Chapter 5 · Section 5.4

5.4 Forced Motion

So far the mass moved only because of how it was started. In practice vibrations are usually driven: an engine shakes its mounts at every revolution, an out-of-balance drum shakes a washing machine, uneven road surface shakes a car, an earthquake shakes a building. In this section an external force $F (t)$ acts on the mass. We will find that, with damping present, the response splits cleanly into two parts: a part that depends on how the motion started and dies away, and a part that is dictated by the force and stays.

Transient and steady state

The equation is

m x^{''} + c x^{'} + k x = F (t) .

By Theorem 4.1.8 its general solution is $x = x_{c} + x_{p}$ , where $x_{c}$ is the general solution of the homogeneous equation (the free motion of Section 5.2 or 5.3) and $x_{p}$ is any one particular solution.

When $c > 0$ , Section 5.3 showed that every solution of the homogeneous equation tends to zero as $t \to \infty$ , whatever the case (over-, critically or underdamped): each term contains a decaying exponential. So after a while $x_{c}$ is negligible and only $x_{p}$ is left.

Definition 5.4.1 (Transient and steady-state solutions)

Let $c > 0$ . In $x = x_{c} + x_{p}$ the part $x_{c}$ , which tends to zero as $t \to \infty$ , is called the transient. If $F$ is periodic (for example $F_{0} cos ω t$ ), the periodic particular solution $x_{p}$ given by undetermined coefficients is called the steady-state solution $x_{ss}$ .

Theorem 5.4.1 (The steady state forgets the initial conditions)

Let $c > 0$ and let $x_{1}$ , $x_{2}$ be any two solutions of $m x^{''} + c x^{'} + k x = F (t)$ (with any initial conditions). Then $x_{1} (t) - x_{2} (t) \to 0$ as $t \to \infty$ .

Why. The difference $x_{1} - x_{2}$ solves the homogeneous equation $m x^{''} + c x^{'} + k x = 0$ (the $F$ 's cancel), and every solution of that equation tends to zero when $c > 0$ . $□$

So the long-term behaviour is decided by the force alone. That is good news for an engineer: to know how much a machine will shake in operation, you need $x_{ss}$ , not the details of how it was switched on. Figures 5.4.1 and 5.4.2 show this for Example 5.4.1.

Figure 5.4.1. Example 5.4.1: the full response

x

(solid) is the sum of a transient (decaying like

e^{- 3 t}

) and the steady state

x_{ss}

. After about 1.5 s the two curves

x

and

x_{ss}

are indistinguishable.

Figure 5.4.2. Three different starts, one equation

x^{''} + 6 x^{'} + 45 x = 5 cos 3 t

. Theorem 5.4.1 in action: all solutions merge into the same steady state (dotted).

Harmonic forcing: the steady-state formula

The most important force is a pure oscillation $F (t) = F_{0} cos ω t$ , with amplitude $F_{0} > 0$ (N) and forcing frequency $ω > 0$ (rad/s). Do not confuse $ω$ , which we choose (for instance the speed of a motor), with $ω_{0} = k / m$ , which belongs to the system.

We find $x_{p}$ by undetermined coefficients (Section 4.3). When $c > 0$ the roots of $m r^{2} + cr + k$ have negative real part, so they are never $\pm iω$ and no modification is needed. Try

x_{p} = a cos ω t + b sin ω t .

Differentiate. $x_{p}^{'} = - aω sin ω t + bω cos ω t, x_{p}^{''} = - a ω^{2} cos ω t - b ω^{2} sin ω t .$
Substitute and collect. In $m x_{p}^{''} + c x_{p}^{'} + k x_{p}$ , the coefficients are $cos ω t : sin ω t : (k - m ω^{2}) a + c ω b, - c ω a + (k - m ω^{2}) b .$
Match with $F_{0} cos ω t$ . Write $D_{1} = k - m ω^{2}$ and $D_{2} = c ω$ to keep the algebra short: $D_{1} a + D_{2} b = F_{0}, - D_{2} a + D_{1} b = 0.$
Solve. The second equation gives $b = D_{2} a / D_{1}$ if $D_{1} \neq = 0$ ; substituting, or simply using Cramer's rule (which also covers $D_{1} = 0$ ), with determinant $D_{1}^{2} + D_{2}^{2} > 0$ : $a = \frac{F _{0} D _{1}}{D _{1}^{2} + D _{2}^{2}}, b = \frac{F _{0} D _{2}}{D _{1}^{2} + D _{2}^{2}} .$
Amplitude–phase form. As in Section 5.2, $a^{2} + b^{2} = F_{0} / D_{1}^{2} + D_{2}^{2}$ , and the angle $θ$ with $cos θ = D_{1} / D_{1}^{2} + D_{2}^{2}$ , $sin θ = D_{2} / D_{1}^{2} + D_{2}^{2}$ gives $x_{p} = M cos (ω t - θ)$ .

