A synthetic mass-spring model satisfies my″+cy′+ky=F(t), with mass m positive, damping coefficient c nonnegative, and spring stiffness k positive. Displacement y is measured from equilibrium. The spring force opposes displacement, and the damping force opposes velocity. Check that all three terms on the left have force units.
Set F=0 and c=0. The equation becomes y″+(k/m)y=0, giving angular frequency ω₀=√(k/m). For m=1 and k=9, ω₀=3 radians per time unit, not three cycles per time unit.
Predict
Will increasing damping always produce oscillation with smaller amplitude? Predict what happens when the characteristic roots stop being complex. Also predict whether energy can increase in an unforced model with positive damping. These qualitative expectations should become checks on the solution, not just descriptions after calculation.
Experiment
For m=1 and k=4, compare several damping values using r²+cr+4=0.
Damping c
Discriminant c²−16
Root type
Behavior class
0
−16
Pure imaginary
Undamped oscillation
2
−12
Complex, negative real part
Underdamped
4
0
Repeated negative real
Critical damping
5
9
Distinct negative real
Overdamped
At critical damping, solve y″+4y′+4y=0 with y(0)=1,y′(0)=0. The result is y=(1+2t)e^(−2t). For t≥0 it stays positive, and y′=−4te^(−2t) is nonpositive. This particular trajectory returns without crossing equilibrium. Do not infer every possible initial condition has exactly the same shape.
Understand
Define mechanical energy E=(m/2)(y′)²+(k/2)y², the sum of kinetic and spring potential energy. Differentiating gives E′=y′(my″+ky). In the unforced equation, my″+ky=−cy′, so E′=−c(y′)². Energy is constant when c=0 and nonincreasing when c is positive. The derivative can be zero instantaneously at a turning point without energy being constant throughout the motion.
The discriminant c²−4mk classifies the homogeneous roots. Under positive damping with a negative discriminant, an exponential envelope multiplies sine and cosine. With a positive discriminant, two decaying exponentials replace the oscillatory basis. At equality, the repeated-root basis includes t times an exponential. These forms follow from the same characteristic equation rather than separate physical laws.
For forcing, the energy balance becomes E′=F(t)y′−c(y′)². Input can add energy or remove it depending on its direction relative to velocity. In the ideal undamped resonant equation y″+y=cos t, a particular solution grows as (t/2)sin t. Adding positive damping changes the sustained sinusoidal response to a bounded amplitude for fixed finite forcing frequency. A large frequency response can still occur; bounded does not mean small or operationally acceptable.
Apply
Use the energy identity to audit a numerical trajectory. If an unforced positive-damping calculation shows a persistent increase in energy, investigate the numerical method or its step size before interpreting it as a new physical effect. Discretization need not preserve the exact continuous energy law.
State which assumptions would fail for nonlinear drag, a spring outside its linear range, or time-dependent forcing. Complete the practice by connecting characteristic roots, trajectory shape, and energy. Each representation checks a different part of the explanation.
Continue
Next rewrite second-order motion as a first-order system in position and velocity. Eigenvectors then explain directions of motion as well as the exponential rates already encoded by characteristic roots.
Continue with gain, phase and practical resonance to derive the forced displacement maximum and distinguish it from the natural and damped free frequencies.
Apply · explain · revise
Practice and reasoning
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For y″+9y=0, what is the undamped angular frequency?
Integers, fractions and finite decimals are accepted. Expressions such as sqrt(2) are not parsed by this checker.
For mass m=2 and spring coefficient k=8, find the positive critical damping coefficient c.
Integers, fractions and finite decimals are accepted. Expressions such as sqrt(2) are not parsed by this checker.
For m,k>0 and c>0, the regime c²<4mk has:
For m y″+c y′+k y=0 with m,k>0,c≥0, derive the energy rate for E=(m/2)y′²+(k/2)y².
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Explain why y″+y=cos t has an unbounded resonant particular response but y″+2y′+y=cos t has a bounded steady sinusoidal response.
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For y″+4y′+4y=0 with y(0)=1,y′(0)=0, solve and decide whether this trajectory oscillates.
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Reflect and continue
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