> = ωc[(N/Nc)2 - 1]½ , N > Nc
<ω> ωc(N/Nc) , N >> Nc.
See figure:
(N/Nc) = (ω0)-2(d2/dt2) + (ωc)-1(d/dt) + sin (8)
Special cases of solutions to (8) in Dynamic Steady State:
3. Damping is present, small ωc << ω0.
More on quasi-static motion: Focus on
points A & B in figure Figure shows cyclic variations in ω
at points A & B. At A, N = 1.2Nc,
From the previous discussion,
0.2Nc < N < 2.2Nc & ω is fast at the bottom & slow at the
top, as in the figure. At B,
N = 2Nc, Nc < N < 3Nc & ω has more regular variations
with time, as in the figure.
(N/Nc) = (ω0)-2(d2/dt2) + (ωc)-1(d/dt) + sin (8)
Special cases of solutions to (8) in Dynamic Steady State:
3. Damping is present, small ωc << ω0.
More on quasi-static motion: In the limit
N >> 2Nc <ω> ωc(N/Nc) >> ωc,
the solution to (8) approximately gives
ω(t) <ω> + αsin(Ωt) (α & Ω constants)
as for point B in the figures.
(N/Nc) = (ω0)-2(d2/dt2) + (ωc)-1(d/dt) + sin (8)
Special cases of solutions to (8) in Dynamic Steady State:
4. Negligible (or small) damping (η 0, ωc >> ω0):
a. If we wait long enough, we can get Static Steady State, at
which the applied torque & the gravitational torque exactly
balance (static equilibrium!) Given by
N = - Ngrav = mgRsin c = Nc sinc Nc
ω = 0, N Nc (a)
b. Also, the solution just discussed, in
which the applied torque balances
the time average damping torque
also applies for all N.
<ω> ωc(N/Nc), all N (b)
See figure:
(N/Nc) = (ω0)-2(d2/dt2) + (ωc)-1(d/dt) + sin (8)
Special cases of solutions to (8) in Dynamic Steady State:
4. Negligible (or small) damping (η 0, ωc >> ω0):
ω = 0, N Nc (a) <ω> ωc(N/Nc), all N (b)
Discussion: See figure. The system has
hysteresis (differing behavior for
increasing & decreasing N): Starting
from N = 0, as N increases for N Nc,
the pendulum is always stopped at
angles c given by N = Ncsinc
(a) gives ω = 0. When N reaches Nc, ω JUMPS
discontinuously: (b) gives <ω> = ωc(N=Nc). Also, <ω>
increases linearly as ωc(N/Nc) as N increases > Nc. For
decreasing N, (b) applies, all N & <ω> decreases linearly to 0.
(N/Nc) = (ω0)-2(d2/dt2) + (ωc)-1(d/dt) + sin (8)
Special cases of solutions to (8) in Dynamic Steady State:
5. General case: Response for ωc << ω0 Response for ωc >> ω0
What is the response for
intermediate conditions where
ωc ω0? To answer this, we must solve the general
equation (8). Must do this numerically. Results for N
vs. <ω> for ωc = 2ω0 are shown here
Increasing N: Get ω = 0 until N = Nc.
At N = Nc, ω JUMPS discontinuously to
<ω> = 2ωc. Then, <ω> increases non-linearly
with N until ωc >> ω0 is reached & it becomes
linear. Decreasing N: Hysteresis with <ω> =0 for N = Nc´ < Nc