Theorem 5.4.2 (Steady-state response to harmonic forcing)

Let $m, c, k > 0$ . The equation $m x^{''} + c x^{'} + k x = F_{0} cos ω t$ has the steady-state solution

x_{ss} (t) = \frac{F _{0}}{Δ} [(k - m ω^{2}) cos ω t + c ω sin ω t] = M cos (ω t - θ),

where $Δ = (k - m ω^{2})^{2} + (c ω)^{2}$ ,

M = \frac{F _{0}}{( k - m ω ^{2} ) ^{2} + c ^{2} ω ^{2}}, cos θ = \frac{k - m ω ^{2}}{Δ}, sin θ = \frac{c ω}{Δ}, 0 < θ < π .

For the force $F_{0} sin ω t$ the steady state is $M sin (ω t - θ)$ with the same $M$ and $θ$ .

The last sentence holds because $F_{0} sin ω t = F_{0} cos (ω t - π /2)$ is the same force delayed by $π / (2 ω)$ , and the response to a delayed force is the delayed response.

Reading the formula

Same frequency. The mass oscillates at the forcing frequency $ω$ , not at its natural frequency.
Amplitude. $M$ depends on $ω$ . As $ω \to 0$ , $M \to F_{0} / k$ : a very slow force just stretches the spring statically. As $ω \to \infty$ , $M \to 0$ like $F_{0} / (m ω^{2})$ : the mass is too heavy to follow a very fast force. In between, $M$ may become much larger than $F_{0} / k$ ; that is resonance, the subject of Section 5.5.
Phase lag. Since $sin θ > 0$ , the response always lags behind the force by $θ$ (a time delay $θ / ω$ ). If $ω < ω_{0}$ then $k - m ω^{2} > 0$ and $θ < π /2$ : mass and force move nearly together. If $ω = ω_{0}$ then $θ = π /2$ exactly. If $ω > ω_{0}$ then $θ > π /2$ : the mass moves nearly opposite to the force (Figure 5.4.3).

It is often convenient to compare $M$ with the static deflection $F_{0} / k$ . Dividing numerator and denominator by $k$ and using $ω_{0}^{2} = k / m$ and $c / k = 2 ζ / ω_{0}$ gives the dimensionless magnification factor

\frac{M}{F _{0} / k} = \frac{1}{( 1 - ( ω / ω _{0} ) ^{2} ) ^{2} + ( 2 ζ ω / ω _{0} ) ^{2}} .

It depends only on the frequency ratio $ω / ω_{0}$ and the damping ratio $ζ$ . We plot it in Section 5.5.

Figure 5.4.3. Example 5.4.3: force (grey) and steady-state response (red), both scaled to amplitude 1. Below the natural frequency the lag is less than a quarter cycle; above it, more than a quarter cycle, so the mass moves almost against the force.

A constant force

If $F (t) = F_{0}$ is constant (switched on at $t = 0$ ), the trial $x_{p} = A$ gives $k A = F_{0}$ , so $x_{p} = F_{0} / k$ . The force simply moves the equilibrium position down by $F_{0} / k$ , and the transient describes how the mass settles there. If the system is underdamped, it overshoots the new equilibrium first (Example 5.4.2). This "step response" is a standard test in control engineering.

Forced motion without damping

If $c = 0$ and $F = F_{0} cos ω t$ with $ω \neq = ω_{0}$ , the trial $x_{p} = a cos ω t$ suffices (there is no $x^{'}$ term to create a sine). Substituting gives $(k - m ω^{2}) a = F_{0}$ , so

x (t) = c_{1} cos ω_{0} t + c_{2} sin ω_{0} t + \frac{F _{0}}{m ( ω _{0}^{2} - ω ^{2} )} cos ω t .

Now nothing decays: the free oscillation at $ω_{0}$ and the forced oscillation at $ω$ coexist forever, and there is no real "steady state". In practice there is always a little damping, which slowly removes the free part. The formula also signals trouble when $ω$ is close to $ω_{0}$ : the coefficient becomes huge, and for $ω = ω_{0}$ it is undefined. That is the resonance case of Section 5.5.

