Damped or Forced Harmonic Motion (Edexcel A-Level Further Mathematics): Revision Notes

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8.3.2 Damped or Forced Harmonic Motion

Introduction to Damped and Forced Harmonic Motion

Harmonic motion describes systems where displacement from equilibrium is opposed by a restoring force proportional to that displacement. When resistance (damping) or external driving forces (forcing) are introduced, the motion changes.

Second-order differential equations model these dynamics.

Damped Harmonic Motion

Damped harmonic motion is governed by the equation:

md2xdt2+cdxdt+kx=0m \frac{d^2x}{dt^2} + c \frac{dx}{dt} + kx = 0

where:

  • mm: mass of the object,
  • cc: damping constant (resistance proportional to velocity),
  • kk: spring constant (restoring force proportional to displacement),
  • x(t)x(t): displacement at time tt

The nature of the motion depends on the discriminant of the characteristic equation:

m2+cmm+km=0m^2 + \frac{c}{m}m + \frac{k}{m} = 0

Cases of Damping

Light Damping (c2<4mkc^2 < 4mk)

Roots are complex (m=α±iβm = \alpha \pm i\beta):

x(t)=e−αt(Acos⁡(βt)+Bsin⁡(βt))x(t) = e^{-\alpha t} \big(A\cos(\beta t) + B\sin(\beta t)\big)

The system oscillates with exponentially decaying amplitude.

Critical Damping (c2=4mkc^2 = 4mk):

Roots are real and equal (m=−c2mm = -\frac{c}{2m}):

x(t)=(A+Bt)e−c2mtx(t) = (A + Bt)e^{-\frac{c}{2m}t}

The system returns to equilibrium without oscillating, as quickly as possible.

Heavy Damping (c2>4mkc^2 > 4mk):

Roots are real and distinct (m1,m2m_1, m_2):

x(t)=Aem1t+Bem2tx(t) = Ae^{m_1t} + Be^{m_2t}

The system slowly returns to equilibrium without oscillating.

Forced Harmonic Motion

Forced harmonic motion occurs when an external periodic force acts on the system, modelled by:

md2xdt2+cdxdt+kx=F(t)m \frac{d^2x}{dt^2} + c \frac{dx}{dt} + kx = F(t)

where F(t)F(t) is the driving force (e.g., F(t)=F0cos⁡(ωt)F(t) = F_0\cos(\omega t)).

The general solution is:

x(t)=xc+xpx(t) = x_c + x_p

where:

  • xcx_c is the complementary function (solution to the homogeneous equation),
  • xpx_p is the particular integral (solution to the non-homogeneous equation).

Worked Examples

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Example 1: Lightly Damped Oscillator

A system with mass m=1 kgm = 1 \, \mathrm{kg}, damping constant c=2 Ns/mc = 2 \, \mathrm{Ns/m}, and spring constant k=10 N/mk = 10 \, \mathrm{N/m} satisfies the equation:

d2xdt2+2dxdt+10x=0\frac{d^2x}{dt^2} + 2\frac{dx}{dt} + 10x = 0

Solve for x(t) x(t) with initial conditions x(0)=1x(0)=1, dxdt(0)=0\frac{dx}{dt}(0) = 0

Step 1: Find the Characteristic Equation

m2+2m+10=0m^2 + 2m + 10 = 0

The discriminant is:

Δ=22−4(1)(10)=−36\Delta = 2^2 - 4(1)(10) = -36

Since Δ<0\Delta < 0, the roots are complex:

m=−1±3im = -1 \pm 3i

Step 2: Write the General Solution

x(t)=e−t(Acos⁡(3t)+Bsin⁡(3t))x(t) = e^{-t}(A\cos(3t) + B\sin(3t))

Step 3: Apply Initial Conditions

At t=0,x(0)=1t = 0, x(0) = 1

1=e0(Acos⁡(0)+Bsin⁡(0))⇒A=11 = e^{0}(A\cos(0) + B\sin(0)) \quad \Rightarrow \quad A = 1

At t=0,dxdt=0t=0, \frac{dx}{dt} = 0

Differentiate x(t)x(t)

dxdt=e−t(−Acos⁡(3t)−Bsin⁡(3t))+e−t(−3Asin⁡(3t)+3Bcos⁡(3t))\frac{dx}{dt} = e^{-t}\big(-A\cos(3t) - B\sin(3t)\big) + e^{-t}\big(-3A\sin(3t) + 3B\cos(3t)\big)

