Unit 5, lesson 28

Forced Oscillations, Beats, and Resonance

How a periodic force drives a mass on a spring, why a drive near the natural angular frequency gives beats, why undamped resonance causes secular growth, and when damping gives a finite amplitude peak.

  • Video 28
  • Damped and Forced Oscillations

Push a child on a swing once per swing and the arcs keep growing. Push at a slightly different rhythm and the motion can swell and fade by turns, and with enough friction it settles into a steady size. Which of these happens depends on how the driving frequency compares with the natural frequency, and on the damping. This lesson adds a periodic force to the mass-spring model and separates three behaviors, beats from two nearby frequencies, the linear growth of undamped resonance and the finite amplitude peak that damping allows.

Key terms

Forced oscillation

Definition Let tt be time and let primes denote derivatives with respect to tt. A forced oscillation is a solution of

mx′′+cx′+kx=F0cos⁡ωt,mx'' + cx' + kx = F_0\cos\omega t,

where m>0m > 0, k>0k > 0 and c≥0c \ge 0 are the mass, stiffness and damping coefficient of the free model, F0>0F_0 > 0 is the force amplitude in newtons and ω>0\omega > 0 is the driving angular frequency in radians per second, so the force completes ω/(2π)\omega/(2\pi) cycles per second. Every solution has the form x=xh+xpx = x_h + x_p, where xpx_p is any one particular solution and xhx_h is a free solution of the homogeneous equation, chosen to match the initial data. Without damping, and with ω≠ω0=k/m\omega \ne \omega_0 = \sqrt{k/m}, one particular solution is

xp=F0cos⁡ωtm(ω02−ω2).x_p = \frac{F_0\cos\omega t}{m(\omega_0^2 - \omega^2)}.

In plain words The drive keeps pushing at its own rhythm. The response is one motion that follows the drive plus whatever free motion the starting conditions leave behind.

Example For x′′+9x=5cos⁡2tx'' + 9x = 5\cos 2t the trial cos⁡2t\cos 2t gives −4cos⁡2t+9cos⁡2t=5cos⁡2t-4\cos 2t + 9\cos 2t = 5\cos 2t, so xp=cos⁡2tx_p = \cos 2t and every solution is x=cos⁡2t+C1cos⁡3t+C2sin⁡3tx = \cos 2t + C_1\cos 3t + C_2\sin 3t. Driving the same system faster, with 5cos⁡4t5\cos 4t, gives xp=59−16cos⁡4t=−57cos⁡4tx_p = \frac{5}{9 - 16}\cos 4t = -\frac{5}{7}\cos 4t. Below the natural frequency the forced motion moves with the force, and above it the forced motion moves against the force.

Natural angular frequency

Definition The natural angular frequency of the oscillator is

ω0=km,\omega_0 = \sqrt{\frac{k}{m}},

measured in radians per second. It is the angular frequency of free undamped motion x=Acos⁡ω0t+Bsin⁡ω0tx = A\cos\omega_0 t + B\sin\omega_0 t, whose period is 2π/ω02\pi/\omega_0 and whose ordinary frequency is ω0/(2π)\omega_0/(2\pi) cycles per second. It depends only on mm and kk. It is different from the driving angular frequency ω\omega, which is set from outside, and from the damped angular frequency ωd=ω02−(c/(2m))2\omega_d = \sqrt{\omega_0^2 - (c/(2m))^2} of an underdamped system, which is smaller.

In plain words A mass on a spring has its own rhythm, the one it swings at when left alone without friction. Every drive is measured against that rhythm.

Example For m=1 kgm = 1\ \text{kg} and k=9 N/mk = 9\ \text{N/m} the natural angular frequency is ω0=3\omega_0 = 3 radians per second. One free cycle takes 2π/3≈2.092\pi/3 \approx 2.09 seconds, about 0.4770.477 cycles per second. Doubling the mass lowers it to 3/2≈2.123/\sqrt{2} \approx 2.12 radians per second.