Method in a nutshell: forced motion

Set up $m x^{''} + c x^{'} + k x = F (t)$ with initial conditions (Section 5.1).
Find $x_{c}$ from $m r^{2} + cr + k = 0$ (Section 5.3).
Find $x_{p}$ by undetermined coefficients. For $F_{0} cos ω t$ with $c > 0$ you may use Theorem 5.4.2 directly; for a constant force $x_{p} = F_{0} / k$ ; for a sum of forces add the particular solutions.
For an IVP: write $x = x_{c} + x_{p}$ , then fit $x (0)$ and $x^{'} (0)$ . Remember $x_{p}$ contributes to both.
Interpret: transient $= x_{c}$ (dies out if $c > 0$ ), steady state $= x_{p}$ ; report $M$ , $θ$ and the time lag $θ / ω$ .

Experiment with the simulator: it starts with the system of Example 5.4.1. Change the initial conditions and check that only the transient changes; change $ω$ and watch the steady-state amplitude and lag respond. The second widget finds particular solutions by undetermined coefficients, step by step (it writes the independent variable as $x$ ).

Worked examples

Example 5.4.1

Solve the problem set up in Example 5.1.2: $2 x^{''} + 12 x^{'} + 90 x = 10 cos 3 t$ , $x (0) = - 0.1$ , $x^{'} (0) = 0$ . Identify the transient and the steady state, and give the steady-state amplitude and phase lag.

Solution

Standard form. $x^{''} + 6 x^{'} + 45 x = 5 cos 3 t$ .
Complementary solution. $r^{2} + 6 r + 45 = 0$ gives $r = - 3 \pm 9 - 45 = - 3 \pm 6 i$ : $x_{c} = e^{- 3 t} (c_{1} cos 6 t + c_{2} sin 6 t) .$
Particular solution. Try $x_{p} = a cos 3 t + b sin 3 t$ . Then $x_{p}^{''} = - 9 x_{p}$ and $x_{p}^{'} = - 3 a sin 3 t + 3 b cos 3 t$ . Substituting: $cos 3 t : sin 3 t : - 9 a + 18 b + 45 a = 36 a + 18 b = 5, - 9 b - 18 a + 45 b = - 18 a + 36 b = 0.$ The second gives $a = 2 b$ ; then $72 b + 18 b = 5$ , so $b = \frac{1}{18}$ and $a = \frac{1}{9}$ .
Initial conditions. $x = x_{c} + x_{p}$ . At $t = 0$ : $c_{1} + \frac{1}{9} = - 0.1$ , so $c_{1} = - \frac{19}{90}$ . Next, $x^{'} (0) = - 3 c_{1} + 6 c_{2} + 3 b = \frac{57}{90} + 6 c_{2} + \frac{1}{6} = 0 ⟹ 6 c_{2} = - \frac{72}{90}, c_{2} = - \frac{2}{15} .$
Solution. $x (t) = transient - e^{- 3 t} (\frac{19}{90} cos 6 t + \frac{2}{15} sin 6 t) + steady state \frac{1}{9} cos 3 t + \frac{1}{18} sin 3 t .$
Steady-state amplitude and phase. $M = \frac{1}{81} + \frac{1}{324} = \frac{5}{324} = \frac{5}{18} \approx 0.124$ m, and $tan θ = \frac{1/18}{1/9} = \frac{1}{2}$ with both coefficients positive, so $θ = arctan \frac{1}{2} \approx 0.464$ rad. The mass lags the force by $θ / ω \approx 0.155$ s.
Check with Theorem 5.4.2. $k - m ω^{2} = 90 - 18 = 72$ , $c ω = 36$ , $7 2^{2} + 3 6^{2} = 365$ , so $M = 10/ (365) = 5 /18$ . ✓
How fast is the transient gone? It is at most about $0.25 e^{- 3 t}$ m; at $t = 2$ s this is below $1$ mm.

Example 5.4.2 (Suddenly applied constant force)

The 0.5 kg mass of Example 5.1.1 ( $k = 50$ N/m) now has a damper with $c = 6$ N·s/m. It hangs at rest when, at $t = 0$ , a constant downward force of 4 N is switched on. Find $x (t)$ , the new equilibrium and the largest displacement.