Substitute t=0t = 0

0=(−A+3B)0 = (-A + 3B)

Substituting A=1A = 1

0=−1+3B⇒B=130 = -1 + 3B \quad \Rightarrow \quad B = \frac{1}{3}

Final Solution

x(t)=e−t(cos⁡(3t)+13sin⁡(3t))x(t) = e^{-t}\left(\cos(3t) + \frac{1}{3}\sin(3t)\right)
infoNote

Example 2: Forced Oscillator

Solve the forced equation:

d2xdt2+3dxdt+2x=cos⁡(2t)\frac{d^2x}{dt^2} + 3\frac{dx}{dt} + 2x = \cos(2t)

Step 1: Solve the Homogeneous Equation

m2+3m+2=0⇒(m+1)(m+2)=0m^2 + 3m + 2 = 0 \quad \Rightarrow \quad (m + 1)(m + 2) = 0

Roots: m=−1,m=−2m = -1, m = -2

Complementary Function:

xc=Ae−t+Be−2tx_c = Ae^{-t} + Be^{-2t}

Step 2: Solve for the Particular Integral

Guess xp=Ccos⁡(2t)+Dsin⁡(2t)x_p = C\cos(2t) + D\sin(2t)

Substitute into the equation:

d2xpdt2=−4Ccos⁡(2t)−4Dsin⁡(2t)\frac{d^2x_p}{dt^2} = -4C\cos(2t) - 4D\sin(2t)dxpdt=−2Csin⁡(2t)+2Dcos⁡(2t)\frac{dx_p}{dt} = -2C\sin(2t) + 2D\cos(2t)

Substitute these into the original equation:

(−4Ccos⁡(2t)−4Dsin⁡(2t))+3(−2Csin⁡(2t)+2Dcos⁡(2t))+2(Ccos⁡(2t)+Dsin⁡(2t))=cos⁡(2t)(-4C\cos(2t) - 4D\sin(2t)) + 3(-2C\sin(2t) + 2D\cos(2t)) + 2(C\cos(2t) + D\sin(2t)) = \cos(2t)

Group terms:

(−4C+6D+2C)cos⁡(2t)+(−4D−6C+2D)sin⁡(2t)=cos⁡(2t)(-4C + 6D + 2C)\cos(2t) + (-4D - 6C + 2D)\sin(2t) = \cos(2t)

Equating coefficients:

−2C+6D=1,−2D−6C=0-2C + 6D = 1, \quad -2D - 6C = 0

Solve simultaneously:

D=110,C=−310D = \frac{1}{10}, \quad C = -\frac{3}{10}

Particular Integral:

xp=−310cos⁡(2t)+110sin⁡(2t)x_p = -\frac{3}{10}\cos(2t) + \frac{1}{10}\sin(2t)

Step 3: General Solution

x(t)=xc+xp=Ae−t+Be−2t−310cos⁡(2t)+110sin⁡(2t)x(t) = x_c + x_p = Ae^{-t} + Be^{-2t} - \frac{3}{10}\cos(2t) + \frac{1}{10}\sin(2t)

Note Summary

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Common Mistakes

  1. Forgetting exponential decay in damping: Ensure the e−αte^{-\alpha t} factor is included in solutions for damped motion.
  2. Incorrect handling of roots: Misidentifying real or complex roots leads to incorrect solutions.
  3. Misapplying particular integral guesses: Choose forms that match the forcing function F(t)F(t)
  4. Errors in initial conditions: Double-check substitutions when finding constants.
infoNote

Key Formulas

  1. Damped Harmonic Motion:
md2xdt2+cdxdt+kx=0m \frac{d^2x}{dt^2} + c \frac{dx}{dt} + kx = 0
  • Light Damping: x(t)=e−αt(Acos⁡(βt)+Bsin⁡(βt))x(t) = e^{-\alpha t}(A\cos(\beta t) + B\sin(\beta t))
  • Critical Damping: x(t)=(A+Bt)e−αtx(t) = (A + Bt)e^{-\alpha t}
  • Heavy Damping: x(t)=Aem1t+Bem2tx(t) = Ae^{m_1t} + Be^{m_2t}
  1. Forced Harmonic Motion:
md2xdt2+cdxdt+kx=F(t)m \frac{d^2x}{dt^2} + c \frac{dx}{dt} + kx = F(t)

General solution: (x(t)=xc+xp)(x(t) = x_c + x_p)

  1. Characteristic Equation:
m2+cmm+km=0m^2 + \frac{c}{m}m + \frac{k}{m} = 0

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