Beats

Definition Beats are the slow periodic swelling and fading of an oscillation that is the sum of two oscillations with nearby frequencies. In an undamped forced oscillator (c=0c = 0, ω≠ω0\omega \ne \omega_0) started from rest, x(0)=0x(0) = 0 and x′(0)=0x'(0) = 0, the response is x=K(cos⁡ωt−cos⁡ω0t)x = K(\cos\omega t - \cos\omega_0 t) with K=F0/(m(ω02−ω2))K = F_0/(m(\omega_0^2 - \omega^2)). The difference of cosines identity turns it into a product,

x=2Ksin⁡(ω0−ω)t2×sin⁡(ω0+ω)t2.\begin{aligned} x &= 2K\sin\frac{(\omega_0 - \omega)t}{2} \\[0.4em] &\qquad \times \sin\frac{(\omega_0 + \omega)t}{2}. \end{aligned}

When ω\omega is close to ω0\omega_0, the second factor oscillates fast and the first changes slowly, so the slow factor sets the size of the fast oscillation. The magnitude envelope 2∣K∣∣sin⁡((ω0−ω)t/2)∣2\lvert K \rvert \lvert \sin((\omega_0 - \omega)t/2) \rvert repeats with the beat period

Tbeat=2π∣ω0−ω∣,T_{\text{beat}} = \frac{2\pi}{\lvert \omega_0 - \omega \rvert},

while the signed slow factor takes twice as long to repeat. For every fixed ω≠ω0\omega \ne \omega_0 the response stays bounded, with ∣x∣≤2∣K∣\lvert x \rvert \le 2\lvert K \rvert.

In plain words The drive and the free motion drift in and out of step. In step their effects add, out of step they cancel, so the size of the motion rises and falls slowly.

Example Take m=1m = 1, k=9k = 9 and c=0c = 0, so ω0=3\omega_0 = 3, and drive with 5cos⁡2t5\cos 2t from rest. The particular solution cos⁡2t\cos 2t starts at x=1x = 1, so subtract the free solution cos⁡3t\cos 3t. Both cosines have zero velocity at t=0t = 0, so the velocity condition holds too, and

x=cos⁡2t−cos⁡3t=2sin⁡t2 sin⁡5t2.\begin{aligned} x &= \cos 2t - \cos 3t \\[0.4em] &= 2\sin\tfrac{t}{2}\,\sin\tfrac{5t}{2}. \end{aligned}

The envelope 2∣sin⁡(t/2)∣2\lvert \sin(t/2) \rvert has period 2π2\pi, and the motion reaches its largest size x=2x = 2 at t=πt = \pi, where cos⁡2π−cos⁡3π=1+1\cos 2\pi - \cos 3\pi = 1 + 1. The frequencies 22 and 33 are not very close, so each beat holds only a few fast oscillations. Closer frequencies give slower beats with more oscillations inside each one.

Undamped resonance

Definition Undamped resonance is the case c=0c = 0 and ω=ω0\omega = \omega_0. The usual trial acos⁡ω0t+bsin⁡ω0ta\cos\omega_0 t + b\sin\omega_0 t solves the homogeneous equation, so substituting it gives 00 and never the force. Multiplying the trial by tt, as the modification rule of undetermined coefficients prescribes, gives the particular solution

xp=F02mω0 tsin⁡ω0t.x_p = \frac{F_0}{2m\omega_0}\,t\sin\omega_0 t.

It also satisfies xp(0)=0x_p(0) = 0 and xp′(0)=0x_p'(0) = 0, so it is the response from rest. Every other solution adds a bounded free oscillation to it, so every solution is unbounded as t→∞t \to \infty.

In plain words Each push arrives at the same point of every swing, so the pushes keep adding up instead of canceling, and without damping nothing removes the energy they supply.

Example Keep m=1m = 1, k=9k = 9 and c=0c = 0, and drive with 6cos⁡3t6\cos 3t. The trial cos⁡3t\cos 3t fails, since (cos⁡3t)′′+9cos⁡3t=0(\cos 3t)'' + 9\cos 3t = 0. For x=tsin⁡3tx = t\sin 3t,

x′=sin⁡3t+3tcos⁡3t,x′′=6cos⁡3t−9tsin⁡3t,\begin{aligned} x' &= \sin 3t + 3t\cos 3t, \\[0.4em] x'' &= 6\cos 3t - 9t\sin 3t, \end{aligned}

so x′′+9x=6cos⁡3tx'' + 9x = 6\cos 3t. Also x(0)=0x(0) = 0 and x′(0)=0x'(0) = 0, so tsin⁡3tt\sin 3t is the response from rest. It matches the formula, since F0/(2mω0)=6/6=1F_0/(2m\omega_0) = 6/6 = 1.