Solution

Equation. $0.5 x^{''} + 6 x^{'} + 50 x = 4$ , i.e. $x^{''} + 12 x^{'} + 100 x = 8$ , with $x (0) = 0$ , $x^{'} (0) = 0$ .
Complementary solution. $r = - 6 \pm 36 - 100 = - 6 \pm 8 i$ : $x_{c} = e^{- 6 t} (c_{1} cos 8 t + c_{2} sin 8 t)$ . ( $ζ = 0.6$ , underdamped.)
Particular solution. $x_{p} = A$ : $100 A = 8$ , so $A = 0.08$ m $= F_{0} / k$ , the new equilibrium.
Initial conditions. $c_{1} + 0.08 = 0$ , so $c_{1} = - 0.08$ . $x^{'} (0) = - 6 c_{1} + 8 c_{2} = 0$ , so $c_{2} = - 0.06$ : $x (t) = 0.08 - e^{- 6 t} (0.08 cos 8 t + 0.06 sin 8 t) .$
Velocity. Differentiating and simplifying, $x^{'} = e^{- 6 t} [(0.48 - 0.48) cos 8 t + (0.36 + 0.64) sin 8 t] = e^{- 6 t} sin 8 t .$
Largest displacement. $x^{'} = 0$ first at $t = π /8 \approx 0.393$ s (where $sin 8 t$ changes sign from $+$ to $-$ ). There $cos 8 t = - 1$ , $sin 8 t = 0$ : $x (\frac{π}{8}) = 0.08 + 0.08 e^{- 3 π /4} \approx 0.08 \times 1.0948 \approx 0.0876 m .$ The overshoot is about $9.5%$ of the final value (Figure 5.4.4). Compare Example 5.1.3, where without damping the overshoot was $100%$ .

Figure 5.4.4. Example 5.4.2: response to a constant force switched on at

t = 0

. The mass overshoots the new equilibrium

F_{0} / k = 8

cm by about

9.5%

and settles within about

0.8

Example 5.4.3 (Amplitude and lag at two frequencies)

For $2 x^{''} + 8 x^{'} + 56 x = 10 cos ω t$ , find the steady-state solution, its amplitude and its phase lag for (a) $ω = 4$ rad/s and (b) $ω = 6$ rad/s. Compare with the natural frequency.

Solution

Natural frequency. $ω_{0} = 56/2 = 28 \approx 5.29$ rad/s. Static deflection $F_{0} / k = 10/56 \approx 0.179$ m.
(a) $ω = 4$ . $k - m ω^{2} = 56 - 32 = 24$ and $c ω = 32$ , so $Δ = 576 + 1024 = 40$ . $M = \frac{10}{40} = 0.25 m, cos θ = \frac{24}{40}, sin θ = \frac{32}{40}, θ = arctan \frac{4}{3} \approx 0.927 rad .$ Explicitly, $x_{ss} = \frac{10}{1600} (24 cos 4 t + 32 sin 4 t) = 0.15 cos 4 t + 0.2 sin 4 t$ . The amplitude is 1.4 times the static deflection.
(b) $ω = 6$ . $k - m ω^{2} = 56 - 72 = - 16$ and $c ω = 48$ , so $Δ = 256 + 2304 = 2560$ , $Δ = 1610 \approx 50.6$ . $M = \frac{10}{16 10} \approx 0.198 m, cos θ < 0, sin θ > 0.$ Second quadrant: $θ = π - arctan \frac{48}{16} = π - arctan 3 \approx 1.893$ rad. Explicitly, $x_{ss} = - \frac{1}{16} cos 6 t + \frac{3}{16} sin 6 t$ .
Comparison. Below $ω_{0}$ the lag is less than $π /2$ ; above $ω_{0}$ it is more than $π /2$ (Figure 5.4.3). Note the quadrant: $arctan (c ω / (k - m ω^{2})) = arctan (- 3) \approx - 1.25$ would be wrong.

Example 5.4.4 (Undamped, away from resonance)

A 0.4 kg mass on a spring with $k = 10$ N/m, without damping, is at rest at equilibrium when a force $2.4 cos 3 t$ N starts to act. Find the motion.

Solution

Equation. $0.4 x^{''} + 10 x = 2.4 cos 3 t$ , i.e. $x^{''} + 25 x = 6 cos 3 t$ , with $x (0) = x^{'} (0) = 0$ . Here $ω_{0} = 5$ and $ω = 3$ .
Particular solution. $x_{p} = a cos 3 t$ : $(- 9 + 25) a = 6$ , so $a = \frac{3}{8}$ .
General solution. $x = c_{1} cos 5 t + c_{2} sin 5 t + \frac{3}{8} cos 3 t$ .
Initial conditions. $x (0) = c_{1} + \frac{3}{8} = 0$ gives $c_{1} = - \frac{3}{8}$ ; $x^{'} (0) = 5 c_{2} = 0$ gives $c_{2} = 0$ .
Answer. $x (t) = \frac{3}{8} (cos 3 t - cos 5 t) m .$ Even though the mass started at rest, the solution contains a free oscillation at $ω_{0} = 5$ that never dies out. The motion is periodic with period $2 π$ (the shortest time after which both $cos 3 t$ and $cos 5 t$ repeat) and never exceeds $\frac{3}{4}$ m in size.

Exercises

Find the steady-state solution and its amplitude for $x^{''} + 2 x^{'} + 17 x = 0.65 cos 3 t$ .
Answer
$x_{ss} = 0.052 cos 3 t + 0.039 sin 3 t$ ; $M = 0.065$
A 0.5 kg mass with $k = 4$ N/m and $c = 2$ N·s/m starts at rest at equilibrium; a force $0.5 sin 2 t$ N acts. Find $x (t)$ .
Answer
$x = 0.1 e^{- 2 t} (cos 2 t + 0.5 sin 2 t) + 0.1 (- cos 2 t + 0.5 sin 2 t)$
For $m = 1$ , $c = 3$ , $k = 12$ and $F = 3 cos 2 t$ , find the steady-state amplitude and phase lag.
Answer
$M = 3/ 8^{2} + 6^{2} = 0.3$ ; $θ = arctan \frac{3}{4} \approx 0.644$ rad
A 1 kg mass with $k = 25$ N/m and $c = 10$ N·s/m (critical damping) is at rest at equilibrium when a constant force of 5 N is switched on. Find $x (t)$ . Does it overshoot?
Answer
$x = 0.2 - (0.2 + t) e^{- 5 t}$ ; no: $x^{'} = 5 t e^{- 5 t} > 0$ , so $x$ rises monotonically to $0.2$
A 2 kg mass with $k = 98$ N/m and no damping starts at rest at equilibrium under the force $6.6 cos 4 t$ N. Find $x (t)$ .
Answer
$x = 0.1 (cos 4 t - cos 7 t)$
Find the steady-state solution of $x^{''} + 3 x^{'} + 5 x = 10 + 5 cos t$ . (Use superposition.)
Answer
$x_{ss} = 2 + 0.8 cos t + 0.6 sin t$
For $x^{''} + x^{'} + 4 x = cos 3 t$ find the steady-state amplitude and the phase lag. Is the forcing frequency above or below the natural frequency?
Answer
$M = 1/ 34 \approx 0.171$ ; $θ = π - arctan \frac{3}{5} \approx 2.60$ rad; above ( $ω = 3 > ω_{0} = 2$ )
The top of a spring ( $k = 16$ N/m) carrying a 1 kg mass is moved up and down as $y (t) = 0.02 cos 2 t$ m. With no damping, the spring force is $- k (x - y)$ , so $x^{''} + 16 x = 16 y$ . Find the steady-state motion of the mass and compare its amplitude with that of the support.
Answer
$x_{ss} = \frac{2}{75} cos 2 t \approx 0.0267 cos 2 t$ m, i.e. $\frac{4}{3}$ times the support amplitude
For $m = 2$ , $c = 4$ , $k = 50$ and $ω = 3$ , how large must $F_{0}$ be to produce a steady-state amplitude of 5 cm?
Answer
$F_{0} = 0.05 3 2^{2} + 1 2^{2} = 0.05 1168 \approx 1.71$ N
Solve $x^{''} + 2 x^{'} + 37 x = 36 e^{- t}$ , $x (0) = 0$ , $x^{'} (0) = 0$ .
Answer
$x = e^{- t} (1 - cos 6 t)$
Show from Theorem 5.4.2 that $M \to F_{0} / k$ as $ω \to 0$ and that $m ω^{2} M \to F_{0}$ as $ω \to \infty$ . Interpret both limits.
Answer
At $ω = 0$ the denominator is $k$ (static stretching). For large $ω$ the denominator behaves like $m ω^{2}$ : the inertia term dominates and the mass barely moves.