Secular growth

Definition Secular growth is growth of an oscillation’s envelope in proportion to a power of tt, here to tt itself. For t≥0t \ge 0 the resonant response from rest satisfies

∣xp(t)∣≤F02mω0 t,\lvert x_p(t) \rvert \le \frac{F_0}{2m\omega_0}\,t,

with equality whenever ∣sin⁡ω0t∣=1\lvert \sin\omega_0 t \rvert = 1, so its envelope is a pair of straight lines through the origin with slopes ±F0/(2mω0)\pm F_0/(2m\omega_0). Unlike beats, which stay below a fixed bound, secular growth passes every bound. A real spring eventually leaves its linear range, and any damping c>0c > 0 keeps the amplitude bounded.

In plain words The swings grow by about the same amount every cycle, without limit.

Example Resonance is the limit of beats. For fixed tt, L’Hôpital’s rule in the variable ω\omega gives

cos⁡ωt−cos⁡ω0tω02−ω2→tsin⁡ω0t2ω0\begin{aligned} &\frac{\cos\omega t - \cos\omega_0 t}{\omega_0^2 - \omega^2} \\[0.4em] &\qquad \to \frac{t\sin\omega_0 t}{2\omega_0} \end{aligned}

as ω→ω0\omega \to \omega_0, so the beat response from rest approaches the resonant one while the beat period 2π/∣ω0−ω∣2\pi/\lvert \omega_0 - \omega \rvert grows without bound. For x=tsin⁡3tx = t\sin 3t the envelope is the pair of lines x=±tx = \pm t, and near t=10t = 10 the swings reach a size of about 1010.

Damped amplitude peak

Definition Let mm, kk and cc be positive. Undetermined coefficients give exactly one particular solution of the form acos⁡ωt+bsin⁡ωta\cos\omega t + b\sin\omega t, and its amplitude a2+b2\sqrt{a^2 + b^2} is

A(ω)=F0D(ω),A(\omega) = \frac{F_0}{\sqrt{D(\omega)}},

where D(ω)=(k−mω2)2+c2ω2D(\omega) = (k - m\omega^2)^2 + c^2\omega^2. Since c>0c > 0, the two squares never vanish together, so A(ω)A(\omega) is finite at every driving frequency. With Ω2=k/m−c2/(2m2)\Omega^2 = k/m - c^2/(2m^2), a number that may be negative, differentiating gives

D′(ω)=4m2ω(ω2−Ω2).D'(\omega) = 4m^2\omega\left(\omega^2 - \Omega^2\right).

So the amplitude has a maximum at a positive frequency if and only if c2<2mkc^2 < 2mk, and the peak is then at

ωpeak=km−c22m2,\omega_{\text{peak}} = \sqrt{\frac{k}{m} - \frac{c^2}{2m^2}},

with largest amplitude F0/(c ωd)F_0/(c\,\omega_d). If c2≥2mkc^2 \ge 2mk, the amplitude decreases for all ω>0\omega > 0 and stays below its zero-frequency value A(0)=F0/kA(0) = F_0/k, the static deflection. The condition c2<2mkc^2 < 2mk is stronger than underdamping, which asks only c2<4mkc^2 < 4mk, and a peak, when it exists, lies below both frequencies, ωpeak<ωd<ω0\omega_{\text{peak}} < \omega_d < \omega_0. Some books call this peak practical resonance.

In plain words Damping caps the response, so the amplitude curve has a finite hump instead of a spike. The hump appears only when the damping is light enough, and its top sits a little below the natural frequency.

Example Take m=1 kgm = 1\ \text{kg}, k=9 N/mk = 9\ \text{N/m}, c=2 N s/mc = 2\ \text{N s/m} and F0=6 NF_0 = 6\ \text{N}. Here c2=4<18=2mkc^2 = 4 < 18 = 2mk, so the peak is at ωpeak=9−2=7≈2.65\omega_{\text{peak}} = \sqrt{9 - 2} = \sqrt{7} \approx 2.65 radians per second, below ω0=3\omega_0 = 3. With ωd=8\omega_d = \sqrt{8} the largest amplitude is 6/(28)=32/4≈1.06 m6/(2\sqrt{8}) = 3\sqrt{2}/4 \approx 1.06\ \text{m}. For comparison, A(3)=6/36=1 mA(3) = 6/\sqrt{36} = 1\ \text{m} and A(0)=6/9≈0.67 mA(0) = 6/9 \approx 0.67\ \text{m}. Because c>0c > 0, every free part decays, so this periodic solution is the long-run motion, the sinusoidal steady state.

Check yourself

  1. A body of mass 2 kg2\ \text{kg} on a spring with k=50 N/mk = 50\ \text{N/m} has no damping. Find its natural angular frequency, and the beat period when it is driven from rest at ω=4.6\omega = 4.6 radians per second.
Show the answer

The natural angular frequency is ω0=50/2=5\omega_0 = \sqrt{50/2} = 5 radians per second. The beat period is

Tbeat=2π∣5−4.6∣=5π≈15.7T_{\text{beat}} = \frac{2\pi}{\lvert 5 - 4.6 \rvert} = 5\pi \approx 15.7

seconds. That is the time between successive maxima of the magnitude envelope. The signed slow factor sin⁡(0.2t)\sin(0.2t) takes 10π≈31.410\pi \approx 31.4 seconds to repeat, and the fast factor oscillates at (5+4.6)/2=4.8(5 + 4.6)/2 = 4.8 radians per second.

  1. Solve x′′+4x=8cos⁡2tx'' + 4x = 8\cos 2t with x(0)=0x(0) = 0 and x′(0)=0x'(0) = 0, where primes are derivatives in tt. Why does the trial acos⁡2t+bsin⁡2ta\cos 2t + b\sin 2t fail?
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The natural angular frequency is 4=2\sqrt{4} = 2, the same as the driving frequency. Every function u=acos⁡2t+bsin⁡2tu = a\cos 2t + b\sin 2t satisfies u′′+4u=0u'' + 4u = 0, so it can never produce 8cos⁡2t8\cos 2t. Try x=tux = tu instead. Then x′′=2u′+tu′′x'' = 2u' + tu'', so x′′+4x=2u′+t(u′′+4u)=2u′x'' + 4x = 2u' + t(u'' + 4u) = 2u', and

2u′=−4asin⁡2t+4bcos⁡2t.2u' = -4a\sin 2t + 4b\cos 2t.

Matching 8cos⁡2t8\cos 2t gives a=0a = 0 and b=2b = 2, so x=2tsin⁡2tx = 2t\sin 2t. This function has x(0)=0x(0) = 0 and x′=2sin⁡2t+4tcos⁡2tx' = 2\sin 2t + 4t\cos 2t, so x′(0)=0x'(0) = 0 and it is the response from rest. Its envelope is ±2t\pm 2t, which is secular growth.

  1. A system with m=1 kgm = 1\ \text{kg}, k=9 N/mk = 9\ \text{N/m} and c=5 N s/mc = 5\ \text{N s/m} is underdamped. Does its amplitude curve A(ω)A(\omega) have a peak at a positive driving frequency?
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It is underdamped because c2=25<36=4mkc^2 = 25 < 36 = 4mk. But c2=25>18=2mkc^2 = 25 > 18 = 2mk, so there is no peak at a positive frequency. Directly,

D′(ω)=2ω(2ω2+7)>0D'(\omega) = 2\omega\left(2\omega^2 + 7\right) > 0

for ω>0\omega > 0, so DD increases and A=F0/DA = F_0/\sqrt{D} decreases. The amplitude stays below its zero-frequency value A(0)=F0/9A(0) = F_0/9, the static deflection. Underdamping decides whether free motion oscillates, while a positive-frequency peak needs the stronger condition c2<2mkc^2 < 2mk.

How this lesson connects

It also uses trigonometric sum-to-product and product-to-sum identities.